Mean And Variance Of A Random Variable

The key point to solve the problem:

A given distribution of probabilities of a random variable X is said to be probability distribution if the sum of probabilities associated with each random variable is equal to 1

i.e. ∑(p_i) = 1

Given distribution is :

Clearly,

Sum of probabilities = P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= 0.3 + 0.2 + 0.4 + 0.1 + 0.05

= 1.05 ≠ 1

∴ The given distribution is not a probability distribution.

The key point to solve the problem:

A given distribution of probabilities of a random variable X is said to be probability distribution if the sum of probabilities associated with each random variable is equal to 1

i.e. ∑(p_i) = 1

Given distribution is :

Clearly,

Sum of probabilities = P(X = 0) + P(X = 1) + P(X = 2)

= 0.6 + 0.4 + 0.2

= 1.2 ≠ 1

∴ The given distribution is not a probability distribution.

The key point to solve the problem:

A given distribution of probabilities of a random variable X is said to be probability distribution if the sum of probabilities associated with each random variable is equal to 1

i.e. ∑(p_i) = 1

Given distribution is :

Clearly,

Sum of probabilities = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 0.1 + 0.5 + 0.2 + 0.1 + 0.1

= 1

∴ The given distribution is a probability distribution.

The key point to solve the problem:

A given distribution of probabilities of a random variable X is said to be probability distribution if the sum of probabilities associated with each random variable is equal to 1

i.e. ∑(p_i) = 1

Given distribution is :

Clearly,

Sum of probabilities = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= 0.3 + 0.2 + 0.4 + 0.1

= 1

∴ The given distribution is a probability distribution.

The key point to solve the problem:

If a probability distribution is given then as per its definition, Sum of probabilities associated with each value of a random variable of given distribution is equal to 1

i.e. ∑(p_i) = 1

∴ P(X = -2) + P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1

0.1 + k + 0.2 + 2k + 0.3 + k = 1

0.6 + 4k = 1

4k = 0.4

K = 0.1

Value of k = 0.1

The key point to solve the problem:

If a probability distribution is given then as per its definition, Sum of probabilities associated with each value of a random variable of given distribution is equal to 1

i.e. ∑(p_i) = 1

∴ Sum of probabilities = 1

∴ a+3a+5a+7a+9a+11a+13a+15a+17a = 1

a(1+3+5+7+9+11+13+15+17) = 1

Thus, a = ………..ans (i)

P(X<3) = P(X = 0) + P(X = 1) + P(X = 2)

= a+3a+5a

= 9a =

P(X≥3) = 1 - P(X<3) {∵ sum of probabilities in distribution is 1}

=

P(0<X<5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 3a + 5a + 7a + 9a

= 24a

= ……..ans (ii)

The key point to solve the problem:

If a probability distribution is given then as per its definition, Sum of probabilities associated with each value of a random variable of given distribution is equal to 1

i.e. ∑(p_i) = 1

Given probability distribution is:

For the probability distribution:

∑(p_i) = 1

∴

Note: It’s better to apply hit and trial method for solving cubic equation, just by putting 1 we get that it satisfies the equation. So ( c - 1 ) will be its one factor. After that divide the polynomial with c-1 which will give a quadratic factor which will be factorized easily.

But we can also proceed as the equation is solved if you can think upto that extent.

(c - 1){3c(c - 2)-1(c - 2)} = 0

(c - 1)(3c - 1)(c - 2) = 0

∴ c = 1 or c = 2 or c = 1/3

As we are given that c > 0

All three values satisfy the given condition, so they must be the values.

BUT BE CAUTIOUS :

As we know that the probability of any event lies between 0 and 1, and we are using c to represent probabilities. So We need to check whether for these values probabilities are defined are valid or not.

X_i : 0 1 2

P_i : 3c³ 4c-10c² 5c-1

As this distribution suggests c = 1 and c = 2 are going to make probabilities invalid.

So, c = 1/3 is the only solution for c … (i)

P(X<2) = 1 – P(X = 2)

= 1 – (5c – 1) = 2 – 5c

= 2 – 5/3 = 1/3

P(1 < X ≤ 3) = P(2)

= 5/3 – 1 = 2/3 ……(ii)

Key point to solve the problem:

If a probability distribution is given then as per its definition, Sum of probabilities associated with each value of random variable of given distribution is equal to 1

i.e. ∑(p_i) = 1

Let,

2P(X = x₁) = 3P (X = x₂) = P (X = x₃) = 5P(X = x₄) = k (say)

∴ P(X = x₁) = k/2

P(X = x₂) = k/3

P(X = x₃) = k

P(X = x₄) = k/5

∵ ∑(p_i) = 1 {∵ it is given that it is a probability distribution }

15k + 30k + 10k + 6k = 30 [ by taace LCM ]

61k = 30

k = 30/61

∴ the required probability distribution is :

The key point to solve the problem:

If a probability distribution is given then as per its definition, Sum of probabilities associated with each value of a random variable of given distribution is equal to 1

i.e. ∑(p_i) = 1

Let, P(X = 0) = k

As sum of probabilities associated with each random variable is 1

∴ P(X<0) + P(X = 0) + P(X>0) = 1

k + k + k = 1 {∵ P(X = 0) = P(X>0) = P(X<0)}

3k = 1

∴ k = 1/3

Thus

P(X<0) = 1/3

P(X = -3) + P(X = -2) + P(X = -1) = 1/3

m+m+m = 1/3 {∵ P(X = -3) = P(X = -2) = P(X = -1) = m (say) }

m = 1/9

Similarly,

P(X>0) = 1/3

P (X = 1) + P(X = 2) + P(X = 3) = 1/3

n+n+n = 1/3 {∵ P(X = 3) = P(X = 2) = P(X = 1) = n (say) }

∴ n = 1/9

∴ the required probability distribution is :

In a deck of 52 cards, there are 4 aces each of one suit respectively.

Let X be the random variable denoting the number of aces for an event when two cards are drawn simultaneously.

∴ X can take values 0, 1 or 2.

P(X = 0) =

[For selecting 0 aces, we removed all 4 aces from the deck and selected out of 48]

P(X = 1) =

[For selecting 1 ace, we need to select and 1 out of 4 and not any other]

P(X = 2) =

[For selecting 2 aces, we need to select and 2 out of 4]

Now we have p_i and x_i.

