Binomial Distribution

Let X be the number of rupees the man won/lost.

Let n be the number of throws required to get a head.

We have two cases, that is

(i). The man tosses once, head comes up, and he quits. (head means he won)

(ii). The man tosses once; tail comes up then he tosses again, tail comes up. (tail means he lost)

(ii). The man tosses, tail comes up then he tosses again, head comes up. (tail means he lost & head means he won)

In Case (i),

The man tosses once, head comes up, and he quits.

Here, number of throws (n) = 1

Amount won/lost (X) = 1

In Case (ii),

The man tosses once; tail comes up then he tosses again, tail comes up.

Here, number of throws (n) = 2

Amount won/lost (X) = -2

In Case (iii),

The man tosses once, tail comes up then he tosses again, head comes up.

Here, number of throws (n) = 2

Amount won/lost (X) = 0

We have the table:

Thus, the probability distribution is

Hence, the answer is obtained.

Let p denote the probability of getting 3, 4 or 5 in a throw of dice.

Let us find out the value of p.

We know, a dice has 6 faces numbered 1, 2, 3, 4, 5 and 6.

So, the probability of getting a 3, 4 or 5 is given as,

If p denotes the probability of getting success, then let q denote the probability of not getting success.

We can say,

p + q = 1

⇒ q = 1 – p

Let X denote the number of successes in the throw of five dice simultaneously.

Let there be total n number of throws of five dice simultaneously.

Then, the probability of getting r successes out of n throws of dice is given by,

P (X = r) = ⁿC_rp^rq^n-r

Now, substitute the value of p and q in the above equation.

Also, put n = 5 (Since there are 5 dice throw)

Now, the probability of getting at least 3 successes is given by

Probability of getting at least 3 successes = P(X = 3) + P(X = 4) + P(X = 5)

Thus,

Thus, the probability of getting 3 successes is 1/2.

Let p be the probability of getting defective items out of 100 items.

And according to the question, the items produced by a company contain 10% defective items.

So, p = 10%

And also, as we know p + q = 1

If p is the probability of getting defective items out of 100, then q is the probability of not getting defective items out of 100.

⇒ q = 1 – p

Let X be the number of defective items drawn out of 8 items. (Since 8 is the sample size)

Then, the probability of getting r defective items out of 8 items is given by,

P (X = r) = ⁿC_rp^rq^n-r

We know that, n = 8

Put all these values in the previous equation.

Then, the probability of getting two defective items out of a sample of 8 is given by putting r = 2.

Now,

Thus, the probability of getting 2 defective items out of a sample of 8 items is .

We know that there are 52 cards in a pack of cards.

And there are 13 cards of each suit.

Let p be the probability of drawing a card of heart from a pack of 52 cards.

Then,

[∵ there are 13 cards of heart in the pack]

Also, p + q = 1

Where if p is the probability of getting a heart out of 52 cards, then q is the probability of not getting a heart out of 52 cards.

⇒ q = 1 – p

Putting the value of p in the above equation, we get

Now, let the card be drawn n times.

And let X denote the number of hearts drawn out of a pack of 52 cards.

Then, the binomial distribution is given by,

P (X = r) = ⁿC_rp^rq^n-r

Putting and above, we get

…(A)

Where r = 0, 1, 2, 3… n

(i). We need to find the number of times a card can be drawn so that at least there is an even chance of drawing a heart.

In simple words, since n is the number of times a card is drawn, we need to find the smallest n for which P (X = 0) is less than 1/4 satisfies.

So,

From equation A, we have

First, put n = 0.

But 1 ≮ 0.25

Now, put n = 1.

⇒ 0.75 < 0.25

But 0.75 ≮ 0.25

Now, put n = 2.

⇒ 0.5625 < 0.25

But 0.5626 ≮ 0.25

Now, put n = 3.

⇒ 0.42 < 0.25

But 0.42 ≮ 0.25

Now, put n = 4.

⇒ 0.31 < 0.25

But 0.31 ≮ 0.25

Now, put n = 5.

⇒ 0.23 < 0.25

Thus, smallest n = 5.

∴, we must draw cards at least 5 times.

(ii). We need to find the number of times a card must be drawn so that the probability of drawing a heart is more than 3/4.

If P (X = 0) is the probability of not drawing a heart at all.

Then, 1 – P (X = 0) is the probability of drawing a heart out of 52 cards.

According to the question,

First, put n = 0.

But 1 ≮ 0.25

Now, put n = 1.

⇒ 0.75 < 0.25

But 0.75 ≮ 0.25

Now, put n = 2.

⇒ 0.5625 < 0.25

But 0.5626 ≮ 0.25

Now, put n = 3.

⇒ 0.42 < 0.25

But 0.42 ≮ 0.25

Now, put n = 4.

⇒ 0.31 < 0.25

But 0.31 ≮ 0.25

Now, put n = 5.

⇒ 0.23 < 0.25

Thus, smallest n = 5.

∴, we must draw cards at least 5 times.

Let X be the number of a graduate assistants.

⇒ X = 8

Let p be the probability of the assistant studying at the office.

Then, q be the probability of the assistant studying at home.

According to the question, each assistant is just likely to study at home as in the office.

Also, we know that (p + q) = 1.

⇒ q = 1 – p

Now, let there be k number of desks in the office.

We need to find the number of desks in the office so that each assistant has a desk at least 90% of the time.

⇒ P (X > k) < 0.10

Now, we must find a value of k that satisfies the above equation.

Clearly,

P (X > 5) = P (X = 6 or X = 7 or X = 8)

⇒ P (X > 5) = 0.14

Also, P (X > 6) = P (X = 7 or X = 8)

⇒ P (X > 6) = 0.035

∴, P (X > 6) < 0.10

Hence, if there are 6 desks, then there is atleast 90% chance for every graduate assistant to get a desk.

Let x be the number of heads in the tosses.

Let n be the total number of tosses.

Then, binomial distribution is given by

P (X = x) = ⁿC_xp^xq^n-x

Where x = 1, 2, 3, …, n

Here, p = probability of getting a head.

And q = probability of getting a tail.

Then,

We need to find the probability of getting at least 6 head.

Then, x = 6, 7, 8 [∵ there are 8 number of tosses]

First, putting n = 8 & x = 6. We get

…(i)

Now, putting n = 8 & x = 7. We get

…(ii)

Now, putting n = 8 & x = 8. We get

…(iii)

The probability of getting at least 6 heads is given by

Probability = P(6) + P(7) + P(8)

Substituting values in (i), (ii) & (iii) in above equation. We get

Thus, the probability of getting at least 6 heads is .

