Question
In a school, 45% of the students play football, 30% play volleyball and 15% both. If a student is selected at random, then the probability that he plays football or volleyball is:
Answer: Option C
Given that, 45% play football; that is, P(F) = $$\frac{{45}}{{100}}$$ = $$\frac{{9}}{{20}}$$
30% play volleyball; that is, P(V) = $$\frac{{30}}{{100}}$$ = $$\frac{{6}}{{20}}$$
And, 15% play both volleyball and football; that is, P(F And V) = $$\frac{{15}}{{100}}$$ = $$\frac{{3}}{{20}}$$
Now, we have to find the probability that 1 student plays football or volleyball;
That we have to find, P(F or V)
We know that, P(F Or V) = P(F) + P(V) - P(F And V)
= $$\frac{{9}}{{20}}$$ + $$\frac{{6}}{{20}}$$ - $$\frac{{3}}{{20}}$$
= $$\frac{{12}}{{20}}$$
= $$\frac{{3}}{{5}}$$
Hence, the required probability $$\frac{{3}}{{5}}$$
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Given that, 45% play football; that is, P(F) = $$\frac{{45}}{{100}}$$ = $$\frac{{9}}{{20}}$$
30% play volleyball; that is, P(V) = $$\frac{{30}}{{100}}$$ = $$\frac{{6}}{{20}}$$
And, 15% play both volleyball and football; that is, P(F And V) = $$\frac{{15}}{{100}}$$ = $$\frac{{3}}{{20}}$$
Now, we have to find the probability that 1 student plays football or volleyball;
That we have to find, P(F or V)
We know that, P(F Or V) = P(F) + P(V) - P(F And V)
= $$\frac{{9}}{{20}}$$ + $$\frac{{6}}{{20}}$$ - $$\frac{{3}}{{20}}$$
= $$\frac{{12}}{{20}}$$
= $$\frac{{3}}{{5}}$$
Hence, the required probability $$\frac{{3}}{{5}}$$
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