As

Thus, ∑(p_i) = 1

∴ Now we are ready to write the probability distribution for X:-

The following table gives probability distribution:

When we toss a coin three times we have the following possibilities:

{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}

Let X be a random variable representing some heads in 3 tosses of a coin.

∵ The probability of getting a head and probability of getting a tail are independent events and P(GETTING A TAIL) = P(GETTING A HEAD) = 1/2

∴ P(Head in the first toss) and P(Head in the second toss) and P(head in the third toss) can be given by their products.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

Thus,

P(X = 0) = P(TTT) = P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 = 1/8

P(X = 1) = P(HTT or THT or TTH) = P(HTT)+P(THT)+P(TTH)

= P(H)P(T)P(T)+ P(T)P(H)P(T)+ P(T)P(T)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X = 2) = P(HHT or HTH or THH) = P(HHT)+P(HTH)+P(THH)

= P(H)P(H)P(T)+ P(H)P(T)P(H)+ P(T)P(H)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X = 3) = P(HHH) = P(H)P(H)P(H) = 1/2 x 1/2 x 1/2 = 1/8

Now we have p_i and x_i.

As

Thus, ∑(p_i) = 1

∴ Now we are ready to write the probability distribution for X:-

The following table gives probability distribution:

In a deck of 52 cards, there are 4 aces each of one suit respectively.

Let X be the random variable denoting the number of aces for an event when 4 cards are drawn simultaneously.

∴ X can take values 0 , 1 , 2 , 3 or 4

P(X = 0) =

[For selecting 0 aces, we removed all 4 aces from the deck and selected out of 48]

P(X = 1) =

[For selecting 1 ace, we selected and 1 out of 4aces and 3 cards from remaining 48]

P(X = 2) =

[For selecting 2 aces, we selected and 2 out of 4aces and 2 cards from remaining 48]

P(X = 3) =

[For selecting 3 aces, we selected and 3 out of 4aces and 1 card from remaining 48]

P(X = 4) =

[For selecting 4 aces, we selected and 4 out of 4 aces]

Now we have p_i and x_i.

∴ Now we are ready to write the probability distribution for X:-

The following table gives probability distribution:

X represents the number of red balls drawn.

∴ X can take values 0,1,2 or 3

∵ there are total 10 balls

n(S) = total possible ways of selecting 2 balls =

P(X = 0) = P(selecting no red balls) =

P(X = 1) = P(selecting 1 red ball and 2 black balls)

=

P(X = 2) = P(selecting 2 red balls and 1 black ball)

=

P(X = 3) = P(selecting 3 red balls and 0 black ball)

=

Now we have p_i and x_i.

Following table represents the probability distribution of X :

X represents the number of defective mangoes drawn.

∴ X can take values 0,1,2 or 3

∵ there are total 20mangoes (15good+5defective) mangoes

n(S) = total possible ways of selecting 5 mangoes =

P(X = 0) = P(selecting no defective mango) =

P(X = 1) = P(selecting 1 defective mango and 3 good mangoes)

=

P(X = 2) = P(selecting 2 defective mangoes and 2 good mangoes)

=

P(X = 3) = P(selecting 3 defective mangoes and 1 good mango)

=

P(X = 4) = P(selecting 4 defective mangoes and no good mango)

=

Now we have p_i and x_i.

Following table represents the probability distribution of X :

When two fair dice are thrown there are total 36 possible outcomes.

∵ X denotes the sum of 2 numbers appearing on dice.

∴ X can take values 2,3,4,5,6,7,8,9,10,11 and 12

As appearance of a number on a fair die is equally likely

i.e. P(appearing of 1) = P(appearing of 2) = P(appearing of 3) = P(appearing of 4) = P(appearing of 5) = P(appearing of 6) = 1/6

And also the appearance of numbers on two different dice is an independent event. So two find conditions like P(1 in the first dice and 2 in the second dice) can be given using multiplication rule of probability.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

P(X = 2) = {∵ (1,1) is the only combination resulting sum = 2}

P(X = 3) =

{∵ (1,2) and (2,1) are the combinations resulting in sum = 3}

P(X = 4) =

{∵ (1,3), (3,1) and (2,2) are the combinations resulting in sum = 4}

P(X = 5) =

{∵ (3,2) (2,3) (1,4) and (4,1) are the combinations resulting in sum = 5}

P(X = 6) =

{∵ (1,5) (5,1) (2,4) (4,2) (3,3) are the combinations resulting in sum = 6}

P(X = 7) =

{∵ (1,6) (6,1) (2,5) (5,2) (3,4) (4,3) are the combinations resulting in sum = 7}

P(X = 8) =

{∵ (3,5) (5,3) (2,6) (6,2) (4,4) are the combinations resulting in sum = 8}

P(X = 9) =

{∵ (3,6) (6,3) (5,4) and (4,5) are the combinations resulting in sum = 9}

P(X = 10) =

{∵ (6,4), (4,6) and (5,5) are the combinations resulting in sum = 10}

P(X = 11) =

{∵ (5,6) and (6,5) are the combinations resulting in sum = 11}

P(X = 12) = {∵ (6,6) is the only combination resulting sum = 2}

Now we have p_i and x_i.

∴ Required probability distribution is:-

Note Many of the times while solving such simple problems we make a mistake in counting. So at first, we should make a frequency table which tells us no of students in the class of the same age.

This makes our interpretation easier.

∴ frequency distribution table for age and number of students is:

X represents the age of a random student.

∴ X can take values 14,15,16,17,18,19,20 or 21

Total No. of students in class = 15

Using the above frequency table, we can easily see a total number of students of a particular age, and hence we can find probability easily.

P(X = 14) = {As probability = }

Similarly,

P(X = 15) = ; P(X = 16) = ; P(X = 17) =

P(X = 18) = ; P(X = 19) = ; P(X = 20) =

P(X = 21) =

Now we have p_i and x_i.

∴ Required probability distribution is:-

X represents the number of defective bolts drawn.

∴ X can take values 0,1,2 or 3

∵ there are total 25 bolts (20 good + 5 defectives) bolts

n(S) = total possible ways of selecting 5 bolts =

P(X = 0) = P(selecting no defective bolt) =

P(X = 1) = P(selecting 1 defective bolt and 3 good bolts)

=

P(X = 2) = P(selecting 2 defective bolts and 2 good bolts)

=

P(X = 3) = P(selecting 3 defective bolts and 1 good bolt)

=

P(X = 4) = P(selecting 4 defective bolts and no good bolt)

=

Now we have p_i and x_i.