Let p be the probability of getting a head in a toss.

If p is the probability of getting a head in a toss, then q is the probability of getting a tail in a toss.

We have p + q = 1

⇒ q = 1 – p

Let X denote a random variable representing the number of heads in 6 tosses of coin. The probability of getting r heads in n tosses of coins is given by

P (X = r) = ⁿC_rp^rq^n-r

We know the values of n, p, and q.

n = 6

We can re-write the binomial distribution as,

…(A)

(i). We need to find the probability of getting 3 heads.

It is given by,

Probability = P (X = 3)

So, putting r = 3 in equation (A). We get

Thus, the probability of getting 3 heads out of tosses of 6 coins is .

(ii). We need to find the probability of getting no head.

It is given by,

Probability = P (X = 0)

So, putting r = 0 in equation (A). We get

Thus the probability of getting 0 heads out of tosses of 6 coins is .

(iii). We need to find the probability of getting at least 1 head.

It is given by,

Probability = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)

Or

Probability = 1 – P (X = 0)

Let us use the shorter formula.

We know from the result of part (ii),

Using this result, let us find out the probability of getting at least one head.

Then,

Thus, the probability of getting at least one head is .

Let p be the probability of the tube to function for more than 500 hours.

This probability is given as 0.2.

⇒ p = 0.2

If p is the probability of the tube to function for more than 500 hours, then q is the probability of the tube to not function for more than 500 hours.

⇒ p + q = 1

⇒ q = 1 – p

Let X denote a random variable that represents the number of the tube that can function for more than 500 hours out of the total 4 tubes.

And let n denote the total number of tube taken in the sample, that is, 4.

Then binomial distribution for r tube to function more than 500 hours out of 4 tubes is given by,

P (X = r) = ⁿC_rp^rq^n-r

Putting n = 4, and above, we get

We need to find the probability that exactly 3 tubes will function for more than 500 hours.

So, put r = 3. We get

Thus, the probability that exactly 3 tubes will function for more than 500 hours is .

Given that, a certain kind of component will survive a given shock

Let p be the probability that component survives the shock test.

Then,

If p is the probability that component survives the shock test, then q is the probability that the component doesn’t survive the shock test.

⇒ p + q = 1

⇒ q = 1 – p

Let X denote a random variable that represents the components that survive the shock test out of the 5 components tested.

The probability that r components out of n components survive the shock test is given by the binomial distribution.

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 5 (sample components tested)

And

We can re-write it as,

…(A)

(i). We need to find the probability that among 5 components tested exactly 2 will survive.

Then, put r = 2 in equation (A). We have

⇒ P (X = 2) = 0.0879

∴, the probability that exactly 2 components out of 5 will survive the test is 0.0879.

(ii). We need to find the probability that among the 5 components tested at most 3 will survive.

So, this can be give in two ways:

(a). Probability that out of 5 components at most 3 will survive = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)

(b). Probability that out of 5 components at most 3 will survive = 1 – [P (X = 4) + P (X = 5)]

Let us solve it by using the formula in (b).

So, let us find out P (X = 4).

Putting r = 4 in equation (A), we get

Now, let us find out P (X = 5).

Putting r = 5 in equation (A), we get

Using the values of P (X = 4) & P (X = 5) in formula (b), we get

⇒ Probability = 0.3672

∴, the probability that out of 5 components at most 3 will survive is 0.3672.

We have been given that, the probability that a bomb dropped from an airplane will strike a certain target is 0.2.

Also, that 6 bombs are dropped.

Let p be the probability that a bomb dropped from an airplane will strike a certain target.

Then, q is the probability that a bomb dropped from an airplane will not strike a certain target.

⇒ p = 0.2

We know that, p + q = 1

⇒ q = 1 – p

Let X be a random variable the represents the number of bombs that strike the target.

Then, the probability that r bombs strike the target out of n bombs is given by,

P (X = r) = ⁿC_rp^rq^n-r

Where n = 6

Let us put the values of n, p, and q in the above equation.

…(A)

(i). We need to find the probability that exactly 2 will strike the target out of 6 bombs.

Probability is given by,

Probability = P (X = 2)

So, put r = 2 in equation (A).

⇒ Probability = 0.24576

∴, the probability that exactly 2 will strike the target out of 6 bombs is 0.24576.

(ii). We need to find the probability that at least 2 will strike the target out of 6 bombs.

Probability is given by,

Probability = P (X ≥ 2)

This can also be written as,

Probability = 1 – P (X < 2)

⇒ Probability = 1 – [P (X = 0) + P (X = 1)]

Let us find the value of P (X = 0).

For this, put r = 0 in equation (A).

Now, let us find the value of P (X = 1).

For this, put r = 0 in equation (A).

Now, putting all these values in 1 – [P (X = 0) + P (X = 1)].

⇒ Probability = 0.345

∴, the probability that at least 2 will strike the target out of 6 bombs is 0.345.

Given that, the probability that the mice are protected from a certain disease is 60%.

Also, the sample size of mice = 5

Let p be the probability of the mice not contracting with a certain disease.

Then, q is the probability of the mice contracting with a certain disease.

⇒ p = 60%

And we know that,

p + q = 1

⇒ q = 1 – p

Let X be a random variable representing a number of mice contracting with the disease.

Then, the probability of r mice contracting with the disease out of n mice inoculated is given by the following binomial distribution.

P (X = r) = ⁿC_rq^rp^n-r

Putting the values,

n = 5,

&

We get

…(A)

(i). We need to find the probability that none of the mice contract with the disease.

For this, put r = 0.

We get the probability as,

Probability = P (X = 0)

From equation (A),

⇒ P (X = 0) = 0.07776

∴, the probability that none of the mice contracts with the disease is 0.07776.

(ii). We need to find the probability that more than 3 mice contract the disease.

So, probability = P (X = 4) + P (X = 5)

⇒ Probability = 0.2125

∴, the probability that more than 3 mice contract the disease is 0.2125.

Let p be the probability of success in the experiment. Then, q is the probability of failure in the experiment.

According to the question,

An experiment succeeds twice as often as it fails.

⇒ p = 2q

But p + q = 1

⇒ 2q + q = 1 [∵ p = 2q]

⇒ 3q = 1

Then, p = 2q

Let X be a random variable representing the number of successes out of 6 experiments.

