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An athlete has 75% of winning the race if he is not injured. If he is injured, his probability of winning the race is only 15%. If he total chances of winning is 51%, what is the probability that he gets injured?

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Could you make your title more specific? – Mike Spivey Nov 18 '11 at 19:45
@MikeSpivey title edited hope it helps – andrei Nov 18 '11 at 19:48
What have you tried so far? Precisely where are you having difficulties? – Austin Mohr Nov 18 '11 at 19:48
@AustinMohr I have no idea how to take into consideration two different events. – andrei Nov 18 '11 at 19:53
@andrei: I still think it could be more specific. – Mike Spivey Nov 18 '11 at 20:08

X = 'is the athlete injured' (X in {0,1})

Y = 'the athlete wins' (Y in {0,1})

  • P(Y=1|X=0) = 0.75
  • P(Y=1|X=1) = 0.15
  • P(Y=1) = 0.51

We are looking for P(X=1) - P(Y=1) = P(Y=1|X=0)*(1-P(X=1)) + P(Y=1|X=1)*P(X=1)

The above equation should provide you with the answer 0.4!

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Thanks for the hint I really didn't know where to start.It's not really a homework just a personal curiosity – andrei Nov 18 '11 at 19:55
There is no Bayes' rule here, only the law of total probability. – Dilip Sarwate Nov 18 '11 at 21:08

The book you are working with undoubtedly provides machinery for tackling such problems. We provide a start, using two different approaches. For this problem, I prefer the second approach, though for more complicated problems the formal first approach may be needed. By carrying out the details of either approach, after a while one concludes that the probability the athlete is injured is $0.4$.

First approach (formal): Let $W$ be the event "the athlete wins."

Let $I$ be the event "the athlete is injured."

We want $P(I)$.

We are told several things:

1) $P(W)=0.51$

2) $P(W|I^c)=0.75$. Here $I^c$ is the complement of $I$, it is a fancy way of writing "not injured." If your course uses a different notation, change what is written here to that notation. Note that in general $P(A|B)$ means the probability of $A$, given the information that $B$ has happened.

3) $P(W|I)=0.15$.

Now the athlete can win in two different ways: a) not injured and wins or b) injured and wins. In symbols, $W=(I^c\cap W)\cup(I\cap W)$.

Let $p=P(I)$. Can you use the "not injured and wins or injured and wins" analysis to find an equation for $p$?

Second approach (tree diagram): Draw a simple tree. Recall that mathematicians draw trees upside down. Label the ends of the two "branches" that come down from the "root" (top!) with the labels "injured" and "not injured." You don't know the relevant probabilities. So write on one branch $p$ for the unknown probability of injured, and therefore on the other $1-p$ for the probability of not injured.

Now each of these two branches splits into two branches. You can label the ends of these branches with "win," "not win," "win," "not win." The probabilities along these new branches have all been supplied to you.

Finally, compute the probability of "win" using your tree diagram. The answer will be in terms of $p$. But the probability of "win" is known. Continue.

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