Following table represents the probability distribution of X :

Note: While reading this question you might have observed that cards are being drawn successively and sometimes in other question you might have get cards are drawn simultaneously

You have to be careful regarding these two words while solving the question. Both have a different meaning.

E.g.:-

When you draw 3 cards out of 52 simultaneously, then total no of ways of drawing is simply 52C₃ = 22100. You draw 3 cards at a time.

When you draw 3 cards out of 52 successively, then total no of ways of drawing is not simply 52C₃, once you have drawn the first card, only 51 cards are remaining and so on. You are drawing only one card at a time

In such case total ways of drawing 3 cards = 52C₁ × 51C₁ × 50C_{1 =} 132600

Hence both are entirely different. So Be cautious regarding it.

Let’s solve the problem now:

Note: In our problem, it is given that after every draw, we are replacing the card. So our sample space will not change

In a deck of 52 cards, there are 4 aces each of one suit respectively.

Let X be the random variable denoting the number of aces for an event when 2 cards are drawn successively.

∴ X can take values 0, 1 or 2

P(X = 0) =

{As we have to select 1 card at a time such that no ace is there so the first probability is 48/52 and as the drawn card is replaced, next time again probability is 48/52 }

Similarly, we proceed for other cases.

P(X = 1) =

{we might get ace in the first card or second card, so both probabilities are added}

Similarly,

P(X = 2) =

Now we have p_i and x_i.

∴ Now we are ready to write the probability distribution for X:-

The following table gives probability distribution:

Note: While reading this question you might have observed that cards are being drawn successively and sometimes in other question you might have get cards are drawn simultaneously

You have to be careful regarding these two words while solving the question. Both have a different meaning.

E.g.:-

When you draw 3 cards out of 52 simultaneously, then total no of ways of drawing is simply 52C₃ = 22100. You draw 3 cards at a time.

When you draw 3 cards out of 52 successively, then total no of ways of drawing is not simply 52C₃, once you have drawn the first card, only 51 cards are remaining and so on. You are drawing only one card at a time

In such case total ways of drawing 3 cards = 52C₁ × 51C₁ × 50C_{1 =} 132600

Hence both are entirely different. So Be cautious regarding it.

Let’s solve the problem now:

Note: In our problem, it is given that after every draw, we are replacing the card. So our sample space will not change

In a deck of 52 cards, there are 4 kings each of one suit respectively.

Let X be the random variable denoting the number of kings for an event when 2 cards are drawn successively.

∴ X can take values 0, 1 or 2

P(X = 0) =

{As we have to select 1 card at a time such that no king is there so the first probability is 48/52 and as the drawn card is replaced, next time again probability is 48/52 }

Similarly, we proceed for other cases.

P(X = 1) =

{we might get a king in the first card or second card, so both probabilities are added}

Similarly,

P(X = 2) =

Now we have p_i and x_i.

∴ Now we are ready to write the probability distribution for X:-

The following table gives probability distribution:

Note: While reading this question you might have observed that cards are being drawn successively and sometimes in other question you might have get cards are drawn simultaneously

You have to be careful regarding these two words while solving the question. Both have a different meaning.

E.g.:-

When you draw 3 cards out of 52 simultaneously, then total no of ways of drawing is simply 52C₃ = 22100. You draw 3 cards at a time.

When you draw 3 cards out of 52 successively, then total no of ways of drawing is not simply 52C₃, once you have drawn the first card, only 51 cards are remaining and so on. You are drawing only one card at a time.

In such case total ways of drawing 3 cards = 52C₁ × 51C₁ × 50C_{1 =} 132600

Hence both are entirely different. So Be cautious regarding it.

Let’s solve the problem now:

Note: In our problem, it is given that after every draw, we are not replacing the card. So our sample space will change in this case

In a deck of 52 cards, there are 4 aces each of one suit respectively.

Let X be the random variable denoting the number of aces for an event when 2 cards are drawn successively.

∴ X can take values 0, 1 or 2

P(X = 0) =

{As we have to select 1 card at a time such that no ace is there so first probability is 48/52 and as the drawn card is not replaced, next time probability is 47/51 because now one card is not in our sample space and that card is not from group of ace, So now ace cards are 47 }

Similarly, we proceed for other cases.

P(X = 1) =

{we might get ace in the first card or in the second card, so both probabilities are added}

Similarly,

P(X = 2) =

Now we have p_i and x_i.

∴ Now we are ready to write the probability distribution for X:-

The following table gives probability distribution:

Let X denote the number of white balls drawn in a random draw of 3 balls.

∴ X can take values 0,1,2 or 3.

Since bag contains 6 red and 4 white ball, i.e. at a total of 10 balls

∴ total no. of ways of selecting 3 balls out of 10 = 10C₃

For selecting 0 white balls, we will select all 3 balls from red

∴ P(X = 0) = P (not selecting any white ball) =

For selecting 1 white ball, we will select all 2 balls from red and 1 from white

∴ P(X = 1) =

For selecting 2 white balls, we will select all 1 ball from red and 2 from white

∴ P(X = 1) =

For selecting 3 white balls, we will select all 0 balls from red and 2 from white

∴ P(X = 1) =

Now we have p_i and x_i.

∴ Now we are ready to write the probability distribution for X:-

The following table gives probability distribution:

When two fair dice are thrown there are total 36 possible outcomes.

In our question, we are throwing two dice 2 times

∵ Y denotes the number of times, a sum of two numbers appearing on dice is equal to 9.

∴ Y can take values 0,1 or 2

Means In both the throw sum of 9 is not obtained, in one of throw of two dice 9 is obtained and in both the throws of two dice a sum of 9 is obtained.

∵ (3,6) (6,3) (5,4) and (4,5) are the combinations resulting in sum = 9

Let A denotes the event of getting a sum of 9 in a throw of 2 dice

∴ P(A) = probability of getting the sum of 9 in a throw of 2 dice is

And P(A’) = probability of not getting the sum of 9 in the throw of 2 dice =

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

P(Y = 0) = P(A’)×P(A’) =

P(Y = 1) = P(A)×P(A’)+ P(A’)×P(A)+ =

P(Y = 2) = P(A)×P(A) =

Now we have p_i and x_i.

∴ The required probability distribution is:-

X represents the number of defective items drawn.

