Question

In class of 40 students, 19 play tennis, 20 play netball and 8 play both. A student is randomly chosen from the class. Determine the probability that the student plays one and only one of those sports ?

Answer 1

Answer:

4/5 or 0.8 or 80%

Step-by-step explanation:

Answer 2

The probability of choosing a student who play only one sport is 0.8

We have a class of 40 students, 19 play tennis, 20 play netball and 8 play both. A student is randomly chosen from the class.

We have to find the probability that the student plays one and only one of those sports.

What is the formula of calculating the probability of any Event 'A' ?

The formula to calculate the probability of the event 'A' is -

P(A) = $\frac{Total\;no.\;of\;favorable\;outcomes}{Total\;no.\;of\;outcomes} = \frac{n(A)}{n(S)}$

Let 'A' be the event representing the probability of choosing a student who play both sports.

According to question -

n ( S ) = 40 students

n ( A ) = 8

P ( A ) = $\frac{n (A)}{n (S)}$

We know that -

P ( A ) + P ( A' ) = 1

P ( A' ) = 1 - P ( A )

where P ( A' ) represents the probability of choosing a student who does not play both sports.

P ( A' ) = 1 - $\frac{n (A)}{n (S)}$ = 1 - $\frac{8}{40}$ = 1 - 0.2 = 0.8

Hence, the probability of choosing a student who play only one sport is 0.8

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