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Of $33$ people, $17$ like red, $14$ like green, and $11$ do not like either. What is the probability that a student likes red and green? What's the probability that exactly one of the following is true: the student likes red (call this event $A$) or the student likes green (call this event $B$).?

So far I have that $P(A) = \frac{17}{33}$ and $P(B) = \frac{14}{33}$. I know we are looking for $P(A \cap B)$ for part one. I am wondering if for part two the formula would be $P(A) + P(B) -2P(A \cap B)$.

soccerfan
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2 Answers2

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Since $11$ did not pass either, we know that $P(A\cup B)=\frac{33-11}{33}=\frac{22}{33}$. We also know that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. That gives us enough information to solve for $P(A\cap B)$

For your second question, we're really asking for $P(A\setminus B)+P(B\setminus A) = P(A)-P(A\cap B) + P(B) - P(A\cap B)$, so your formula should work

G Tony Jacobs
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Computing that for part two is correct.

Guide:

$$P(A^c \cap B^c) =1-P(A \cup B)=1-P(A)-P(B)+P(A \cap B)$$

we can solve for $P(A \cap B)$.

Siong Thye Goh
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