∴ X can take values 0,1,2 or 3

∵ there are total 25 items (20 good+5 defectives) items

n(S) = total possible ways of selecting 5 items =

P(X = 0) = P(selecting no defective item) =

P(X = 1) = P(selecting 1 defective item and 3 good items)

=

P(X = 2) = P(selecting 2 defective items and 2 good items)

=

P(X = 3) = P(selecting 3 defective items and 1 good item)

=

P(X = 4) = P(selecting 4 defective items and no good item)

=

Now we have p_i and x_i.

Following table represents probability distribution of X :

Note: While reading this question you might have observed that cards are being drawn successively and sometimes in other question you might have get cards are drawn simultaneously

You have to be careful regarding these two words while solving question. Both have a different meaning.

E.g.:-

When you draw 3 cards out of 52 simultaneously, then total no of ways of drawing is simply 52C₃ = 22100. You draw 3 cards at a time.

When you draw 3 cards out of 52 successively, then total no of ways of drawing is not simply 52C₃, once you have drawn the first card, only 51 cards are remaining and so on. You are drawing only one card at a time

In such case total ways of drawing 3 cards = 52C₁ × 51C₁ × 50C_{1 =} 132600

Hence both are entirely different. So Be cautious regarding it.

Let’s solve the problem now:

Note: In our problem, it is given that after every draw, we are replacing the card. So our sample space will not change

In a deck of 52 cards, there are 13 hearts.

Let X be the random variable denoting the number of hearts drawn for an event when 3 cards are drawn successively with replacement.

∴ X can take values 0 , 1, 2 or 3

P(X = 0) =

{As we have to select 1 card at a time such that no heart is there so the first probability is 39/52 and as the drawn card is replaced, next time again probability is 39/52 and again the same thing }

Similarly, we proceed for other cases.

P(X = 1) =

{we might get a heart in the first card or second card or third, so probabilities in all 3 cases are added as they are mutually exclusive events}

Similarly,

P(X = 2) =

{First 2 cards are heart and 3^rd is non-heart, last 2 are hearts and so on cases. As these cases are mutually exclusive. Hence they are added }

Similarly,

P(X = 3) =

Now we have p_i and x_i.

∴ Now we are ready to write the probability distribution for X:-

The following table gives probability distribution:

Let X denote the number of blue balls drawn in a random draw of 3 balls.

∴ X can take values 0,1,2 or 3.

Since bag contains 4 red and 3 blue balls i.e. at a total of 7 balls

∵ balls are drawn with replacement

For selecting 0 blue balls, we will select all 3 red balls

∴ P(X = 0) = P (not selecting any blue ball) =

For selecting 1 blue ball, we will select all 2 red balls and 1 blue

∴ P(X = 1) =

For selecting 2 blue balls, we will select all 1 red ball and 2 blue

∴ P(X = 2) =

For selecting 3 blue balls, we will select all 3 blue and 0 red balls

∴ P(X = 3) = P (selecting all blue ball) =

Now we have p_i and x_i.

∴ Now we are ready to write the probability distribution for X:-

The following table gives probability distribution:

In a deck of 52 cards, there are 13 spades.

Let X be the random variable denoting the number of success and success here is getting a spade for an event when two cards are drawn simultaneously.

∴ We can say that the number of successes is equal to some spades obtained in each draw.

∴ X can take values 0, 1 or 2.

P(X = 0) =

[For selecting 0 spades, we removed all 13 spades from the deck and selected out of 39]

P(X = 1) =

[For selecting 1 spade, we need to select and 1 out of 13 spade and not any other spade]

P(X = 2) =

[For selecting 2 spades, we need to select and 2 out of 13 spades]

Now we have p_i and x_i.

∴ Now we are ready to write the probability distribution for X:-

The following table gives probability distribution:

A fair dice is tossed twice.

Every time a throwing dice is an independent event.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

Let A denote the event of getting a number less than 3 on a single throw of dice.

∴ P(A) = {out of 6 outcomes 1 and 2 are favourable}

P(not A) = P(A’) =

As success is considered when number appearing on dice is less than 3.

Let X denotes the success.

As we are throwing two dice so that we can get success in both throws or either one of them, or we may even not get succeed.

∴ X can take values 0,1 or 2

P(X = 0) = P(not success)× P(not success) = P(A’)× P(A’) =

P(X = 1) = P(A) ×P(A’)+ P(A’)×P(A)

=

P(X = 2) = P(success)× P(success) = P(A)× P(A) =

Now we have p_i and x_i.

∴ Now we are ready to write the probability distribution for X:-

The following table gives probability distribution:

The key point to solve the problem:

A variable X is said to be a random variable if the sum of probabilities associated with each value of X gets equal to 1

i.e. ∑(p_i) = 1 where p_i is probability associated with x_i

X represents the number of black balls drawn.

∴ X can take values 0,1 or 2 as both balls drawn can be black which corresponds to X = 2, either of 2 balls can be black which corresponds to X = 1, and if neither of balls drawn is black it corresponds to X = 0

∵ there are a total of 7 balls

n(S) = total possible ways of selecting 2 balls =

P(X = 0) = P(selecting no black balls) =

P(X = 1) = P(selecting 1 black ball and 1 red)

=

P(X = 0) = P(selecting all black balls) =

Now we have p_i and x_i.

Clearly, ∑(p_i) =

∴ X is a random variable, and the above table represents its probability distribution.

Since X represents the difference between some heads and number of tails when a coin is tossed 6 times.

As the coin is tossed 6 times following outcomes are possible:

Let H denotes heads and T denotes tails

Outcomes: 6H0T( 6 heads 0 Tails), 5H1T, 4H2T, 3H3T, 2H4T and 1H5T

∴ differences can be 6,4,2,0,-2 and -4

∴ X can take values -4,-2,0,2,4 or 6 …

X represents the number of defective bulbs drawn.

∴ X can take values 0,1 or 2 (as maximum bulbs drawn are 2)

∵ there are total 10 bulbs (7 good+3 defectives)

n(S) = total possible ways of selecting 2 items from sample =

P(X = 0) = P(selecting no defective bulb) =

P(X = 1) = P(selecting 1 defective bulb and 1 good bulb)

=

P(X = 1) = P(selecting 0 defective bulb and 2 good bulb)

=

Now we have p_i and x_i.

Following table represents the probability distribution of X :

Let X denote the number of red balls drawn in a random draw of 4 balls.

∴ X can take values 0,1,2,3 or 4.

