Answer to Question #245646 in Statistics and Probability for Shweta

Question

In a workshop, three robots, Q, R and S, are employed to make chairs

Robot Q makes 25% of the chairs

Robot R makes 45% of the chairs

The remaining chairs are made by Robot S

Evidence has shown that 2 percent of the chairs made by robot Q are defective, 3 percent of the chairs made by robot R, and 5 percent of the chairs made by robot S are defective

1. Construct a tree diagram that illustrates all possible outcomes and probabilities

A chair is randomly selected.

1. What is the probability that the chair that robot Q made is defective (1 marks)

(c) What is the probability of findings a broken chair (2 marks)

(d) Given that a chair is defective, what is the probability that it was not made by robot R (2 marks)

Answer 1

Let Q, R & S represent the event that a chair is made by robot Q, R & S respectively.

Let D & ND represent the event that the selected chair is defective & non-defective respectively.

P(Q) = 0.25

P(R) = 0.45

P(S) = 1 - (0.25 + 0.45) = 0.3

P(D|Q) = 0.02, P(ND|Q) = 1 - 0.02 = 0.98

P(D|R) = 0.03, P(ND|R) = 1 - 0.03 = 0.97

P(D|S) = 0.05, P(ND|S) = 1 - 0.05 = 0.95

a) Tree diagram :

b) From the tree diagram, it can be seen that P(D|Q) = 0.02.

The probability that a chair made by robot Q is defective is equal to 0.02 .

c) By total probability theorem,

P(D) = P(D|Q)P(Q) + P(D|R)P(R) + P(D|S)P(S)

P(D) = 0.02 × 0.25 + 0.03 × 0.45 + 0.05 × 0.3

P(D) = 0.005 + 0.0135 + 0.015

P(D) = 0.0335

The probability of finding a broken chair is 0.0355 .

d)

$P(R|D) = \frac{P(R \cap D)}{P(D)} \\ = \frac{P(R) \times P(D|R)}{P(D)} \\ = \frac{0.3 \times 0.05}{0.0335} \\ = \frac{150}{335} \\ = 0.44776 \\ P(R^C|D) = 1 -P(R|D) \\ = 1 -0.44776 \\ = 0.55224$

Given that a chair is defective, the probability that it was not made by robot R is equal to 0.55224 .