Find the mean (𝓊), variance (σ2) and standard deviation (σ) for each of the following probability distributions:
(i)
(ii)
(iii)
(iv)
(i) Given :
To find : mean (𝓊), variance (σ2) and standard deviation (σ)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Standard deviation =
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3) + x4P(x4)
Mean = E(X) = 0() + 1() +2() +3() = 0 + + + = = =
Mean = E(X) = = 1.2
= = 1.44
E(X2) = = P(x1) + P(x2) + P(x3) + P(x4)
E(X2) = () + () + () + () = 0 + + + = =
E(X2) = 2
Variance = E(X2) - = 2 – 1.44 = 0.56
Variance = E(X2) - = 0.56
Standard deviation = = = 0.74
Mean = 1.2
Variance = 0.56
Standard deviation = 0.74
(ii) Given :
To find : mean (𝓊), variance (σ2) and standard deviation (σ)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Standard deviation =
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3) + x4P(x4)
Mean = E(X) = 1(0.4) + 2(0.3) +3(0.2) +4(0.1) = 0.4 + 0.6 + 0.6 + 0.4 = 2
Mean = E(X) = 2
= = 4
E(X2) = = P(x1) + P(x2) + P(x3) + P(x4)
E(X2) = (0.4) + (0.3) + (0.2) + (0.1) = 0.4 + 1.2 + 1.8 + 1.6 = 5
E(X2) = 5
Variance = E(X2) - = 5 – 4 = 1
Variance = E(X2) - = 1
Standard deviation = = = 1
Mean = 2
Variance = 1
Standard deviation = 1
(iii) Given :
To find : mean (𝓊), variance (σ2) and standard deviation (σ)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Standard deviation =
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3) + x4P(x4)
Mean = E(X) = -3(0.2) + (-1)(0.4) + 0(0.3) + 2(0.1)= -0.6 - 0.4 + 0 + 0.2 = -0.8
Mean = E(X) = -0.8
= = 0.64
E(X2) = = P(x1) + P(x2) + P(x3) + P(x4)
E(X2)= (0.2) + (0.4) + (0.3) + (0.1) = 1.8 + 0.4 + 0+ 0.4 = 2.6
E(X2) = 2.6
Variance = E(X2) - = 2.6 – 0.64 = 1.96
Variance = E(X2) - = 1.96
Standard deviation = = = 1.4
Mean = -0.8
Variance = 1.96
Standard deviation = 1.4
(iv) Given :
To find : mean (𝓊), variance (σ2) and standard deviation (σ)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Standard deviation =
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3) + x4P(x4) + x5P(x5)
Mean = E(X) = -2(0.1) + (-1)(0.2) + 0(0.4) + 1(0.2) + 2(0.1)
Mean = E(X) = -0.2 - 0.2 + 0 + 0.2 + 0.2 = 0
Mean = E(X) = 0
= = 0
E(X2) = = P(x1) + P(x2) + P(x3) + P(x4) + P(x5)
E(X2) = (0.1) + (0.2) + (0.4) + (0.2) + (0.1)
E(X2) = 0.4 + 0.2 + 0 + 0.2 +0.4 = 1.2
E(X2) = 1.2
Variance = E(X2) - = 1.2 – 0 = 1.2
Variance = E(X2) - = 1.2
Standard deviation = = = 1.095
Mean = 0
Variance = 1.2
Standard deviation = 1.095
Find the mean and variance of the number of heads when two coins are tossed simultaneously.
Given : Two coins are tossed simultaneously
To find : mean (𝓊), variance (σ2)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
When two coins are tossed simultaneously,
Total possible outcomes = TT , TH , HT , HH where H denotes head and T denotes tail.
P(0) = (zero heads = 1 [TT] )
P(1) = (one heads = 2 [HT , TH] )
P(2) = (two heads = 1 [HH] )
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() +2() = 0 + + = = 1
Mean = E(X) = 1
= = 1
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () = 0 + + = = = 1.5
E(X2) = 1.5
Variance = E(X2) - = 1.5 – 1 = 0.5
Variance = E(X2) - = 0.5
Mean = 1
Variance = 0.5
Find the mean and variance of the number of tails when three coins are tossed simultaneously.