Since bag contains 8 red and 4 white balls i.e. at a total of 12 balls

∴ total no. of ways of selecting 4 balls out of 12 = 12C₄

For selecting 0 red balls we will select all 3 balls from red

∴ P(X = 0) = P (not selecting any red ball) =

For selecting 1 red ball, we will select 1 ball from 8 red and 3 balls from 4 white

∴ P(X = 1) =

For selecting 2 red ball, we will select 2 balls from 8 red and 2 balls from 4 white

∴ P(X = 2) =

For selecting 3 red ball we will select 3 balls from 8 red and 1 ball from 4 white

∴ P(X = 3) =

∴ P(X = 4) = P (selecting all red ball) =

Now we have p_i and x_i.

∴ Now we are ready to write the probability distribution for X :-

The following table gives probability distribution:

The key point to solve the problem:

If a probability distribution is given then as per its definition, Sum of probabilities associated with each value of a random variable of given distribution is equal to 1

i.e. ∑(p_i) = 1

Given distribution is :

(i)

∴

∴ (i)

(ii) P(X>2) = P(X = 3) =

∴ P(X>2) =

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = = …(ii)

(iii) ∴ P(X>2) + P(X≤ 2) =

Given,

Our variable is X and from equation we see that it is taking values X = 0,1,2,3,4 …… (any whole number)

And it represents the number of colleges in which you are going to get admission.

According to equation given we have :

∴ P(X = 0) = k×0 = 0

P(X = 1) = k×1 = k

P(X = 2) = 2k×2 = 4k

P(X = 3) = k(5-3) = 2k

P(X = 4) = k(5-4) = k

P(X>4) = 0

As in question, it is not given either X is random variable or the given distribution is a probability distribution, we can’t apply any thing directly.

But we have a hint in the question,

As P(X = 0) = 0

It means the chance of not getting admission in any college 0

∴ admission is sure

Hence the sum of all probabilities must be equal to 1 as getting admission has become a sure event.

This makes the given distribution a probability distribution an X a random variable also.

∴ k + 4k+2k+k = 1

8k = 1

k = 1/8

i) P(getting admission in exactly one college) = P(1) = k = 1/8

ii) P(getting admission in atmost 2 colleges) = P(X≤2)

= P(X = 1)+P(X = 2)

= k+4k = 5k = 5/8

iii) P(getting admission in atleast 2 colleges) = P(X≥2)

= P(X = 2)+P(X = 3)+P(X = 4)

= 4k+2k+k = 7k = 7/8

Mean of any probability distribution is given by Mean = ∑x_ip_i

Standard Deviation is given by SD = √Variance where variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ mean = 0.4 + 1.5 + 1.2 = 3.1

And variance = 0.8 + 4.5 + 4.8 – (3.1)² = 0.49

∴ Standard deviation = √ 0.49 = 0.7

Mean of any probability distribution is given by Mean = ∑x_ip_i

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ mean = 0.4 + 0.3+0.8+1.5= 3.0

And variance =0.4+0.9+3.2+7.5 – (3.0)² = 3

∴ Standard deviation = √ 3= 1.732

Mean of any probability distribution is given by Mean = ∑x_ip_i

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ mean = -1.25-0.5+0.5+0.25 = -1

And variance = 6.25+2+0.5+0.5 – (-1)² = 8.25

∴ Standard deviation = √8.25= 2.9

Mean of any probability distribution is given by Mean = ∑x_ip_i

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ mean = -0.3 + 0 + 0.1 + 0.6 + 0.6 = 1.0

And variance =0.3 + 0 + 0.1 + 1.2 + 1.8 – (1)² = 2.4

∴ Standard deviation = √2.4 = 1.5

Mean of any probability distribution is given by Mean = ∑x_ip_i

Standard Deviation is given by SD = √Variance where variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ mean = 0.4+0.6+0.6+0.4 = 2.0

And variance = 0.4 +1.2 + 1.8 + 1.6– (2)² = 1.0

∴ Standard deviation = √1 = 1

Mean of any probability distribution is given by Mean = ∑x_ip_i

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ mean = 0+0.5+0.6+0.5 = 1.6

And variance = 0 +0.5 + 1.8 + 2.5– (1.6)² = 2.24

∴ Standard deviation = √2.24 = 1.497

Mean of any probability distribution is given by Mean = ∑x_ip_i

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ mean = -0.2-0.2+0+0.2+0.2 = 0

And variance = 0 +0.4+0.2+0.2+0.4– (0)² = 1.2

∴ Standard deviation = √1.2 = 1.095

Mean of any probability distribution is given by Mean = ∑x_ip_i

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ mean = -0.15-0.45+0+0.25+0.15 = -0.2

And variance = 0 +0.45+0.25+0.45+0.45– (-0.2)² = 1.56

∴ Standard deviation = √1.56 = 1.248

Mean of any probability distribution is given by Mean = ∑x_ip_i

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ Mean =

Variance =

∴ standard deviation = √ =

To find the value of k we will be using the very basic idea of probability.

Note: We know that the sum of the probabilities of all random variables taken from a given sample space is equal to 1.

∴ P(X=0.5) + P(X=1) + P(X=1.5) + P(X=2) = 1

∴ k + k² + 2k² + k = 1

⇒ 3k² + 2k – 1 = 0

⇒ 3k² + 3k - k – 1 = 0

⇒ 3k(k+1) – (k+1) = 0

⇒ (3k-1)(k+1) = 0

∴ k = 1/3 or k = -1

∵ k represents probability of an event. Hence 0≤P(X)≤1

∴ k = 1/3

Mean of any probability distribution is given by- Mean = ∑x_ip_i

Now we have,

X: 0.5 1 1.5 2

P(X): 1/3 1/9 2/9 1/3

∴ first we need to find the product i.e. p_ix_i and add them to get mean.

∴ Mean = 0.5 x (1/3) + 1 x (1/9) + 1.5 x (2/9) +2 x (1/3)

Mean of any probability distribution is given by Mean = ∑x_ip_i

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

∴ p₁x₁ = ap and p₂x₂ = bq Similarly p₁x₁²= a²p and p₂x₂²= b²q

∴ Mean = ap + bq

Variance = a²p + b²q – (ap + bq)²

[∵ p+q = 1 ……given]

=

∴ SD = √{pq(a-b)² } = |a-b|√pq

When we toss a coin three times we have the following possibilities:

{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}

Let X be a random variable representing number of tails in 3 tosses of a coin.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) can be given by their individual products.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

Thus,

P(X=0) = P(HHH) = P(H)P(H)P(H) = 1/2 x 1/2 x 1/2 = 1/8

P(X=1) = P(HHT or HTH or THH) = P(HHT)+P(HTH)+P(THH)

= P(H)P(H)P(T)+ P(H)P(T)P(H)+ P(T)P(H)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=2) = P(HTT or THT or TTH) = P(HTT)+P(THT)+P(TTH)

= P(H)P(T)P(T)+ P(T)P(H)P(T)+ P(T)P(T)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=3) = P(TTT) = P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 = 1/8

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ Mean =

Variance =

In a deck of 52 cards there are 4 kings each of one suit respectively.

