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At a particular school with 160 male students, 50 play football, 46 play basketball, and 16 play both. What is the probability that a randomly selected male student plays basketball or football?

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Final answer:

To find the probability of a randomly selected male student playing either basketball or football, we apply the principle of inclusion-exclusion to the given data. The calculation results in a probability of 1/2, or 50%, that a student plays either sport.

Step-by-step explanation:

The question asks about probability, specifically the probability that a randomly chosen male student from a school plays football or basketball. To answer this, we must apply the principle of inclusion-exclusion for two sets: let's define A as the event that a student plays football, and B as the event that a student plays basketball. The school has 160 male students, 50 of which play football, 46 play basketball, and 16 play both. The principle of inclusion-exclusion tells us to add the number of students in each individual group and then subtract the number who are counted twice because they are in both groups:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculating the probabilities gives us:

  • P(A) = number of students playing football / total number of students = 50/160
  • P(B) = number of students playing basketball / total number of students = 46/160
  • P(A ∩ B) = number of students playing both / total number of students = 16/160

Therefore:

P(A ∪ B) = (50/160) + (46/160) - (16/160)
= 80/160
= 1/2

So, the probability that a randomly selected male student plays basketball or football is 1/2 or 50%.

User GrahamS
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