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My question here is more about the equations than the answer.

The question in the text is if a student is selected at random, what is the probability that the person is left handed OR participates in a team sport?

So, I know the equation is: $P(A \cup B)= P(A) + P(B) - P(A \cap B)$ where $P(A)$ is prob of being left handed and $P(B)$ = prob of playing team sport.

So, I can easily see that $P(A \cup B) = 24/63 + 39/63 -P(A \cap B)$

Given the constructed table I can easily see that the intersection of "Team" and "Left" is $13/63$. Therefore the answer is: $24/63 + 39/63 - 13/63 = 79.4$%. Yes, this matches the book.

However, I then tried to use my formula of $P(A \cap B)= P(A)P(B|A)$ instead of visually looking for the intersection. In words I would say this is equal to the probability of being left handed, which is $24/63$ times the probability of being a team player GIVEN that you are left handed which would seem to me to be $13/63$. {this must be the error but I don't see it}

Now I get: $24/63+39/63-[(24/63)(13/63)]$. Clearly NOT the correct answer.

Why doesn't my formula of $P(A \cap B)=P(A)P(B|A)$ produce the correct answer?

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    ${\rm P}(B|A)=\frac{13}{24}$, not $\frac{13}{63}$. – 2017-01-14
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    Can you show me how I would see 13/24 as $P(B|A)$? – 2017-01-14
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    Can you show me the math of 13/24? – 2017-01-14
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    In this context, $P(B\mid A)$ (*read aloud as the probability of $B$ given $A$*) is the probability that a person plays a teamsport given that they are left-handed. Look at all of the left handed people and ignore everyone else. We essentially restrict our sample space to just those people who are left handed. $13$ of the $24$ left-handed people play a teamsport, each of which are equally likely to be selected, so the probability is $\frac{13}{24}$. – 2017-01-14
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    Oh, I see this now. Your explanation helped me to see this. Thank you. – 2017-01-14

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The definition of a conditional probability is $P(B|A) = \frac{P(A\cap B)}{P(A)}$, so your formula $P(A\cap B) = P(A)P(B|A)$ is alright, you only need to compute $P(B|A)$ correctly. In this particular case, however, it is not very sensible to compute $P(B|A)$ at all because you can take $P(A\cap B)$ directly from the table.