Then, the probability of getting r success out of n experiments is given by,

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 6,

&

…(i)

Then, probability that there will be at least 4 successes out of 6 is given by,

Probability = P (X = 4) + P (X = 5) + P (X = 6)

Put r = 4, 5 and 6 respectively for P (X = 4), P (X = 5) and P (X = 6) in equation (i), we have

⇒ Probability = 0.68

∴, the required probability that in the next 6 trials there will be at least 4 successes is 0.68.

Given that, the chance of any one of the 20 kidney dialysis machines to be out of service during a day is 0.02.

Let p be the probability of a kidney dialysis machine to get out of service.

⇒ p = 0.02

Then, q is the probability of a kidney dialysis machine to not be out of service during a day.

And we know that, p + q = 1

⇒ q = 1 – p

Let X be a random variable that represents a number of machines out of service during a day out of n machines.

Then, the probability of r machines out of total n machines taken in the sample to get out of service is given by,

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 20 (as given in the question)

Putting the values of n, p & q in the above formula, we get

…(i)

We need to find the probability that exactly 3 machines will be out of service on the same day.

So, for this, just put r = 3 in the formula (i),

Observe the formula so obtained after substituting the value of r.

The calculation will be huge since the values of p and q are very small.

So, in this case of low probability events, we use Poisson’s distribution rather than Binomial distribution.

Then,

Poisson’s constant can be found out as,

λ = np

where n = 20 & p = 0.02.

⇒ λ = 20 × 0.02

⇒ λ = 0.4

Poisson’s distribution is given as,

Put r = 3,

Look up in the table for the value of .

⇒ P (X = 3) = 0.0071

∴, the required probability that exactly 3 machines will be out of service on one day is 0.0071.

Given that, the probability that a student entering a university will graduate is 0.4.

Let p be the probability that a student entering a university will graduate.

⇒ p = 0.4

Then, q is the probability of a student not graduating.

But, p + q = 1

⇒ q = 1 – p

Let X be a random variable that represents the number of students out of n students graduating after entering a university.

Then, the probability of r students out of n students graduating after entering a university is given by,

P (X = r) = ⁿC_rp^rq^n-r

Here, a sample size of students, n = 3

Putting the value of n, p, and q in the above formula, we get

…(A)

(i). We need to find the probability that out of 3 students entering a university, none will graduate.

Put r = 0 in equation (A). We get

∴, the probability that none will graduate is 0.216.

(ii). We need to find the probability that out of 3 students only 1 will graduate.

So, put r = 1 in equation (A). We get

⇒ P (X = 1) = 0.432

∴, the probability that exactly one will graduate is 0.432.

(iii) We need to find the probability that out of 3 students all will graduate.

So, put r = 3 in equation (A). We get

⇒ P (X = 3) = 0.064

∴, the probability that all will graduate is 0.064.

Given that, 10 eggs are drawn successively.

Defective eggs in the lot = 10%

Let p be the probability that the eggs drawn from the lot are defective.

⇒ p = 10%

Then, q is the probability that the eggs drawn from the lot is not defective.

And, p + q = 1

⇒ q = 1 – p

Let X be a random variable that represents defective eggs picked out of n eggs from the lot.

Then, the probability of taking r defective eggs out of n eggs from the lot is given by,

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 10

Putting the values of n, p, and q in the above formula, we get

…(i)

We need to find the probability of getting at least one defective egg from the lot.

This is represented as,

Probability = P (X ≥ 1)

This is also written as,

P (X ≥ 1) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) + P (X = 9) + P (X = 10)

Or

P (X ≥ 1) = 1 – P (X < 1)

⇒ P (X ≥ 1) = 1 – P (X = 0)

Putting r = 0 in P (X = r) formula in (i), we get

∴, the probability of getting at least one defective egg from the lot is .

Given that, there are 20 questions of true-false exam.

If the coin falls head, he answers “true.”

If the coin falls tail, he answers “false.”

Let p be the probability of a correct answer.

That is, p = getting a head and a right answer to be “true” for a question or getting a tail and a right answer to be “false” for a question.

Then, q is the probability of the answer to be incorrect.

And, p + q = 1

⇒ q = 1 – p

Let X denote a random variable representing the number of correct answers out of 20 questions.

Then, the probability of getting r correct answers out of n answered questions is given by,

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 20 [∵ there are 20 questions in total]

Putting values of n, p, and q in the above equation, we get

…(i)

We need to find the probability that he answers at least 12 questions correctly.

This can be represented as,

Probability = P (X ≥ 12)

It can be written as,

P (X ≥ 12) = P (X = 12) + P (X = 13) + P (X = 14) + P (X = 15) + P (X = 16) + P (X = 17) + P (X = 18) + P (X = 19) + P (X = 20) …(ii)

Put r = 12, 13, 14, 15, 16, 17, 18, 19, 20 in equation (i) subsequently and substitute in (ii), we get

∴, we have got the required probability.

Given that, X has a binomial distribution.

Also, n = 6 & p = 1/2

And as we know that, p + q = 1

⇒ q = 1 – p

Let us define a binomial distribution with x as a variable out of other n variables.

P (X = x) = ⁿC_xp^xq^n-x

Putting values of n, p and q, we get

From the derived result, it can be seen that P (X = x) will be maximum if ⁶C_x is maximum.

Let us check at what value of x would ⁶C_x be maximum.

Check at x = 0.

⇒ ⁶C₀ = 1

Check at x = 6.

⇒ ⁶C₆ = 1

Check at x = 1.

⇒ ⁶C₁ = 6

Check at x = 5.

⇒ ⁶C₅ = 6

Check at x = 2.

⇒ ⁶C₂ = 15

Check at x = 4.

⇒ ⁶C₄ = 15

Check at x = 3.

⇒ ⁶C₃ = 20

Note that at x = 3, ⁶C_x is maximum.

Hence, we have shown that x = 3 is the most likely outcome.

We have been given that, there are 3 possible answers of the total 5 questions out of which only 1 answer is correct.

Let p be the probability of getting a correct answer out of the 3 alternative answers.

Then, q is the probability of not getting a correct answer out of 3 alternatives.

And, p + q = 1

⇒ q = 1 – p

Let X be any random variable representing a number of correct answers just be guessing out of 5 questions.

Then, the probability that the candidate would get r answers correct by just guessing out of 5 questions is given by this Binomial distribution.

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 5.

Substitute the value of n, p, and q in the above formula.

…(i)

We need to find the probability that a candidate would get four or more correct answers just by guessing.

The probability can be expressed as,

Probability = P (X ≥ 4)

This is in turn can be written as,

P (X ≥ 4) = P (X = 4) + P (X = 5)

Put r = 4, 5 subsequently in equation (i) and then substitute in the above formula.