Given : Three coins are tossed simultaneously
To find : mean (𝓊) and variance (σ2)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
When three coins are tossed simultaneously,
Total possible outcomes = TTT , TTH , THT , HTT , THH , HTH , HHT , HHH where H denotes head and T denotes tail.
P(0) = (zero tails = 1 [HHH] )
P(1) = (one tail = 3 [HTH , THH , HHT ] )
P(2) = (two tail = 3 [HTT , THT , TTH ] )
P(3) = (three tails = 1 [TTT] )
The probability distribution table is as follows,
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3) + x4P(x4)
Mean = E(X) = 0() + 1() +2() +3() = 0 + + + = = =
Mean = E(X) = = 1.5
= = 2.25
E(X2) = = P(x1) + P(x2) + P(x3) + P(x4)
E(X2) = () + () + () + () = 0 + + + = = = 3
E(X2) = 3
Variance = E(X2) - = 3 – 2.25 = 0.75
Variance = E(X2) - = 0.75
Mean = 1.5
Variance = 0.75
A die is tossed twice. ‘Getting an odd number on a toss’ is considered a success. Find the probability distribution of a number of successes. Also, find the mean and variance of the number of successes.
Given : A die is tossed twice and ‘Getting an odd number on a toss’ is considered a success.
To find : probability distribution of the number of successes and mean (𝓊) and variance (σ2)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
When a die is tossed twice,
Total possible outcomes =
{(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6)
(2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6)
(3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6)
(4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6)
(5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6)
(6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6)}
‘Getting an odd number on a toss’ is considered a success.
P(0) = = (zero odd numbers = 9 )
P(1) = = (one odd number = 18 )
P(2) = = (two odd numbers = 9 )
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() +2() = 0 + + = = 1
Mean = E(X) = 1
= = 1
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () = 0 + + = = = 1.5
E(X2) = 1.5
Variance = E(X2) - = 1.5 – 1 = 0.5
Variance = E(X2) - = 0.5
The probability distribution table is as follows,
Mean = 1
Variance = 0.5
A die is tossed twice. ‘Getting a number greater than 4’ is considered a success. Find the probability distribution of a number of successes. Also, find the mean and variance of the number of successes.
Given : A die is tossed twice and ‘Getting a number greater than 4 ’ is considered a success.
To find : probability distribution of the number of successes and mean (𝓊) and variance (σ2)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
When a die is tossed twice,
Total possible outcomes =
{(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6)
(2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6)
(3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6)
(4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6)
(5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6)
(6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6)}
‘Getting a number greater than 4’ is considered a success.
P(0) = = (zero numbers greater than 4 = 16 )
P(1) = = (one number greater than 4= 16 )
P(2) = = (two numbers greater than 4= 4 )
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() +2() = 0 + + = = =
Mean = E(X) =
= =
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () = 0 + + =
E(X2) =
Variance = E(X2) - = – =
Variance = E(X2) - =
The probability distribution table is as follows,
Mean =
Variance =
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of a number of successes, find the probability distribution of the number of successes. Also, find the mean and variance of a number of successes. [CBSE 2008]
Given : A die is tossed twice and ‘Getting a number greater than 4 ’ is considered a success.
To find : probability distribution of the number of successes and mean (𝓊) and variance (σ2)
Formula used :
Mean = E(X) =
Variance = E(X2) -
When a die is tossed 4 times,
Total possible outcomes = 62 = 36
Getting a doublet is considered as a success
The possible doublets are (1,1) , (2,2) , (3,3) , (4,4) , (5,5) , (6,6)
Let p be the probability of success,
p = =
q = 1 – p = 1 - =
q =
since the die is thrown 4 times, n = 4
x can take the values of 1,2,3,4
P(x) = nCx
P(0) = 4C0 =
P(1) = 4C1 = =
P(2) = 4C2 = =
P(3) = 4C3 = =
P(4) = 4C4 =
The probability distribution table is as follows,
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3) + x4P(x4) + x5P(x5)
Mean = E(X) = 0() + 1() + 2() + 3() + 4()
Mean = E(X) = 0 + + + + = = =
Mean = E(X) =
= =
E(X2) = = P(x1) + P(x2) + P(x3) + P(x4) + P(x5)
E(X2) = () + () + () + () + ()
E(X2) = 0 + + + + = =
E(X2) = 1
Variance = E(X2) - = 1 – =
Variance = E(X2) - =
The probability distribution table is as follows,
Mean =
Variance =
A coin is tossed 4 times. Let X denote the number of heads. Find the probability distribution of X. also, find the mean and variance of X.
