Question

In a certain Algebra 2 class of 30 students, 8 of them play basketball and 10 of them play baseball. There are 18 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?

Answer 1

The probability that a student chosen randomly from the class plays both basketball and baseball = 1/5

Given :

• In a certain Algebra 2 class of 30 students, 8 of them play basketball and 10 of them play baseball.

• There are 18 students who play neither sport

To find :

The probability that a student chosen randomly from the class plays both basketball and baseball

Solution :

Step 1 of 3 :

Find total number of possible outcomes

In the certain Algebra 2 class there are 30 students

So the total number of possible outcomes = 30

Step 2 of 3 :

Find total number of possible outcomes for the event

Let E be the event that a student chosen randomly from the class plays both basketball and baseball

Let A = The set of students who play basketball and B = The set of students who play baseball

A ∩ B = The set of students who play both sports

A ∪ B = The set of students who plays basketball or baseball

In a certain Algebra 2 class of 30 students, 8 of them play basketball and 10 of them play baseball.

There are 18 students who play neither sport

So by the given condition

n(A) = 8 , n(B) = 10 , n(A' ∩ B') = 18

Now ,

n(A' ∩ B') = 18 gives

n(S) - n(A ∪ B) = 18

⇒ 30 - n(A ∪ B) = 18

⇒ n(A ∪ B) = 12

From set theory

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

⇒ n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

⇒ n(A ∩ B) = 8 + 10 - 12

⇒ n(A ∩ B) = 6

So total number of possible outcomes for the event E is 6

Step 3 of 3 :

Find the probability

Hence the required probability

= P(E)

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: A }{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf = \frac{6}{30}$

$\displaystyle \sf = \frac{1}{5}$

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