⇒ P (X ≥ 4) = 0.0453

Hence, the probability that a candidate would get four or more correct answers just by guessing is 0.0453.

Given that, a person buys a lottery ticket in 50 lotteries.

The probability of winning a prize is 1/100.

Let p be the probability of winning a prize.

Then,

And q be the probability of not winning a prize.

We can write,

p + q = 1

⇒ q = 1 – p

Let X be a random variable representing the number of times the person wins the lottery out of n lotteries.

Then, the probability of the person winning the lottery r times out of n times is given by this Binomial distribution.

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 50

So, putting the value of n, p, and q in the formula of P (X = r), we get

…(A)

(i). We need to find the probability that he will win the prize at least once.

The probability is given by,

Probability = P (X ≥ 1)

Or this can be written as,

P (X ≥ 1) = 1 – P (X < 1)

⇒ P (X ≥ 1) = 1 – P (X = 0) …(B)

So, put r = 0 in equation (A) and then substitute in equation (B), we get

Thus, the probability that he will win the prize at least once is .

(ii). We need to find the probability that he will win the prize exactly once.

The probability is given by,

Probability = P (X = 1)

Put r = 1 in equation (A), we get

Thus, the probability that he will win the prize exactly once is .

(iii). We need to find the probability that he will win at least twice.

The probability is given by,

Probability = P (X ≥ 2)

Or can be expressed as,

P (X ≥ 2) = 1 – P (X < 2)

⇒ P (X ≥ 2) = 1 – [P (X = 0) + P (X = 1)] …(B)

Put r = 0 and r = 1 in equation (A) one by one and then substitute it in the equation (B), we get

Thus, the probability that he will win the prize at least twice is .

Given that, the probability of a shooter hitting a target is 3/4.

Let p be the probability of hitting a target and q be the probability of not hitting a target.

Then,

But, we know p + q = 1

⇒ q= 1 – p

Let the shooter shoot n times in total.

Let X be a random variable representing the number of times the shooter hits the target out of total n times.

Then, the probability of hitting the target r times out of total n times is given by Binomial distribution as,

P (X = r) = ⁿC_rp^rq^n-r

Substitute the value of p and q in the above formula, we get

…(i)

We need to find the minimum number of times the shooter must fire so that the probability of hitting the target at least once is more than 0.99.

It can be represented as,

P(hitting the target atleast once) > 0.99

⇒ P (X ≥ 1) > 0.99

⇒ 1 – P (X < 1) > 0.99

⇒ 1 – P (X = 0) > 0.99

Put r = 0 in equation (i) and then substitute in the above equation, we have

⇒ 4ⁿ > 100

We need to find the minimum value of n to satisfy this inequality.

Take n = 0.

⇒ 4⁰ > 100

⇒ 1 > 100

But 1 ≯ 100.

Take n = 1.

⇒ 4¹ > 100

⇒ 4 > 100

But 4 ≯ 100.

Take n = 2.

⇒ 4² > 100

⇒ 16 > 100

But 16 ≯ 100.

Take n = 3.

⇒ 4³> 100

⇒ 64 > 100

But 64 ≯ 100.

Take n = 4.

⇒ 4⁴ > 100

⇒ 256 > 100

It is true.

Hence, the minimum value of n to satisfy the inequality is 4.

∴, the shooter must fire 4 times.

Let the man toss the coin n times.

Let p be the probability of getting a head in a toss.

Then, q is the probability of getting a tail in a toss.

Since the coin has only two outcomes, so the probability of getting a head = 1/2

Also, p + q = 1

⇒ q = 1 – p

Let X be a random variable that represents a number of occurrence of the head in n tosses of a fair coin.

Then, the probability of getting r number of heads out of total n tosses is given by this Binomial distribution.

P (X = r) = ⁿC_rp^rq^n-r

Substituting the value of p and q in the above equation, we get

…(i)

We need to find the number of times the man must toss a fair coin so that the probability of having at least one head is more than 90%.

We can represent it as,

P (getting atleast one head) > 90%

⇒ P (X ≥ 1) > 90%

⇒ 1 – P (X < 1) > 90% [∵ P (X ≥ 1) = 1 – P (X < 1)]

⇒ 1 – P (X = 0) > 90% [∵ P (X < 1) = P (X = 0)]

Put r = 0 in equation (i) and then, substituting it in the above equation.

⇒ 2ⁿ > 10

Now, we need to find the minimum value of n that satisfy this inequality.

Put n = 0.

⇒ 2⁰ > 10

⇒ 1 > 10

But 1 ≯ 10.

Put n = 1.

⇒ 2¹ > 10

⇒ 2 > 10

But 2 ≯ 10.

Put n = 2.

⇒ 2² > 10

⇒ 4 > 10

But 4 ≯ 10.

Put n = 3.

⇒ 2³ > 10

⇒ 8 > 10

But 8 ≯ 10.

Put n = 4.

⇒ 2⁴ > 10

⇒ 16 > 10

It is true.

Thus, the minimum n that satisfy this inequality is 4.

Hence, the man should toss the coin 4 or more times.

Let the man toss the coin n times.

Let p be the probability of getting a head in a toss.

Then, q is the probability of getting a tail in a toss.

Since the coin has only two outcomes, so the probability of getting a head = 1/2

Also, p + q = 1

⇒ q = 1 – p

Let X be a random variable that represents a number of occurrence of the head in n tosses of a fair coin.

Then, the probability of getting r number of heads out of total n tosses is given by this Binomial distribution.

P (X = r) = ⁿC_rp^rq^n-r

Substituting the value of p and q in the above equation, we get

…(i)

We need to find the number of times the man must toss a fair coin so that the probability of having at least one head is more than 80%.

We can represent it as,

P (getting atleast one head) > 80%

⇒ P (X ≥ 1) > 80%

⇒ 1 – P (X < 1) > 80% [∵ P (X ≥ 1) = 1 – P (X < 1)]

⇒ 1 – P (X = 0) > 80% [∵ P (X < 1) = P (X = 0)]

Put r = 0 in equation (i) and then, substituting it in the above equation.

⇒ 2ⁿ > 5

Now, we need to find the minimum value of n that satisfy this inequality.

Put n = 0.

⇒ 2⁰ > 5

⇒ 1 > 5

But 1 ≯ 5.

Put n = 1.

⇒ 2¹ > 5

⇒ 2 > 5

But 2 ≯ 5.

Put n = 2.

⇒ 2² > 5

⇒ 4 > 5

But 4 ≯ 5.