Let X be the random variable denoting the number of kings for an event when two cards are drawn simultaneously.

∴ X can take values 0 , 1 or 2.

P(X=0) =

[For selecting 0 kings, we removed all 4 kings from deck and selected out of 48]

P(X=1) =

[For selecting 1 king, we need to select and 1 out of 4 and not any other]

P(X=2) =

[For selecting 2 king, we need to select and 2 out of 4]

Now we have p_i and x_i.

Let’s proceed to find mean and standard deviation.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Standard Deviation is given by SD = √ Variance where variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ mean =

Variance =

∴ Standard deviation = √variance = √(400/2873)

=

When we toss a coin three times we have the following possibilities:

{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}

Let X be a random variable representing number of tails in 3 tosses of a coin.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) can be given by their individual products.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

Thus,

P(X=0) = P(HHH) = P(H)P(H)P(H) = 1/2 x 1/2 x 1/2 = 1/8

P(X=1) = P(HHT or HTH or THH) = P(HHT)+P(HTH)+P(THH)

= P(H)P(H)P(T)+ P(H)P(T)P(H)+ P(T)P(H)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=2) = P(HTT or THT or TTH) = P(HTT)+P(THT)+P(TTH)

= P(H)P(T)P(T)+ P(T)P(H)P(T)+ P(T)P(T)P(H)

= 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2 + 1/2 x 1/2 x 1/2

= 3/8

P(X=3) = P(TTT) = P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 = 1/8

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

Standard Deviation is given by SD = √ Variance

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

∴ Mean =

Variance =

Standard Deviation = √(3/4) = = 0.87

As there are total of two bad eggs. Therefore while drawing 3 eggs we can draw 1 bad egg or 2 or 0 bad eggs.

Let X be the random variable denoting number of bad eggs that can be drawn in each draw.

Clearly X can take values 0,1 or 2

P(X=0) = P(all 3 are good eggs) =

[Since there are 10 good eggs so for selecting all good we took all three from 10 and 0 eggs from 2 bad ones. Total sample points are no of ways of selecting 3 eggs from total of 12 eggs]

Similarly,

P(X=1) = P(1 bad and 2 good eggs) =

P(X=2) = P(2 Bad eggs and 1 good egg) =

Now we have p_i and x_i.

Let’s proceed to find mean

Mean of any probability distribution is given by Mean = ∑x_ip_i

∴ first we need to find the products i.e. p_ix_i and add them to get mean.

Following table gives the required products :

∴ mean =

When a pair of fair dice is thrown there are total 36 possible outcomes.

X denotes the minimum of two numbers which appear

∴ X can take values 1,2,3,4,5 and 6

P(X=1) = 11/36

[Possible Pairs: (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1)]

P(X=2) = 9/36

[Possible Pairs: (2,2),(3,2),(4,2),(5,2),(6,2),(2,6),(2,5),(2,4),(2,3)]

P(X=3) = 7/36

[Possible Pairs: (3,3),(3,4),(4,3),(5,3),(3,5),(3,6),(6,3)]

P(X=4) = 5/36

[Possible Pairs: (4,4),(5,4),(4,5),(4,6),(6,4)]

P(X=5) = 3/36

[Possible Pairs (5,5),(5,6),(6,5)]

P(X=6) = 1/36

[Possible Pairs: (6,6)]

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

Standard Deviation is given by SD = √ Variance

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

Required Probability distribution table:-

∴ Mean =

Variance =

Standard deviation = √variance =

Say, H represents event of getting a head and T represents getting a tail.

When we toss a coin 4 times we have the following possibilities:

{HHHH,HHHT,HHTH,THHH,HTHH,THHT,TTHH,HHTT,THTH…………,TTTT}

A total of 2⁴ = 16 possibilities.

Let X be a random variable representing number of heads occuring in 4 tosses of a coin.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) and P(tail in 4^th toss) can be given by their individual products.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

Thus,

P(X=0) = P(TTTT) = P(T)P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 x 1/2 = 1/16

Selecting a coin out of 4 which will show head rest all showing tail

∴

P(X=1) =

Similarly,

P(X=2) =

P(X=3) =

P(X=4) = P(HHHH) = 1/16

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

∴ Mean =

Variance =

When a fair dice is thrown there are total 6 possible outcomes.

∵ X denote twice the number appearing on die

∴ X can take values 2,4,6,8,10 and 12

As appearance of a number on a fair die is equally likely

i.e. P(appearing of 1) = P(appearing of 2) = P(appearing of 3) = P(appearing of 4) = P(appearing of 5) = P(appearing of 6) = 1/6

∴ appearance of twice of the number is also equally likely with a probability of 1/6.

P(X=2)=P(X=4)=P(X=6)=P(X=8)=P(X=10)=P(X=12)=1/6

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

Required Probability distribution table:-

∴ Mean =

Variance =

When a fair dice is thrown there are total 6 possible outcomes.

∵ X denote 1 or 3 according as an odd or an even number appears.

P(appearing of even number on a die) = 3/6 [favourable outcomes {2,4,6}]

P(appearing of an odd number on a die) = 3/6 [favourable outcomes {1,4,3}]

P(X=1) = 3/6 = 1/2

P(X=3) = 3/6 = 1/2

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table gives the required products :

Required Probability distribution table:-

∴ Mean =

Variance =

Say, H represents event of getting a head and T represents getting a tail.

When we toss a coin 4 times we have the following possibilities:

{HHHH,HHHT,HHTH,THHH,HTHH,THHT,TTHH,HHTT,THTH…………,TTTT}

A total of 2⁴ = 16 possibilities.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) and P(tail in 4^th toss) can be given by their individual products.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

As X is a random variable representing longest string of head occurring in 4 tosses.

∴ X can take following values:

X = 0 [ all tails (TTTT) ]

X = 1 [Longest string contains only 1 head e.g. (HTTT),(TTTH),(HTHT)..]

