WAEC 2020 · Paper 2 · Q5

1. There are 25 equally likely numbers.

(a)

1. The even numbers are $2, 4, \ldots, 24$: 12 of them.
2. $P(\text{even}) = \frac{12}{25}$.

(b)

1. The primes up to 25 are $2, 3, 5, 7, 11, 13, 17, 19, 23$: 9 of them.
2. $P(\text{prime}) = \frac{9}{25}$.

(c)

1. Only 2 is both even and prime, so it's in both lists.
2. Add the two lists and take it away once: $\frac{12}{25} + \frac{9}{25} - \frac{1}{25} = \frac{20}{25}$
   $= \frac45$.

(d)

1. Both even and prime: only 2.
2. $P = \frac{1}{25}$.