Put n = 3.

⇒ 2³ > 5

⇒ 8 > 5

It is true.

Thus, the minimum n that satisfy this inequality is 3.

Hence, the man should toss the coin 3 or more times.

Given that, a pair of dice is thrown 4 times.

And a doublet is considered as a success.

Let p be the probability of getting a doublet in a throw of a pair of dice.

Since, there are 36 possible outcomes in total. {(1, 1), (1, 2), (1, 3), …, (1, 6), ..., (2, 6), …, (6 ,6)}

And the 6 possible doublets in 36 outcomes. {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

So,

And also, p + q = 1

⇒ q = 1 – p

Let X denote a random variable representing a number of doublets (successes) out of 4 throws.

So, Binomial distribution of getting r successes out of 4 throws is given by

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 4.

Now, substituting values of n, p and q in the formula P (X = r). We get

…(i)

We need to find the probability distribution of the number of successes.

The probability of 0 success in 4 throws is given by,

Probability = P (X = 0)

Put r = 0 in (i),

The probability of 1 success in 4 throws is given by,

Probability = P (X = 1)

Put r = 1 in (i),

The probability of 2 successes in 4 throws is given by,

Probability = P (X = 2)

Put r = 2 in (i),

The probability of 3 successes in 4 throws is given by,

Probability = P (X = 3)

Put r = 3 in (i),

The probability of 4 successes in 4 throws is given by,

Probability = P (X = 4)

Put r = 4 in (i),

Thus, the probability distribution is

Given that, there are 30 bulbs, which include 6 defective bulbs.

A sample of 4 bulbs is drawn at random with replacement.

Let p be the probability of defective bulbs.

Since 6 bulbs are defective out of 30 bulbs.

Then, let q be the probability of fine bulbs.

And we know, p + q = 1

⇒ q = 1 – p

Let X denote a random variable representing a number of defective bulbs out of 4 bulbs drawn at random.

So, Binomial distribution of getting r successes out of 4 bulbs drawn at random is given by

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 4.

Now, substituting values of n, p and q in the formula P (X = r). We get

…(i)

We need to find the probability distribution of the number of successes.

The probability of 0 defective bulb in 4 sample bulbs is given by,

Probability = P (X = 0)

Put r = 0 in (i),

The probability of 1 defective bulb in 4 sample bulbs is given by,

Probability = P (X = 1)

Put r = 1 in (i),

The probability of 2 successes in 4 throws is given by,

Probability = P (X = 2)

Put r = 2 in (i),

The probability of 3 successes in 4 throws is given by,

Probability = P (X = 3)

Put r = 3 in (i),

The probability of 4 successes in 4 throws is given by,

Probability = P (X = 4)

Put r = 4 in (i),

Thus, the probability distribution is

Given that, obtaining multiple of 3 in a throw of a die is a success.

There are total 10 throws of a fair die.

Let p be the probability of getting multiple of 3 in a throw of a die.

Since there can be a total 6 outcomes in a throw of a die. That is, {1, 2, 3, 4, 5, 6}.

And a multiple of 3 in a die is {3}, {6}.

Then, let q be the probability of not getting a multiple of 3 in a throw of a die.

And we know, p + q = 1

⇒ q = 1 – p

Let X be a random variable representing a number of successes (getting multiple of 3 in die) out of 10 throws of a die.

Then, the probability of getting r successes out of n throws of die is given by,

P (X = r) = ⁿC_rp^rq^n-r

Here n = 10.

Now, put values of n, p, and q in the above equation.

…(i)

We need to find the probability of getting a multiple of 3 in at least 8 of the throws out of 10 throws of a fair die.

It is given,

Probability = P (X ≥ 8)

This can be written as,

P (X ≥ 8) = P (X = 8) + P (X = 9) + P (X = 10)

Just out r = 8, 9, 10 in equation (i) to find the value of P (X = 8), P (X = 9), P (X = 10) respectively, then substitute in the above equation.

Thus, the probability of getting a multiple of 3 in at least 8 of the throws out of 10 throws of a fair die is 201/3¹⁰.

Given that, a die is thrown 5 times.

Let p be the probability of getting an odd number in a throw.

Since there are 6 possible outcomes in a throw of a die. That is, {1, 2, 3, 4, 5, 6}

And there are 3 odd numbers out of these 6 outcomes. That is, {1, 3, 5}

Then, let q be the probability of getting an even number in a throw.

And we know that, p + q = 1

⇒ q = 1 – p

Let X be a random variable representing a number of odd numbers in n throw of a die.

Then, the probability of getting r odd numbers out of n throw of the die can be given as,

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 5

Putting values of n, p, and q in the above formula. We get

…(i)

We need to find the probability that an odd number will come up exactly three times.

It is given by,

Probability = P (X = 3)

Put r = 3 in equation (i) to find P (X = 3), we get

Thus, the probability of getting an odd number to come up exactly three times is 5/16.

Given that, the probability of a man hitting a target is 0.25.

And he shoots 7 times in total.

Let p be the probability of hitting the target.

Then,

p = 0.25

Then, q be the probability of not hitting the target.

And we know that,

p + q = 1

⇒ q = 1 – p

Let X be a random variable representing the number of times the man hits the target out of n shoots.

Then, the probability of hitting the target r times out of n times is given by,

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 7

Putting the values of n, p, and q in the above equation, we get

…(i)

We need to find the probability of hitting the target at least twice.

This can be expressed as,

Probability = P (X ≥ 2)

Or

P (X ≥ 2) = 1 – P (X < 2)

⇒ P (X ≥ 2) = 1 – [P (X = 0) + P (X = 1)]

So, put r = 0, 1 in equation (i) to get P (X = 0) and P (X = 1) respectively and then, substitute in the above formula.

We get

Thus, the probability of hitting the target at least twice is 4547/8192.

Given that, the probability that one bulb is defective is 1/50.

The bulbs are packed in boxes of 10.

Let p be the probability of bulb being defective.

Then,

Let q be the probability of the bulb not being defective.

Also, we know that

p + q = 1

⇒ q = 1 – p

Let X be a random variable representing a number of defective bulbs out of n bulbs.

Then, the probability of r bulbs to be defective out of n bulbs is given by,

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 10

Putting the values of n, p, and q in the above equation. We get

…(A)

(i). We need to find the probability that none of the bulbs is defective.

The probability is given by,

Probability = P (X = 0)

Put r = 0 in equation (A),

∴, the probability that none of the bulbs is defective is (49/50)¹⁰.

