1
$\begingroup$

Two chefs A and B are challenged to separately create a new dish in 2 hours. From past experience, we know that:

  1. The probability that chef A’s dish is a hit is 2/3
  2. The probability that chef B’s dish is a hit is 1/2
  3. The probability that at least one of their dishes is a hit is 3/4

Assuming that only one of the dishes can be labeled a hit, what is the probability that it was created by chef A?

My approach:

Given, P(A) = 2/3 P(B) = 1/2 P(A,B) = 3/4

       P(A|B) = P(A) + P(B) - P(A,B) = 5/12

       P(A only wins) = P(A) - P(A|B) = 1/4
       P(B only wins) = P(B) - P(A|B) = 1/12

Therefore

       P(A only | one chef wins) = 1/4 / ( 1/4 + 1/12 ) = 3/4

Is this correct ?

  • 0
    Basically, just asking for answer is not considered to be a good practice. Please consider sharing your work progress (or any ideas which you have have) so that people here can help you better. – 2017-01-18
  • 0
    To me the question doesn't make sense. "only one dish can be labelled a hit" implies $P( A \cap B) = 0$ but from the supplied probabilities you have $P( A \cap B) = 5/12$ – 2017-01-18

1 Answers 1

1

P(A) = 2/3 P(B) = 1/2 P(A,B) = 3/4

No, it is $P(A) = 2/3~,~ P(B) = 1/2~,~ P(A\cup B) = 3/4$

"At least one" means "A or B".

Then $P(A\cap B)=5/12$ because $P(A\cap B)=P(A)+P(B)- P(A\cup B)$

If the condition is that both do not win, we are looking for $$\begin{align}P(A\mid (A\cap B)^\complement)~=&~\dfrac{P(A\cap (A\cap B)^\complement)}{P((A\cap B)^\complement)} \\[2ex] ~=&~\dfrac{P(A)-P(A\cap B)}{1-P(A\cap B)}\end{align}$$

If the condition is that one or the other but not both wins, we are looking for: $$\begin{align}\mathsf P(A\mid A\oplus B) ~=&~ \dfrac{P(A\cap B^\complement)}{P(A\cap B^\complement)+P(A^\complement\cap B)} \\[1ex] =&~ \dfrac{P(A\cup B)-P(B)}{2 P(A\cup B)-P(B)-P(A)}\end{align}$$