Given : A coin is tossed 4 times
To find : probability distribution of X and mean (𝓊) and variance (σ2)
Formula used :
Mean = E(X) =
Variance = E(X2) -
A coin is tossed 4 times,
Total possible outcomes = 24 = 16
X denotes the number of heads
Let p be the probability of getting a head,
p =
q = 1 – p = 1 - =
q =
since the coin is tossed 4 times, n = 4
X can take the values of 1,2,3,4
P(x) = nCx
P(0) = 4C0 =
P(1) = 4C1 = =
P(2) = 4C2 = =
P(3) = 4C3 = =
P(4) = 4C4 =
The probability distribution table is as follows,
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3) + x4P(x4) + x5P(x5)
Mean = E(X) = 0() + 1() + 2() + 3() + 4()
Mean = E(X) = 0 + + + + = = = 2
Mean = E(X) = 2
= = 4
E(X2) = = P(x1) + P(x2) + P(x3) + P(x4) + P(x5)
E(X2) = () + () + () + () + ()
E(X2) = 0 + + + + = = = 5
E(X2) = 5
Variance = E(X2) - = 5 – 4 = 1
Variance = E(X2) - = 1
The probability distribution table is as follows,
Mean = 2
Variance = 1
Let X denote the number of times ‘a total of 9’ appears in two throws of a pair of dice. Find the probability distribution of X. Also, find the mean, variance and standard deviation of X.
Given : Let X denote the number of times ‘a total of 9’ appears in two throws of a pair of dice
To find : probability distribution of X ,mean (𝓊) and variance (σ2) and standard deviation
Formula used :
Mean = E(X) =
Variance = E(X2) -
Standard deviation =
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
When a die is tossed twice,
Total possible outcomes =
{(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6)
(2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6)
(3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6)
(4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6)
(5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6)
(6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6)}
Let X denote the number of times ‘a total of 9’ appears in two throws of a pair of dice
p = =
q = 1 - =
Two dice are tossed twice, hence n = 2
P(0) = 2C0 =
P(1) = 2C1 =
P(2) = 2C2 =
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() +2() = 0 + + = = =
Mean = E(X) =
= =
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () = 0 + + =
E(X2) =
Variance = E(X2) - = – =
Variance = E(X2) - =
Standard deviation = = =
The probability distribution table is as follows,
Mean =
Variance =
Standard deviation =
There are 5 cards, numbers 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Find the mean and variance of X.
Given : There are 5 cards, numbers 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn.
To find : mean (𝓊) and variance (σ2) of X
Formula used :
Mean = E(X) =
Variance = E(X2) -
There are 5 cards, numbers 1 to 5, one number on each card. Two cards are drawn at random without replacement.
X denote the sum of the numbers on two cards drawn
The minimum value of X will be 3 as the two cards drawn are 1 and 2
The maximum value of X will be 9 as the two cards drawn are 4 and 5
For X = 3 the two cards can be (1,2) and (2,1)
For X = 4 the two cards can be (1,3) and (3,1)
For X = 5 the two cards can be (1,4) , (4,1) , (2,3) and (3,2)
For X = 6 the two cards can be (1,5) , (5,1) , (2,4) and (4,2)
For X = 7 the two cards can be (3,4) , (4,3) , (2,5) and (5,2)
For X = 8 the two cards can be (5,3) and (3,5)
For X = 9 the two cards can be (4,5) and (4,5)
Total outcomes = 20
P(3) = =
P(4) = =
P(5) = =
P(6) = =
P(7) = =
P(8) = =
P(9) = =
The probability distribution table is as follows,
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3) + x4P(x4) + x5P(x5) + x6P(x6) + x7P(x7)
Mean = E(X) = 3() + 4() + 5() + 6() + 7() + 8() + 9()
Mean = E(X) = + + + + + + = = = 6
Mean = E(X) = 6
= = 36
E(X2) = = P(x1) + P(x2) + P(x3) + P(x4) + P(x5) + P(x6) + P(x7)
E(X2) = () + () + () + () + () + () + ()
E(X2) = + + + + + + = = = 39
E(X2) = 39
Variance = E(X2) - = 39 – 36 = 3
Variance = E(X2) - = 3
Mean = 6
Variance = 3
Two cards are drawn from a well-shuffled pack of 52 cards. Find the probability distribution of a number of kings. Also, compute the variance for the number of kings. [CBSE 2007]
Given : Two cards are drawn from a well-shuffled pack of 52 cards.
