Question #245571

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% of the chairs made by robot Q are defective, 3% of the chairs made by robot R, and 5% of the chairs made by robot S are defective.

(a) construct a tree diagram that illustrates all possible outcomes and possibilities?

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 finding a broken chair?

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



Expert's answer

Part a




Part b

P(Q)= 25% i.e. 0.25

P(R)= 45% i.e. 0.45

P(S)= 1-(0.25+0.45)= 0.30

Let, 'D' be the event that represents defective. So, we have:-

P(D|Q)= 0.02


Part c

P(D|R)= 0.03

P(D|S)= 0.05

Now, we have to find:- P(Rc|D).

As we know that, P(Rc|D)= 1-P(R|D).

Using Bayes theorem, we have:-

P(R∣D)=[P(D∣R).P(R)]/[P(D∣Q).P(Q)+P(D∣R).P(R)+P(D∣S).P(S)].Puttingthevalues,weget:−P(R∣D)=(0.03×0.45)/[(0.2×0.25)+(0.3×0.45)+(0.05×0.3)].So,P(R∣D)=0.0675P(R|D)= [P(D|R).P(R)]/[P(D|Q).P(Q)+P(D|R).P(R)+ P(D|S).P(S)].\\ Putting the values, we get:-\\ P(R|D)= (0.03×0.45)/[(0.2×0.25)+(0.3×0.45)+(0.05×0.3)].\\ So, P(R|D)= 0.0675


Part d

P(Rc∣D)=1−0.0675=0.9375So,required probability=0.9375P(Rc|D)= 1-0.0675= 0.9375\\ So, required \space probability= 0.9375\\


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