(ii). We need to find the probability that exactly two bulbs are defective.

The probability is given by,

Probability = P (X = 2)

Put r = 2 in equation (A),

∴, the probability that none of the bulbs is defective is .

(iii). We need to find the probability of more than 8 bulbs working properly.

This can also be interpreted as, the probability that at most 2 bulbs are defective.

This can be represented as,

Probability = P (X ≤ 2)

Or

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

Put r = 0, 1, 2 to find P (X = 0), P (X = 1) and P (X = 2) and then, substitute in the above equation.

∴, the probability of more than 8 bulbs working properly is .

Given that, a box has 20 pens out of which 2 are defective.

Let p be the probability of a number of pens being defective out of 20 pens.

Then, q be the probability of a number of pens not being defective.

Also, p + q = 1

⇒ q = 1 – p

Let X be a random variable representing a number of defective pens out of 5 pens.

Then, the probability of getting r defective pens out of n pens is given by,

P (X = r) = ⁿC_rp^rq^n-r

Here, n = 5

Putting all the values of n, p, and q in the above equation, we get

…(i)

We need to find the probability that at most 2 pens are defective out of 5 pens.

This can be represented as,

Probability = P (X ≤ 2)

Or

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

Put r = 0, 1, 2 in equation (i) to find P (X = 0), P (X = 1) and P (X = 2), and then substitute in the above equation. We get,

Thus, the probability that at most 2 pens are defective out of 5 pens is .

Let X be a binomial variate with parameters in and p.

Mean = np

Variance = npq

Mean - Variance = np – npq

= np(1 - q)

= np² > 0

So, Mean – Variance > 0

⇒ Mean >Variance

So, mean can never be less than variance.

Let X denote the variance with parameters n and p

p + q = 1 ⇒ q = 1 - p

Now, np = 9

⇒ n = 12

The required binomial distribution is given by,

Let X denote the variance with parameters n and p

p + q = 1 ⇒ q = 1 - p

Given, mean = np = 9 and variance = npq = 6

npq = 6

⇒9q = 6

Now, np = 9

⇒ n = 27

The required binomial distribution is given by,

Given that,

n = 5

Also, Mean+ Variance = 4.8

⇒np + npq = 4.8

⇒np(1+q) = 4.8

⇒5p(1+q) = 4.8

⇒5(1 - q)(1+q) = 4.8 [Since, p + q = 1]

⇒5(1 - q²) = 4.8

The required binomial distribution is given by,

Let X denote the variance with parameters n and p

p + q = 1 ⇒ q = 1 - p

Given, mean = np = 20 and variance = npq = 16

npq = 16

⇒20q = 16

Now, np = 20

⇒ n = 100

∴ n = 100 and

The required binomial distribution is given by,

Let n and p be the parameters of the required binomial distribution. So,

1 - p = q

Also,

⇒625q = 150[(1+q)²]

⇒25q = 6[(1+q)²]

⇒6 + 6q² + 12q – 25q = 0

⇒6q² – 13q + 6 = 0

⇒6q² – 9q - 4q + 6 = 0

⇒3q(2q - 3) - 2(2q - 3) = 0

⇒(2q - 3)(3q - 2) = 0

⇒(2q - 3) = 0 or (3q - 2) = 0

As, q≤1

Now, p = 1 - q

⇒n = 15

The required binomial distribution is given by,

Standard deviation =

⇒ npq = 16

Therefore, np = 20, npq = 16

Let X denote the variance with parameters n and p

p + q = 1 ⇒ q = 1 - p

mean = np = 20 and variance = npq = 16

npq = 16

⇒20 q = 16

Now, np = 20

⇒ n = 100

Let p = probability of selecting defective bolt, so

p = 0.1

q = 1 – p

Given, n = 400,

(i) mean = np

(ii) Standard deviation =

=

=

∴ Standard deviation = 6

So, mean = 40, standard deviation = 6

Let X denote the variance with parameters n and p

p + q = 1 ⇒ q = 1 - p

Now, np = 5

⇒ n = 15

The required binomial distribution is given by,

Let p be the probability of a ship returning safely to ports, so

Given, n = 500

Mean = np

=

= 450

∴ Mean = 450

Standard deviation =

= 6.71

So, mean = 450, standard deviation = 6.71

Given that parameters for binomial distribution are n, p

Now, np = 16

⇒ n = 32

The required binomial distribution is given by,

P(X≥2) = 1 - [P(X = 0) + P(X = 1)]

So,

Let p be the probability of success

∴ the p = probability of getting 5 or 6

Given, n = 8

Mean = np

= 2.66

Standard deviation =

= 1.33

Therefore, mean = 2.66, standard deviation = 1.33.

Let n and p be the parameters of the binomial distribution.

Let, p = probability of having a boy in the family

Given, p = q

Since, p + q = 1

⇒ p + p = 1

⇒ 2p = 1

Given, n = 8

The expected number of boys = np

= 4

The expected number of boys = 4

Let p be the probability of a defective item produced in the factory, so,

p = 0.02

As, p + q = 1

Given, n = 10,000

Expected number of defective items = np

= 200

Standard deviation =

= 14

So, the expected number of defective items = 200

Standard deviation = 14

Let p be the probability of success

∴ the p = probability of getting 1 or 6

Given, n = 3

Mean = np

= 1

Variance = npq

= 0.66

Therefore, mean = 1, standard deviation = 0.66.

Let parameters for binomial distribution be n, p

Now, np = 3

⇒ n = 6

The required binomial distribution is given by,

Now, P(X≤5) = 1 - P(X = 6)

Given that parameters for binomial distribution are n, p

Now, np = 4

⇒ n = 8

The required binomial distribution is given by,

Now, P(X≥5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

Given that parameters for binomial distribution are n, p

⇒ n = 4

The required binomial distribution is given by,

Now, P(X ≥ 1) = 1 - P(X = 0)

Let, n and p be the parameters of the binomial distribution,

Given, n = 6

Now, p = 1 - q

The required binomial distribution is given by,

Throwing doublet: {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Total outcomes = 6×6 = 36

Let p = probability of success

So, q = 1 - p

Given, n = 4

The required binomial distribution is given by,

Let, X, be the random variable getting doublet, so,

Throwing doublet: {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Total outcomes = 6×6 = 36

Let p = probability of success

So, q = 1 - p

Given, n = 3

The required binomial distribution is given by,

Let, X, be the random variable getting doublet, so,

Out of 15 bulbs, five are defective.

Let p = probability that drawn bulb is defective

Also, the q = probability that drawn bulb is not defective.

