WAEC 2022 · Paper 2 · Q12

A basket contains 12 fruits: orange, apple and avocado pear, all of the same size. The numbers of oranges, apples and avocado pears form three consecutive integers. Two fruits are drawn one after the other without replacement. Calculate the probability that:

(a)
the first is an orange and the second is an avocado pear;
(b)
both are of the same fruit;
(c)
at least one is an apple.

Worked solution (try it first)

The numbers are $x$ , $x + 1$ and $x + 2$ , with $3x + 3 = 12$ , so $x = 3$ : 3 oranges, 4 apples and 5 avocado pears.

(a)

$P(\text{orange, then avocado}) = \dfrac{3}{12} \times \dfrac{5}{11}$
$= \dfrac{15}{132}$
$= \dfrac{5}{44}$
$\approx 0.1136$ .

(b)

Same fruit: $\dfrac{3 \times 2 + 4 \times 3 + 5 \times 4}{12 \times 11} = \dfrac{38}{132}$
$= \dfrac{19}{66}$
$\approx 0.2879$ .

(c)

At least one apple is everything except no apple: $1 - \dfrac{8}{12} \times \dfrac{7}{11} = 1 - \dfrac{56}{132}$
$= \dfrac{19}{33}$
$\approx 0.5758$ .

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