To find : probability distribution of the number of kings and variance (σ2)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
Two cards are drawn from a well-shuffled pack of 52 cards.
Let X denote the number of kings in the two cards
There are 4 king cards present in a pack of well-shuffled pack of 52 cards.
P(0) = = =
P(1) = = =
P(2) = = =
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() +2() = 0 + + = =
Mean = E(X) =
= =
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () = 0 + + =
E(X2) =
Variance = E(X2) - = – = = =
Variance = E(X2) - =
The probability distribution table is as follows,
Variance =
A box contains 16 bulbs, out of which 4 bulbs are defective. Three bulbs are drawn at random from the box. Let X be the number of defective bulbs drawn. Find the mean and variance of X.
Given : A box contains 16 bulbs, out of which 4 bulbs are defective. Three bulbs are drawn at random
To find : mean (𝓊) and variance (σ2)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
A box contains 16 bulbs, out of which 4 bulbs are defective. Three bulbs are drawn at random
Let X denote the number of defective bulbs drawn
There are 4 defective bulbs present in 16 bulbs
P(0) = = =
P(1) = = =
P(2) = = =
P(3) = = =
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() + 2() + 3() = 0 + + + =
Mean = E(X) = =
= =
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () + () = 0 + + + =
E(X2) =
Variance = E(X2) - = – = = =
Variance = E(X2) - =
Mean = E(X) =
Variance =
20% of the bulbs produced by a machine are defective. Find the probability distribution of the number of defective bulbs in a sample of 4 bulbs chosen at random. [CBSE 2004C]
Given : 20% of the bulbs produced by a machine are defective.
To find probability distribution of a number of defective bulbs in a sample of 4 bulbs chosen at random.
Formula used :
The probability distribution table is given by ,
Where P(x) = nCx
Here p is the probability of getting a defective bulb.
q = 1 – p
Let the total number of bulbs produced by a machine be x
20% of the bulbs produced by a machine are defective.
Number of defective bulbs produced by a machine = =
X denotes the number of defective bulbs in a sample of 4 bulbs chosen at random.
Let p be the probability of getting a defective bulb,
p = =
p =
q = 1 – p = 1 - =
q =
since 4 bulbs are chosen at random, n = 4
X can take the values of 0,1,2,3,4
P(x) = nCx
P(0) = 4C0 =
P(1) = 4C1 =
P(2) = 4C2 =
P(3) = 4C3 =
P(4) = 4C4 =
The probability distribution table is as follows,
Four bad eggs are mixed with 10 good ones. Three eggs are drawn one by one without replacement. Let X be the number of bad eggs drawn. Find the mean and variance of X.
Given : Four bad eggs are mixed with 10 good ones. Three eggs are drawn one by one without replacement.
To find : mean (𝓊) and variance (σ2)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
Four bad eggs are mixed with 10 good ones. Three eggs are drawn one by one without replacement.
Let X denote the number of bad eggs drawn
There are 4 bad eggs present in 14 eggs
P(0) = = =
P(1) = = =
P(2) = = =
P(3) = = =
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() + 2() + 3() = 0 + + + =
Mean = E(X) = =
= =
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () + () = 0 + + + =
E(X2) =
Variance = E(X2) - = – = =
Variance = E(X2) - =
Mean = E(X) =
Variance =
Four rotten oranges are accidentally mixed with 16 good ones. Three oranges are drawn at random from the mixed lot. Let X be the number of rotten oranges drawn. Find the mean and variance of X.
Given : Four rotten oranges are mixed with 16 good ones. Three oranges are drawn one by one without replacement.
To find : mean (𝓊) and variance (σ2)
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
Four rotten oranges are mixed with 16 good ones. Three oranges are drawn one by one without replacement.
