WAEC 2019 · Paper 2 · Q13

(a)
A fair coin is tossed 6 times. Calculate the probability of obtaining at most three heads.
(b)
A pool of jurors consists of 1 British, 3 Americans and 2 Africans. If 2 jurors are selected one after the other to sit on a jury for a trial, find the probability that they are: (i) 1 American and 1 African; (ii) of different nationalities.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

$X \sim B\left(6, \frac12\right)$ , so each outcome has probability $\left(\frac12\right)^6 = \frac{1}{64}$ .
$P(X \le 3) = \dfrac{{}^6C_0 + {}^6C_1 + {}^6C_2 + {}^6C_3}{64}$
$= \dfrac{1 + 6 + 15 + 20}{64}$
$= \dfrac{42}{64}$
$= \dfrac{21}{32}$ .

(b)

There are 6 jurors and $\,{}^6C_2 = 15$ pairs.

(i)

1 American and 1 African: $\,{}^3C_1 \times {}^2C_1 = 6$ , so $P = \dfrac{6}{15} = \dfrac25$ .

(ii)

Pairs of the same nationality: 2 Americans, $\,{}^3C_2 = 3$ , or 2 Africans, $\,{}^2C_2 = 1$ .
So $P(\text{different}) = 1 - \dfrac{4}{15}$
$= \dfrac{11}{15}$ .

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