WAEC 2010 · Paper 2 · Q14

(a)
A box $P$ contains 3 white and 5 green identical balls. Another box $Q$ contains 6 white and 4 green identical balls. A ball is drawn at random from $P$ and dropped into $Q$ . A ball is then drawn at random from $Q$ . Find the probability that the ball drawn from $Q$ is green.
(b)
A fair die is thrown five times. Find, correct to three decimal places, the probability of obtaining at least two sixes.
(c)
If $\dfrac{{}^xC_3}{{}^xP_2} = \dfrac23$ , find $x$ .

Worked solution (try it first)

After the transfer, $Q$ holds 11 balls.
If a green ball moves ( $\frac58$ ), $Q$ has 5 green: probability $\frac58 \times \frac5{11} = \frac{25}{88}$ .
If a white ball moves ( $\frac38$ ), $Q$ has 4 green: probability $\frac38 \times \frac4{11} = \frac{12}{88}$ .
Add: $\dfrac{25 + 12}{88} = \dfrac{37}{88}$
$\approx 0.42$ .

The number of sixes is binomial with $n = 5$ , $p = \frac16$ , $q = \frac56$ .
$P(0) = \left(\frac56\right)^5$
$= \frac{3125}{7776}$ and $P(1) = 5 \times \frac16 \times \left(\frac56\right)^4$
$= \frac{3125}{7776}$ .
$P(\text{at least } 2) = 1 - \dfrac{6250}{7776}$
$= \dfrac{1526}{7776}$
$\approx 0.196$ .

${}^xC_3 = \dfrac{x!}{3!(x - 3)!}$ and ${}^xP_2 = \dfrac{x!}{(x - 2)!}$ .
Divide: $\dfrac{{}^xC_3}{{}^xP_2} = \dfrac{(x - 2)!}{3!(x - 3)!}$
$= \dfrac{x - 2}{6}$ .
Solve $\dfrac{x - 2}{6} = \dfrac23$ : $x - 2 = 4$ , so $x = 6$ .