Let X denote the number of rotten oranges drawn
There are 4 rotten oranges present in 20 oranges
P(0) = = =
P(1) = = =
P(2) = = =
P(3) = = =
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() + 2() + 3() = 0 + + + =
Mean = E(X) = =
= =
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () + () = 0 + + + =
E(X2) = =
Variance = E(X2) - = – = =
Variance = E(X2) - =
Mean = E(X) =
Variance =
Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls. Let X be the number of red balls drawn. Find the mean and variance of X.
Given : Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls.
To find : mean (𝓊) and variance (σ2) of X
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls.
Let X be the number of red balls drawn.
P(0) = = =
P(1) = = =
P(2) = = =
P(3) = = =
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() + 2() + 3() = 0 + + + =
Mean = E(X) = =
= =
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () + () = 0 + + + =
E(X2) = =
Variance = E(X2) - = – = =
Variance = E(X2) - =
Mean = E(X) =
Variance =
Two cards are drawn without replacement from a well-shuffled deck of 52 cards. Let X be the number of face cards drawn. Find the mean and variance of X.
Given : Two cards are drawn without replacement from a well-shuffled deck of 52 cards.
To find : mean (𝓊) and variance (σ2) of X
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
Two cards are drawn without replacement from a well-shuffled deck of 52 cards.
Let X denote the number of face cards drawn
There are 12 face cards present in 52 cards
P(0) = = =
P(1) = = =
P(2) = = =
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() + 2() = 0 + + = = =
Mean = E(X) =
= =
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () = 0 + + =
E(X2) =
Variance = E(X2) - = – = =
Variance = E(X2) - =
Mean = E(X) =
Variance =
Two cards are drawn one by one with replacement from a well-shuffled deck of 52 cars. Find the mean and variance of the number of aces.
Given : Two cards are drawn with replacement from a well-shuffled deck of 52 cards.
To find : mean (𝓊) and variance (σ2) of X
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
Two cards are drawn with replacement from a well-shuffled deck of 52 cards.
Let X denote the number of ace cards drawn
There are 4 face cards present in 52 cards
X can take the value of 0,1,2.
P(0) = =
P(1) = = =
P(2) = =
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() + 2() = 0 + + = = =
Mean = E(X) =
= =
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () = 0 + + =
E(X2) =
Variance = E(X2) - = – =
Variance = E(X2) - =
Mean = E(X) =
Variance =
Three cards are drawn successively with replacement from a well – shuffled deck of 52 cards. A random variable X denotes the number of hearts in the three cards drawn. Find the mean and variance of X.
Given : Three cards are drawn successively with replacement from a well – shuffled deck of 52 cards.
To find : mean (𝓊) and variance (σ2) of X
Formula used :
Mean = E(X) =
Variance = E(X2) -
Mean = E(X) = = x1P(x1) + x2P(x2) + x3P(x3)
Three cards are drawn successively with replacement from a well – shuffled deck of 52 cards.
Let X be the number of hearts drawn.
Number of hearts in 52 cards is 13
P(0) = =
P(1) = =
P(2) = =
P(3) = =
The probability distribution table is as follows,
Mean = E(X) = 0() + 1() + 2() + 3() = 0 + + + = =
Mean = E(X) =
= =
E(X2) = = P(x1) + P(x2) + P(x3)
E(X2) = () + () + () + () = 0 + + + = =
E(X2) =
Variance = E(X2) - = – = =
Variance = E(X2) - =
Mean = E(X) =
Variance =
Five defective bulbs are accidently mixed with 20 good ones. It is not possible to just look at a bulb and tell whether or not it is defective. Find the probability distribution from this lot.
Given : Five defective bulbs are accidently mixed with 20 good ones.
To find : probability distribution from this lot
Formula used :
Five defective bulbs are accidently mixed with 20 good ones.
Total number of bulbs = 25
X denote the number of defective bulbs drawn
X can draw the value 0 , 1 , 2 , 3 , 4.
since the number of bulbs drawn is 4, n = 4
P(0) = P(getting a no defective bulb) = = =
P(1) = P(getting 1 defective bulb and 3 good ones) = =
P(1) = =
P(2) = P(getting 2 defective bulbs and 2 good one) =
P(2) = = =
P(3) = P(getting 3 defective bulbs and 1 good one) =
P(3) = =
P(4) = P(getting all defective bulbs) = = =
P(4) =
The probability distribution table is as follows,