Let X be the probability that the drawn bulb is defective out of 4.

Then, X follows a binomial distribution with,

The required binomial distribution is given by,

Let, X, be the random variable getting defective ball, so,

Let p = probability of getting a 2 when a die is thrown.

X follows a binomial distribution with,

Let p be the probability of getting an even number on the toss when a die is thrown.

Let q be the probability of not getting an even number on the toss when a die is thrown.

As X follows a binomial distribution with,

∴ Variance = npq

Let p be the probability of getting a spade card.

Let q be the probability of getting a spade card.

X follows a binomial distribution with,

The probability distribution is given by,

The required binomial distribution is given by,

So, mean = np

Let p = probability of drawing a red ball

Let number of red balls drawn be X.

X follows a binomial distribution with,

The required binomial distribution is given by,

So, mean = np

Total oranges = 5+20 = 25

Let p = probability of drawing a bad orange

Let some bad oranges drawn be X.

X follows a binomial distribution with,

The required binomial distribution is given by,

So, mean = np

Let p = probability of drawing a red card

Let a number of the red cards drawn be X.

X follows a binomial distribution with,

The required binomial distribution is given by,

So, mean = np

Given:

n=20

q=0.75

p=?

q=(1-p)

=0.75

p=1-0.75

=0.25

Mean = np

= 20(0.75)

= 15

Given:

Mean=5=np

Variance=4

=np(1-p)

n=?

5(1-p)=4

P=

np=5

n= 25

Given:

n=200

p=0.2

mean=?

Mean=np

= 200(0.2)

=40

Given:

Mean=20

Standard deviation=np(1-p)

=4

P=?

20(1-p)=4

Given:

Mean=10=np

Standard deviation=2

=npq

q=?

10q=4

Given:

Mean=4

Variance=2

P(x=1)=?

P(x=r)= ⁿC_r p^r q^n-r

P(x=1)= ⁿC₁ pq^n-1

= npq^n-1

np(1-p)=2

4(1-p)=2

=q

np=4

n=8

P(x=1)

Given:

Mean= 2

Variance=1

P(x>1)=?

P(x>1)=1-p(x1)

= 1-[p(x=0)+p(x=1)]

=1-[ 11qⁿ +npq^n-1]

2(1-p)=1

=q

np=2

n=4

Given:

n = 4

p(x=0)=

q=?

P(x=r)= ⁿC_r p^r q^n-r

p(x=0)= 11q⁴

Given:

Mean=4

Variance=3

P(x=0)=?

P(x=r)= ⁿC_r p^r q^n-r

p(x=0)= 11qⁿ

4(1-p)=3

=q

np=4

n=16

Given:

P (X = 1) = P (X = 2) = α

P (X= 4)=?

P(x=r)= ⁿC_r p^r q^n-r

P(x=1)= ⁿC₁ p¹ q^n-1

=npq^{n-1 (1)}

P(x=2)= ⁿC₂ p² q^n-2

⁽²⁾

Equating both equations ,we have:

2q = (n-1) p

4q² = (n-1)² p²

P(x=4)= ⁿC₄ p⁴ q^n-4

For large n; n

0

So,

Given:

n=4

p(getting atleast one head)=?

• 0 head : HHHH

• 1 head : HTTT

THTT

TTHT

TTTH

• 2 heads :HHTT

HTHT

HTTH

THHT

THTH

TTHH

• 3 heads: HHHT

HHTH

HTHH

THHH

• 4 heads : HHHH

Total outcomes = 16

Outcomes containing at least 1 head = 15

P(getting at least 1 head)=

Given:

n=5

p(x=2)=9p(x=3)

p=?

P(x=r)= ⁿC_r p^r q^n-r

P(x=1)= ⁵C₁ p¹ q⁴

= 5p q⁴

P(x=3)= ⁵C₃ p³ q²

= 20 p³ q²

Put values in equation, we have:

5p q⁴= 9(20 p³ q²)

q²= 9(4 p²)

(1-p)²= 36p²

1+p²-2p= 36p²

35p²+2p-1=0

Given:

Total bulbs=100

Defective bulb=10

n=5

P(non-defective bulbs)=?

P(defective bulbs)

=p

P(non-defective bulbs)=p(x=0)

= ⁵C₀ p⁰ q⁵

Given:

n = 4

p(x=4)=?

P(x=r)= ⁿC_r p^r q^n-r

p(x=0)= 11q⁴

p=1-q

P(x=4)= ⁴C₄ p⁴ q⁰

=11p⁴

Given:

Chances of hitting a target in a trial=10%=

Chances of not hitting the target = 90%=

Total probability = 50%=5

Let number of trials=n

So number of trials missed=n-1 to hit on nth trial

Total probability

= 5

Take log on both sides

n=6.58

n7

So one must hit it seven times.

Given:

P(getting a head)=

Probability of getting 7 heads = ⁿC₇ (1/2)ⁿ

Probability of getting 9 heads = ⁿC₉ (1/2)ⁿ

Both probabilities are equal if n=9+7=16

So probability of getting 4 heads =¹⁶C₄¹⁶

Given:

P(getting a head)=p(getting a tail)=

=0.20

Given:

Total outcomes= 6¹⁰

Fourth 6 is at 10^th place.

So other three 6 have to be placed at any of the remaining 9 places=⁹C₃5⁶

For remaining 5 places we have 1,2,3,4&5

So total number of favorable cases

Given:

P=?

P(x=r)= ⁿC_r p^r q^n-r

=

P(x=n-r)= ⁿC_n-r p^n-r q^n-(n-r)

= ⁿC_r p^n-r q^r

=

Put values in equation, we have:

1-p=p

1=2p

P=1/2

Given:

P(H)=1/2=p

q=1/2

P(x=r)= ⁿC_r p^r qⁿ

P(x=4)= ⁿC₄ p⁴ q^n-4 = ⁿC₄ pⁿ

P(x=5)= ⁿC₅ p⁵ q^n-5 = ⁿC₅ pⁿ

P(x=6)= ⁿC₆ p⁶ q^n-6 = ⁿC₆ pⁿ

Since they are in AP;

So,

p(x=5)=

2 p(x=5) = p(x=4)+ p(x=5)

2ⁿC₅ pⁿ = ⁿC₄ pⁿ +ⁿC₆ pⁿ

4n – 16 = 55n – 225

51n = 209

n= 4

Given:

n=100

p(H)=P(T)=1/2

P(x=r)= ⁿC_r p^r q^n-r

P(x=50)= ¹⁰⁰C₅₀ p⁵⁰ q⁵⁰

(1)

P(x=51)= ¹⁰⁰C₅₁ p⁵¹ q⁴⁹

(2)

Equate both equations;

51-51p = 50 p

101 p = 51

Given:

n=99

p(H)=P(T)=1/2

P(x=r)= ⁿC_r p^r q^n-r

P(x=r)= ⁿC_r pⁿ

ⁿC_r =

Its maximum value is closest near n/2

Since n is an odd number, so it have two values:

i.e.

