Is this conditional probability or independent events?

Question

Jack and Ben have been tracking their success rate of converting goals in their rugby games. Jack converts $70\%$ of his goals and Ben converts $60\%$. At a recent home game, both get a kick, but only one converts his goal. What is the probability that it was Ben?

I tried using the conditional probability formula and it didn't work. Someone suggested using the addition formula $P(J \text{ xor } B) = P(J) + P(B) - 2P(J \text{ and } B)$ and I thought the answer it gave seemed correct ($0.46$) but the answer at the back of the textbook says otherwise.

Answer 1

Based on bayes' theorem:

$$P(Ben \ converted \ a \ kick \ | \ both \ get \ a \ kick \ and \ one \ converted \ it ) = \frac{60\% \cdot 30\%}{60\% \cdot 30\% + 70\% \cdot 40\%}$$

Answer 2

Let's start right after they both get a kick, since that even doesn't have any numbers associated with it. At this point, there's a 70% chance that Jack converts his kick and a (presumably independent) 60% chance that Ben converts his. So the four possible outcomes, each with its probability, are as follows:

• Both convert their kicks: $0.7 \times 0.6 = 0.42$,
• Jack converts his but not Ben: $0.7 \times 0.4 = 0.28$,
• Ben converts his but not Jack: $0.3 \times 0.6 = 0.18$,
• Neither converts their kick: $0.3 \times 0.4 = 0.12$.

(Notice that these four probabilities add up to 100%, as they should.)

So far, it's been independent events, but now we bring in conditional probability. We are told that one and only converts their kick, so the first and last possibility are ruled out. We're conditioning on an event with $0.28 + 0.18 = 0.46$ probability, and we're looking at the event with $0.18$ probability, so we divide to get the conditional probability: $0.18 \div 0.46 \approx 0.39$ (or you could give an exact rational answer).

This is the same result as @Tortar gave, but perhaps the added detail will help you understand it.

Answer 3

Yes, that is the formula for exclusive or, and since $B$ and $J$ are independent:$$\def\P{\mathop{\sf P}}\P(J\text{ xor } B)~{=\P(J)+\P(B)-2\P(J\text{ and }B)\\=\P(B)+\P(J)-2\P(B)\P(J)\\=\P(B)(1-\P(J))+\P(J)(1-\P(B))}$$

However, you use this formula to get the answer; it is not the answer by itself.

The conditional probability you seek is: $$\P(B\mid J\text{ xor } B)~{=\dfrac{\P(B\text{ and not }J)}{\P(J\text{ xor } B)}\\=\dfrac{\P(B)(1-\P(J))}{\P(B)(1-\P(J))+\P(J)(1-\P(B))}\\=\dfrac{\P(B)-\P(B)\P(J))}{\P(B)+\P(J)-2\P(B)\P(J)}}$$