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What is the probability that a randomly chosen student likes either soccer or basketball?

In a class of 30 students:
- 12 like soccer
- 15 like basketball
- 5 like both sports

A) [tex]\frac{12}{30} + \frac{15}{30}[/tex]

B) [tex]\frac{12}{30} + \frac{15}{30} - \frac{5}{30}[/tex]

C) [tex]\frac{5}{30}[/tex]

D) [tex]\frac{27}{30} + \frac{5}{30}[/tex]

Answer :

Let's go through the solution step-by-step:

In the class, there are 30 students. We have the following information:

- 12 students like soccer.
- 15 students like basketball.
- 5 students like both soccer and basketball.

We are asked to find the probability that a randomly chosen student likes either soccer or basketball.

To solve this, we can use the principle of Inclusion-Exclusion. The formula for the probability that a student likes either soccer or basketball is:

[tex]\[
P(\text{Soccer or Basketball}) = P(\text{Soccer}) + P(\text{Basketball}) - P(\text{Both})
\][/tex]

1. First, calculate the number of students who like either soccer or basketball:
- Add the number of students who like soccer and basketball, then subtract the number of students who like both to avoid double-counting them.

[tex]\[
\text{Students who like either} = 12 + 15 - 5 = 22
\][/tex]

2. Next, calculate the probability by dividing the number of students who like either sport by the total number of students:

[tex]\[
\text{Probability} = \frac{22}{30} = 0.7333\ldots
\][/tex]

The correct probability is approximately 0.73, which corresponds to option B in the question.

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