S0, r= 49,50

Given:

P(atleast 1 head) = 0.8

n=?

p(H)=p(T) = 1/2

Required probability = p= 1-

P0.8

For n= 2

For n=3

So, n=3

Given:

Mean= 2

Variance=1

P(x>1)=?

P(x>1)=1-p(x1)

= 1-[p(x=0)+p(x=1)]

=1-[ 11qⁿ +npq^n-1]

2(1-p)=1

=q

np=2

n=4

Given:

Probability of getting head to appear first time(r=1), n should be even.

So we have to check probability for n=2,4,6…

P(x=r)= ⁿC_r p^r q^n-r = 2/5

For n=2;

P(X=1)= pq

For n=4

P(X=1) = pq³

Required probability = pq + pq³ + pq⁵ + … =2/5

2/5= pq(1+ q² +q⁴ + …)

4-2p=5-5p

3p= 1

P =1/3

Given:

n=8

p= 1/2

P (|X – 4)| ≤ 2)=?

Since |X – 4)| ≤ 2

i.e. (x-4) ≤ 2

OR (4-X) ≤ 2

x ≤ 6

OR 2 ≤ x

P (|X – 4)| ≤ 2)= P (2≤x≤ 6)

=p(x=2)+ p(x=3)+ p(x=4)+ p(x=5)+ p(x=6)

P(x=r)= ⁿC_r p^r q^n-r

P (2≤x≤ 6) = ⁸C₂ p² q⁶ + ⁸C₃ p³ q⁵ + ⁸C₄ p⁴ q⁴ + ⁸C₅ p⁵ q³ + ⁸C₆ p⁶ q²

=p⁸ [⁸C₂ +⁸C₃ +⁸C₄ + ⁸C₅ + ⁸C₆ ]

Given:

n=100

p = 1/3

P(x=r)= ⁿC_r p^r q^n-r

Its maximum value is

Since is not an integer ; its value is

= 33.67

i.e. 33

Given:

Total outcomes= 6⁸

Fix Third 6 is at 8^th row.

So other two 6 have to be placed at any of the remaining 7 places=⁷C₂

For remaining 5 places we have 1,2,3,4&5

So total number of favourable cases=

Given:

P(selected coupon bears number 9)= p

p=9/15

= 3/5

P(selected coupon bears number 9) = p(all coupons bear number 9) – p( selected coupon bears number 8)

n=7

p(that all coupon bear number 9)

P(x=7)= ⁷C₇ p⁷ q⁰

= p⁷

P(that the selected coupon bears number 9)

P= 8/15

P(x=7)= ⁷C₇ p⁷ q⁰

= p⁷

Required probability

Given:

A 5-digit number is divisible by 5 if it ends with 0 0r 5.

So at ones place we have two digits out of 10 (one at a time)

Remaining places we have 9 digits out of 10

For first place ; we have only 8 digits out of 9 because 0 cannot be at first place

So probability

Given:

P(H)=p(T)=1/2

n=10

p(x=6)=?

P(x=r)= ⁿC_r p^r qⁿ

P(x=6)

Given:

Mean=4

Variance=3

P(x=6)=?

P(x=r)= ⁿC_r p^r q^n-r

4(1-p)=3

=q

np=4

n=16

P(x=r)= ⁿC_r p^r q^n-r

P(x=6)= ¹⁶C₆

Given:

P(x=1)=1/4=p

q=3/4

Standard deviation =3

Mean=?

= 3

mean = np

= 12

Given:

n=4

p(getting atleast one head)=?

• 0 head : HHHH

• 1 head : HTTT

THTT

TTHT

TTTH

• 2 heads :HHTT

HTHT

HTTH

THHT

THTH

TTHH

• 3 heads: HHHT

HHTH

HTHH

THHH

• 4 heads : HHHH

Total ouctomes = 16

Outcomes containing atleast 1 head = 15

P(getting atleast 1 head)=

Given:

n=3

P (X = 1) = 8 P (X = 3)

P =?

P(x=r)= ⁿC_r p^r q^n-r

P(x=1)= ⁿC₁ p¹ q^n-1

=3pq^{2 (1)}

P(x=3)= ³C₃ p³ q⁰

= p³⁽²⁾

Put values in the equations ,we have:

3pq² =8 p³

3 (1-p)² = 8p²

3+3p²-6p = 6p²

5p²+6p-3=0

Given:

P(H)=P(T)=1/2

P(H)= H¹T^n-1+H²T^n-2+H³T^n-3+…..+H^n-1T¹+Hⁿ

Required probability = p= 1-

P0.8

For n= 2

For n=3

So, n=3

Given:

P(females)=p(males)=1/2

p(x=n-1) =

n=?

p(x=n-1) = ⁿC_n-1 p^n-1 q¹

Since p=q

p(x=n-1)= n p^n-1q

=n p^n-1p

=npⁿ

=n

comparing LHS anr RHS

we have,

n=12

Given:

Total pens=100

Defective pens=10

n=5

P(atmost one non-defective pen)=?

P(defective pens

=p

P(atmost one non-defective pen)=p(x<10)

=p(x=0)+p(x=1)

= ⁵C₀ p⁰ q⁵+5pq⁴

Given:

P=?

P(x=r)= ⁿC_r p^r q^n-r

=

P(x=n-r)= ⁿC_n-r p^n-r q^n-(n-r)

= ⁿC_r p^n-r q^r

=

Put values in equation, we have:

1-p=p

1=2p

P=1/2

Given:

n=5

q=0.3

P(x=4)=?

P=1-0.3=0.7

P(x=r)= ⁿC_r p^r q^n-r

P(x=4)= ⁵C₄ p⁴ q

=⁵C₄ (0.7)⁴ (0.3)

.

Given:

n=10

p(true)=p(false)=1/2

p(x8) = p(x=8) + p(x=9) + p(x=10)

= ¹⁰C₈ p⁸ q² + ¹⁰C₉ p⁹ q¹ + ¹⁰C₁₀ p¹⁰ q⁰

= p¹⁰ (45+ 10+1)