Answer to Question #245919 in Statistics and Probability for Mirmir

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

 

(a) Construct a tree diagram that illustrates all possible outcomes and probabilities

 

A chair is randomly selected.

 

(b) What is the probability that the chair that robot Q made is defective                                          

 

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

 

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

Answer 1

Solution:

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)

$P(R|D)=\dfrac{P(R \cap D)}{P(D)}=\dfrac{P(R) \times P(D|R)}{P(D)} \\=\dfrac{0.3\times0.05}{0.0335}=0.44776$

Now, $P(R^c|D)=1-P(R|D)=1-0.44776$

$\therefore P(R^c|D)=0.55224$