In a certain Algebra 2 class of 27 students, 11 of them play basketball and 10 of them play baseball. There are 8 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball?
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1HINT. First figure out how many play basketball but not baseball, and vice versa. A Venn diagram might help as well – Bram28 Nov 15 '19 at 12:28
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Agree with Bram that a Venn diagram is the way to go: start by filling in the middle. – Ben Collister Nov 15 '19 at 12:59
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Hint
Inclusion-exclusion principle (for two sets) can be stated as
$$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$$
That is to say, the probability of $A$ or $B$ being true is going to be the sum of the probability of $A$ being true and the probability of $B$ being true minus the probability of them both being simultaneously true.
Here, letting $A$ be "plays basketball" and $B$ be "plays baseball" and using the numbers provided in the question, we can reach our conclusion.
$$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B) = \frac{11}{27}+\frac{10}{27}-\frac{8}{27}$$
As a way to build intuition and to graphically organize your thoughts, as alluded to in the comments above you can use Venn Diagrams and try to figure out how many students are in (and as a result the probabilities corresponding to) each particular region of the diagram. Here, you are told the combined total of the regions corresponding to being in the "basketball" circle as well as the amount in the region inside of both the basketball and the baseball circles. From that, you should be able to find how many play basketball only.
That line of thinking leads to the similar identity:
$$Pr(A\cup B) = Pr(A\setminus B)+Pr(A\cap B)+Pr(B\setminus A)$$
That is to say, the probability of being in $A$ or $B$ is the probability of being in $A$ but not $B$ plus the probability of being in both $A$ and $B$ plus the probability of being in $B$ but not $A$, a simple consequence of the Law of Total Probability.
JMoravitz
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