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 $$\frac{3}{5}$$