Q. 16. The probability that a certain kind of component will survive a given shock test is 3/4. Find the probability that exactly two of the next four components tested will survive. (Jan. 2001, Dec. 2000, June 2000)

Solution. Here n = 4, x = 2, p = 3/4
q = 1 - p
or q = [1- (3/4)]
or q = 1/4
Using Binomial Distribution
P[X = x] = (ⁿC_x) X (p^x) X (q^{n - x})
P[X = 2] = (⁴C₂) X (3/4)² X (1/4)²
= 0.21
Hence, the required probability is 0.21.

Q. 17. Let X be the number of 1's obtained in 15 throws of an unbiased dice. Find its mean and variance. (June 2001)

Solution. X is a binomial variance with p = 1/6, q = [1 - (1/6)] = 5/6, and n = 15
Mean = n X p
or Mean = 15 X (1/6) = 2.5

Variance = n X p X q
or Variance = 15 X (1/6) X (5/6) = 2.09
Therefore, mean = 2.5 and variance = 2.09.

Q. 18. The average number of radioactive particles through a counter during 1 milli second in a laboratory experiment is 3. What is the probability that five particles enter the counter in a given millisecond? (Dec. 2001)

Solution. Given, x = 5 and l = 3
Using Poisson Distribution
Probability = [e^-l(l)^x]/x!
= [e^-3(3)⁵]/5!
= 0.1008

Q. 19. In a bulb making factory, three machines A, B and C manufacture respectively 15, 35 and 50 percent of the total. Out of their total outputs 4, 5 and 3 percent are defective. A bulb is drawn from the produce at random and is found to be defective. What is the probability that it is manufactured by (i) factory A (ii) factory C? (Dec. 2002)

Solution. This problem is based on Bayes theorem.
Let A = an event that the output is defective
B₁ = an event that bulb is manufactured by machine A
B₂ = an event that bulb is manufactured by machine B
B₃ = an event that bulb is manufactured by machine C

Therefore, P(B₁) = 15/100, P(B₂) = 35/100, P(B₃) = 50/100
P(A | B₁) = 4/100, P(A | B₂) = 5/100, P(A | B₃) = 3/100

or P(B₁ | A) = 12/77

or P(B₃ | A) = 30/77