Question

The baseball team has a double-header on Saturday. The probability that they will win both games is 18%. The probability that they will win just the first game is 60%, What is the probability that the team will win the 2nd game given that they have already won the first game?

Answer 1

I got 30%
60% as a decimal is 0.6
18% as a decimal is 0.18
0.6 times 30% or 0.3 is 0.18

Answer 2

Let the event to win the first game is F and to win the second game is S.

This question is from conditional probability and hence we use the below formula.

$P(S/F)=\frac{P(F and S) }{P(F)}$

Now, we have been given that

$P(F and S)=0.18\\ P(F)=0.60$

On substituting these given values in the formula, we get

$P(S/F)=\frac{0.18}{0.60} \\ \\ P(S/F)=0.3=30\%$

Therefore, the probability for winning the second game is 30%