X = 2 [ Longest string contain only 2 head e.g. (HHTT),(HHTH),(THHT)…]

X = 3 [Longest string contain only 3 head e.g. ( HHHT) And (THHH)]

X = 4 [ Longest string contain 4 heads i.e. (HHHH) ]

Thus,

P(X=0) = 1/16

P(X=1) = 7/16 [by counting number of favourable outcomes as explained]

P(X=2) = 5/16

P(X=3) = 2/16

P(X=4) = 1/16

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

∴ Mean =

Variance =

As box contains cards numbered as 1,1,2,2 and 3

∴ possible sums of card numbers are 2,3,4 and 5

Hence, X can take values 2,3,4 and 5

X=2 [ when drawn cards are (1,1)]

X=3 [when drawn cards are (1,2) or (2,1)]

X=4 [when drawn cards are (2,2) or (3,1) or (1,3)]

X=5 [when drawn cards are (2,3) or (3,2)]

As Y is a random variable representing maximum of the two numbers drawn

∴ Y can take values 1,2 and 3.

Y=1 [when drawn cards are 1 and 1]

Y=2 [when drawn cards are (1,2) or (2,2) or (2,1)]

Y=3 [when drawn cards are (1,3) or (3,1) or (2,3) or (3,2)]

Note : P(1) represents probability of drawing card numbered as 1, similarly P(2) and P(3)

∴ P(X=2) = P(1)P(1) =

[For drawing first card we had 2 favourable outcomes as 1,1 out of total 5 ,in second time of drawing ,as we drew a card numbered as 1 we are having 1 favourable outcome out of total remaining of 4]

Similarly,

P(X=3) = P(2)P(1) + P(1)P(2) =

P(X=4) = P(2)P(2)+P(3)P(1)+P(1)P(3) =

P(X=5) = P(2)P(3)+P(3)P(2) =

Similarly,

P(Y=1) = P(1)P(1) =

P(Y=2) = P(1)P(2)+P(2)P(1)+P(2)P(2) =

P(Y=3) = P(2)P(3)+P(3)P(2)+ P(1)P(3)+P(3)P(1)

=

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

∴ Mean for(X) = 0.2+1.2+1.2+1 = 3.6

Variance for(X) = 0.4+3.6+4.8+5.0-3.6² = 13.8-3.6² = 0.84

Similarly probability distribution for Y is given below:

∴ Mean for(Y) = 0.1+1.0+1.2 = 2.3

Variance for(Y) = 0.1+2.0+3.6-2.3² = 5.7-2.3² = 0.41

As success is considered when we get an odd number when we roll a die.

As die is rolled twice , so we can get no success or a single success or we can get odd both the times an odd number.

If X is the random variable denoting the success then X can take value 0,1 or 2

∵ P(getting an odd number in a single rolling of die) = 3/6 = 1/2

As rolling a die is an independent event:

∴ P(getting an odd on first roll and probability of getting odd on second roll)=P(getting an odd on first roll) x P(getting an odd on second roll)

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

∴ P(X=0) = P(even number on first throw) x P(even on second throw)

=

P(X=1) = P(even number on first throw) x P(odd on second throw) +

P(odd number on first throw) x P(even on second throw)

=

P(X=2) = P(odd number on first throw) x P(odd on second throw)

=

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

∵ Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ Variance =

Let X be the random variable denoting the number of defective bulbs drawn in each draw. Since we are drawing a maximum of 3 bulbs at a time, So we can get at max 3 defective bulbs as total defective bulbs are 5.

∴ X can take values 0,1,2 and 3

P(X=0) = P(drawing no defective bulbs)

As we are finding probability for 0 defective bulbs ,so we will select all 3 bulbs

from 9 good bulbs.

n(s) = total possible ways =

∴ P(X=0) =

P(X=1) = P(drawing 1 defective bulbs and 2 good bulbs)

As we are finding probability for 1 defective bulbs ,so we will select 2 bulbs

from 9 good bulbs and 1 from 5 defective ones

∴ P(X=1) =

Similarly,

P(X=2) =

P(X=3) =

So, Probability distribution is given below:

As player sets Rs 10 on a number ,if he wins he get Rs 100

∴ his profit is Rs 90.

If he loses, he suffers a loss of Rs 10

He gets a profit when ball comes to rest in his selected slot.

Total possible outcome = 13

Favourable outcomes = 1

∴ probability of getting profit = 1/13

And probability of loss = 12/13

If X is the random variable denoting gain and loss of player

∴ X can take values 90 and -10

P(X=90) = 1/13

And P(X=-10) = 12/13

Now we have p_i and x_i.

Let’s proceed to find mean

Mean of any probability distribution is given by Mean = ∑x_ip_i

∴ first we need to find the products i.e. p_ix_i and add them to get mean

Following table representing probability distribution gives the required products :

E(X) = Mean =

We have total 26 red cards in a deck of 52 cards.

As we are drawing maximum 3 cards at a time so we can get maximum 3 red cards.

If X denotes the number of red cards ,then X can take values from 0,1,2 and 3

P(X=0) = probability of drawing no red cards

We need to select all 3 cards from remaining 26 cards

Total possible ways of selecting 3 cards =

∴ P(X=0) =

P(X=1) = P(selecting one red and 2 black cards) =

P(X=2) = P(selecting 2 red and 1 black cards) =

P(X=3) =

So, Probability distribution is given below:

Mean =

X represents the number of black balls drawn.

∴ X can take values 0,1 and 2

∵ there are total 7 balls

n(S) = total possible ways of selecting 2 balls =

P(X=0) = P(selecting no black balls) =

P(X=1) = P(selecting 1 black ball and 1 red ball)

=

P(X=2) = P(selecting 2 black ball and 0 red ball) =

X is said to be a random variable if some of the probabilities associated with each value of X is 1

Here,

P(X=0) + P(X=1) + P(X=2) =

∴ X is a random variable.

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

∴ Mean =

∵ Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ Variance =

∵ two numbers are selected at random like {(2,3) or (5,4) or (4,5)..etc}

Total ways of selecting two numbers without replacement = 6 x 5 =30

As X denote the larger of two numbers selected

∴ X can take values 3,4,5,6 and 7

P(X=3) = P(larger number is 3) = [{2,3},{3,2}]

P(X=4) = P(larger number is 4) = [{2,4},{4,2},{3,4},{4,3}]

P(X=5) = P(larger number is 5) = [{2,5},{3,5},{4,5} and their reverse order]

P(X=6) = P(larger number is 6) = [{2,6},{3,6},{4,6},{5,6} and their reverse order]

P(X=7) = P(larger number is 7) = [{2,7},{3,7},{4,7},{5,7},{6,7} and their reverse order]

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

∴ Mean =

∵ Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ Variance =

We are asked to find the expected amount he wins or lose i.e we have to find the mean of probability distribution of random variable X denoting the win/loss.

