The probability of a certain horse finishing in the top 5 of a race is 20%. The probability of him finishing in the top 10 is 45%.
A. What is the probability of him finishing anywhere from 6-10th? B. What is the probability of the horse finishing anywhere from 6-10th given that he finished in the top 10?
Got some comments saying to post what I have tried so far so I will do that. For part B here is what I tried - Let P(T5) and P(T10) be the probability of finishing top 5 and top 10. We know P(T5 and T10) = .2 We also know P(T5 and T10)= P(10) * P(T5|T10) So .2 = .45 * P(T5|T10) Then P(T5|T10) = .2/.45 Therefore the probability of finishing anywhere from 6-10th given that finished in top 10 is 1-P(T5|T10)= 1-.2/.45
If there is not enough information to answer - what other info do we need?
Thanks, Jack
Let's suppose that the question is "probability to finish in the top $[6;10]$
Thus the probability is
A. $45\%-20\%=25\%$
B. $\frac{25}{45}=\frac{5}{9}\approx 55.56\%$
@Jack: the lack of information is that you do not know the probability to arrive exactly 10th. In any case the results will not be very far from the ones I showed you.
Edit: further explanation on B.
Set the events:
A: your horse arrives in the interval $[6;10]$
B: your horse arrives in the top 10, say in the interval $[1;10]$
The question is to calculate
$$\mathbb{P}[A|B]=\frac{\mathbb{P}[A \cap B]}{\mathbb{P}[B]}=\frac{\mathbb{P}[A]}{\mathbb{P}[B]}=\frac{0.25}{0.45}$$
It is self evident that $\mathbb{P}[A \cap B]=\mathbb{P}[A]$ as $A\subset B$