Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.5 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.5

Question 1.

Compute P(X = k) for the binomial distribution, B(n, p) where

(i) n = 6, p = \(\frac { 1 }{ 3 }\), k = 3

(ii) n = 10, p = \(\frac { 1 }{ 5 }\), k = 4

(iii) n = 9, p = \(\frac { 1 }{ 2 }\), k = 7

Solution:

Question 2.

The probability that Mr. Q hits a target ait any trial is \(\frac { 1 }{ 4 }\). Suppose he tries at the target 10 times. Find the probability that he hits the target (i) exactly 4 times (ii) atleast one time

Solution:

Let p be the probability of hitting the target

n = 10, p = \(\frac { 1 }{ 4 }\), X ~ B(n, p)

P(X = x) = ⁿC_{x} p^{x} q^{n-x}, x = 0, 1, 2, …. n

q = 1 – p = \(\frac { 3 }{ 4 }\)

(i) Probability of hitting the target exactly 4 times

(ii) Probability of hitting atleast one time

P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0)

= 1 – ^{10}C_{0} × (\(\frac { 1 }{ 4 }\))° × (\(\frac { 3 }{ 4 }\))^{10}

= 1 – \(\frac { 3^{10} }{ 4^{10} }\)

Question 3.

Using binomial distribution find the mean and variance of X for the following experiments

(i) A fair coin is tossed 100 times and X denotes the number of heads.

(ii) A fair die is tossed 240 times and X denotes the number of times that four appeared.

Solution:

(i) Let p be the probability of getting heads

⇒ p = \(\frac { 1 }{ 2 }\)

n = 100, p = \(\frac { 1 }{ 2 }\), X ~ B(n, p)

q = 1 – p = \(\frac { 1 }{ 2 }\)

Mean = np

= 100 × \(\frac { 1 }{ 2 }\)

= 50

Variance = npq

= 50 × \(\frac { 1 }{ 2 }\)

= 25

(ii) Let p be the probability of getting 4 when a die is thrown

n = 240, p = \(\frac { 1 }{ 6 }\), X ~ B(n, p)

q = 1 – p = \(\frac { 5 }{ 6 }\)

Mean = np

= 240 × \(\frac { 1 }{ 6 }\) = 40

Variance = npq

= 40 × \(\frac { 5 }{ 6 }\) = \(\frac { 100}{ 3 }\)

Question 4.

The probability that a certain kind of component will survive a electrical test is \(\frac { 3 }{ 4 }\). Find the probability that exactly 3 of the 5 components tested survive.

Solution:

Let p be the probability of a certain component survive in an electrical test.

n = 5, p =\(\frac { 3 }{ 4 }\), X ~ B(n, p), k = 3

q = 1 – p = \(\frac { 1 }{ 4 }\)

P(X = x) = ⁿC_{x} p^{x} q^{n-x}, x = 0, 1, 2, …. n

Question 5.

A retailer purchases a certain kind of electronic, device from a manufacturer. The manufacturer indicates that the defective rate of the device 5 is 5%. The inspector of the retailer randomly picks 10 items from a shipment. What is the probability that there will be

(i) atleast one defective item

(ii) exactly two defective items?

Solution:

Let p be the probability that indicates the defective rate of an electronic device

p = 5%

= 0.05

q = 1 – p = 0.95

n = 10, p = 0.05, X ~ B(n, p)

P(X = x) = ⁿC_{x} p^{x} q^{n-x}, x = 0, 1, 2, …. n

(i) Atleast 1 defective item

P(X ≥ 1) = 1 – (X < 1)

= 1 – P(X = 0)

= 1 – ^{10}C_{0} (0.05)° (0.95)^{10}

= 1 – (0.95)^{10}

(ii) Exactly two defective items

P(X = 2) = ^{10}C_{2} (0.05)² (0.95)^{8}

Question 6.

If the probability that fluorescent light has a useful life of atleast 600 hours is 0.9, find the probabilities that among 12 such lights

(i) exactly 10 will have a useful life of atleast 600 hours

(ii) At atleast 11 will have a useful life of atleast 600 hours

(iii) At atleast 2 will not have a useful life of at least 600 hours.