As he decided to throw the dice thrice but to quit at the instant he loses

∴ if he wins in all throw he can make earning of Rs 15

If he wins in first two throw and lose in last, he earns Rs (10-1) = Rs 9

If he wins in first throw and then loses ,he earns = Rs 4

If he loses in first throw itself, he earns Rs = -1

Thus X can take values -1,4,9 and 15

P(getting a number greater than 4 in a throw of die) = 2/6 = 1/3

P(getting a number not greater than 4 in a throw of die) = 4/6 = 2/3

P(X=-1) = P(getting number less than or equal to 4) = 2/3

P(X=4) = P(getting > 4) x P(getting ≤ 4) =

P(X=9) = P(getting > 4) x P(getting > 4) x P(getting ≤ 4) =

P(X=15) = P(getting > 4) x P(getting > 4) x P(getting > 4) =

So, Probability distribution is given below:

∵ Mean = ∑x_ip_i

Mean =

He can win around Rs 1.45

The probability distribution of X is:

We Know that

Therefore

4=4

a = 0

The probability distribution of X is:

We Know that

Therefore,

2k⁴+3k²-5k³+2k-3k²+3k-1=1

2k⁴-5k³+5k-2=0

2(k⁴-1)-5k²-1)=0

2(k²-1)(k²+1)-5k(k-1)(k+1)=0

2(k-1)(k+1)(k²+1)-5k(k-1)(k+1)=0

(k-1)(k+1)(2(k²+1)-5k=0

(k-1)(k+1)(2k²-5k+2)=0

(k-1)(k+1)(2k²-k-4k+2)=0

(k-1)(k+1)(k(2k-1)-2(2k-1))=0

(k-1)(K+1)(2k-1)(k-2)

k-1=0

K+1=0

2k-1=0

k-2=0

.

The sample space of the experiment consist of 6 elementary events where, x_i=1, 2, 3,4,5,6

The random variable X i.e. number on the upper face of a cubical die when it is thrown is-1,2,3,4,5,6

P(X=1) =P (1) =

P(X=2) =P (2) =

P(X=3) =P (3) =

P(X=4) =P (4) =

P(X=5) =P (5) =

P(X=6) =P (6) =

The probability distribution of X is

_{E(X) = 3.5}

The probability distribution of X is:

We Know that

Therefore

⇒ 2k+4k+3k+k=1

⇒ 10k=1

The probability distribution of X is:

We Know that

Therefore

The probability distribution of X is:

We Know that

Therefore

⇒ c+2c+3c+4c=1

P(X≤2)=P(X=1)+P(X=2)

⇒ c+2c

⇒ 3c

The probability distribution of X is:

We Know that

Therefore

⇒ k+2k+3k+4k=1

⇒ 10k=1

P(X≥3)=P(X=3)+P(X=4)

⇒ 3k+4k

The probability distribution of X is:

We Know that

Therefore,

⇒ a+3a+5a+7a+9a+11a+13a+15a+17a=1

⇒ 81a=1

Option (A) is incorrect because by putting , the sum total of pxi is not equal to 1

Option (B) is incorrect because by putting , the sum total of pxi is not equal to 1

Option () is incorrect because by putting , the sum total of pxi is not equal to 1

E=(X:X is a prime number)=(2,3,5,7)

P(E)=P(X=2)+(X=3)+(X=5)+(X=7)

P(E)=0.23+0.12+0.20+0.07=0.62

F=(X:X<4)=(1,2,3)

P(F)=P(X=1)+(X=2)+(X=3)

P(F)=0.15+0.23+0.12=0.5

E∩F=(X is a prime number as well as <4)=(2,3)

P(E∩F)=P(X=2)+P(X=3)

⇒ 0.23+0.12=0.35

P(E∪F)=P(E)+P(F)-P(E∩F)

⇒ 0.62+0.50-0.35

Let the value P(X=0) be x and the value of P(X=1) be A

GIVEN- P(X=3) =2P(X=1)

=2(A) =2A

P(X=2) =0.3

So, The Probability distribution of X is

Mean of E(X) =1.3

⇒ (0×x)+(1×A)+(2×0.3)+(3×2A)=1.3

⇒ A+0.6+6A=1.3

⇒ 7A+0.6=1.3

⇒ 7A=0.7

⇒ A=0.1

By putting the value of A in probability distribution, we get

We Know that

Therefore,

⇒ x+0.1+0.3+0.2=1

x=1-0.6

x=0.4

The probability distribution of X is:

We Know that

Therefore,

⇒ 2p+2p+3p+p² +2p² +7p² +2p=1

⇒ 10p² +9p-1=0

⇒ 10p² +10p-p-1=0

⇒ 10p(p+1)-1(p+1)=0

⇒ (10p-1)(p+1)

10p-1=0 or p+1=0

We considered because p is a probability and probability cannot be in negative

Option (B) is incorrect because p is a probability and probability cannot be in negative

Option (C) is incorrect because p is a probability and probability cannot be in negative

Option (D) is incorrect because by putting , the sum total of pxi is not equal to 1

The probability distribution of X is:

We Know that

Therefore

⇒ k+3k+3k+k=1

⇒ 8k=1

⇒ (1×k)+(2×3k)+(3×3k)+(4×k)

⇒ k+6k+9k+4k=20k

⇒ (1² ×k)+(2² ×3k)+(3² ×3k)+(4² ×k)

⇒ k+12k+27k+16k=56k

Var(X)=E(X²) - (E(X))²

The probability distribution of X is:

We Know that

Therefore

⇒ k=32

=3.8125

The probability distribution of X is:

We Know that

Therefore,

(-4×0.1)+(-3×0.2)+(-2×0.3)+(-1×0.2)+(0×0.2)

⇒ -0.4-0.6-0.6-0.2

The probability distribution of X is:

Let X be a random variable whose possible values

Possibilities p(x₁), p(x₂),…………p(x_n)respectively.

Let = E (X) be the mean of X. The variance of X, denoted by Var (X) or defined as