Solution:

Let p be the probability of the useful life hours of a fluorescent light.

n = 12, p = 0.9, X ~ B(n, p)

q – 1 – p = 0.1

P(X = x) = ⁿC_{x} p^{x} q^{n-x}, x = 0, 1, 2, …. n

(i) Exactly 10

P(X = 10) = ^{12}C_{10} (0.9)^{10} (0.1)²

(ii) Atleast 11

P(X ≥ 11) = P(X = 11) + P(X = 12)

= ^{12}C_{11} (0.9)^{11} (0.1)^{1} + ^{12}C_{12} (0.9)^{12} (0.1)°

= 12 × (0.9)^{11} × 0.1 + 1 × (0.9)^{12} × 1

= (0.9)^{11} (12 × 0.1 × 0.9)

= (0.9)^{11} (1.2 + 0.9)

= (2.1) (0.9)^{11}

(iii) Atleast 2 will not have a useful

P(X ≤ 10) = 1 – P(X > 10)

= 1 – [P(X = 11) + P(X = 12)]

= 1 – (2.1)(0.9)^{11}

Question 7.

The mean and standard deviation of a binomial variate X are respectively 6 and 2. Find

(i) The probability mass function

(ii) P(X = 3)

(iii) P(X ≥ 2).

Solution:

X ~ B(n, p)

Given, Mean = 6, Standard deviation = 2

np = 6, \(\sqrt { npq }\) = 2

npq = 4

\(\frac { npq }{ np }\) = \(\frac { 4 }{ 6 }\)

q = \(\frac { 2 }{ 3 }\)

p = 1 – q = \(\frac { 1 }{ 3 }\)

npq = 4

n × \(\frac { 1 }{ 3 }\) × \(\frac { 2 }{ 3 }\) = 4

n = 18

(i) Probability mass function

P(X = x) = ⁿC_{x} p^{x} q^{n-x}, x = 0, 1, 2, …. n

Question 8.

If X ~ B(n, p) such that 4P(X = 4) = P(x = 2) and n = 6. Find the distribution, mean and standard deviation of X.

Solution:

X ~ B(n, p)

Given, 4P(X = 4) = P(X = 2), n = 6

P(X = x) = ⁿC_{x} p^{x} q^{n-x}, x = 0, 1, 2, …. n

⇒ 4 ^{6}C_{4} p^{4} q² = ^{6}C_{2} p² q^{4}

⇒ 4 p² = q²

4 (1 – q)² = q²

4(1 – 2q + q²) = q²

3q² – 8q + 4 = 0

(q – 2) (3q – 2) = 0

q = \(\frac { 2 }{ 3 }\) (q ≠ 2)

Distribution

P(X = x) = ^{6}C_{x} (\(\frac { 1 }{ 3 }\))^{x} (\(\frac { 2 }{ 3 }\))^{6-x}

x = 0, 1, 2, 3, 4, 5, 6

Mean = np = 6 × \(\frac { 1 }{ 3 }\) = 2

Variance = npq = 2 × \(\frac { 2 }{ 3 }\) = \(\frac { 4 }{ 3 }\)

Question 9.

In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. Find the mean and variance of the random variable.

Solution:

n = 5, X ~ B(n, p)

Given, P(X = 1) = 0.4096

P(X = 2) = 0.2048

P(X = x) = ⁿC_{x} p^{x} q^{n-x}, x = 0, 1, 2, …. n

⇒ ^{5}C_{1} p q^{4} = 0.4096

⇒ ^{5}C_{2} p² q³ = 0.2048

5 p q^{4} = 0.4096 …….. (1)

10 p² q³ = 0.2048 ………. (2)

(1) divided by (2),

⇒ \(\frac { 5pq^4 }{ 10p^2q^3 }\) = 2

\(\frac { p }{ q }\) = 4

q = 4p

q = 4 (1 – q)

q = 4 – 4q

5q = 4

q = \(\frac { 4 }{ 5 }\)

p = 1 – q = \(\frac { 1 }{ 5 }\)

Mean = np

= 5 × \(\frac { 1 }{ 5 }\) = 1

Variance = npq

= 1 × \(\frac { 4 }{ 5 }\) = \(\frac { 4 }{ 5 }\)

Distribution

P(X = x) = ^{5}C_{x} (\(\frac { 1 }{ 5 }\))^{x} (\(\frac { 4 }{ 5 }\))^{5-x}

x = 0, 1, 2, 3, 4, 5.