2017 H2 Maths Paper 2 Q5 — Worked Solution

Part (i)

3 marks

Approach

$T$ is the number of counters removed when the 2nd red counter is drawn. Since there are only 3 yellow counters in the bag, at most 3 yellows can be drawn before the 2nd red, so $T$ ranges from $2$ (both reds drawn first) to $5$ (all 3 yellows drawn before the 2nd red). For $T = t$ , the first $t - 1$ counters removed must contain exactly 1 red and $t - 2$ yellow (in any order), and the $t$ -th counter removed must be red.

P(T = t) = \frac{\binom{6}{1}\binom{3}{t - 2}}{\binom{9}{t - 1}} \times \frac{5}{10 - t},\quad t = 2,3,4,5

By GC (or direct calculation),

P(T = 2) = \frac{6}{9} \times \frac{5}{8} = \frac{5}{12}

P(T = 3) = \frac{\binom{6}{1}\binom{3}{1}}{\binom{9}{2}} \times \frac{5}{7} = \frac{18}{36} \times \frac{5}{7} = \frac{5}{14}

P(T = 4) = \frac{\binom{6}{1}\binom{3}{2}}{\binom{9}{3}} \times \frac{5}{6} = \frac{18}{84} \times \frac{5}{6} = \frac{5}{28}

P(T = 5) = \frac{\binom{6}{1}\binom{3}{3}}{\binom{9}{4}} \times \frac{5}{5} = \frac{6}{126} \times 1 = \frac{1}{21}

To check your answer: the four probabilities sum to

\frac{5}{12} + \frac{5}{14} + \frac{5}{28} + \frac{1}{21} = \frac{35 + 30 + 15 + 4}{84} = 1

. ✓

$\therefore$

$t$	$2$	$3$	$4$	$5$
$P(T = t)$	$\frac{5}{12}$	$\frac{5}{14}$	$\frac{5}{28}$	$\frac{1}{21}$

Part (ii)

2 marks

Using the distribution in (i),

E(T) = \sum t\, P(T = t) = \frac{20}{7}

Var(T) = E\left( T^{2} \right) - \left\lbrack E(T) \right\rbrack^{2} = \frac{125}{14} - \left( \frac{20}{7} \right)^{2} = \frac{75}{98}

$\therefore$ $E(T) = \frac{20}{7} = 2.86$ (3 s.f.), $Var(T) = \frac{75}{98} = 0.765$ (3 s.f.).

Part (iii)

2 marks

Approach

“At least 4 counters” means $T \geq 4$ , i.e. $T = 4$ or $T = 5$ . Let $W$ be the number of Lee’s 15 games (out of 15) in which he has to take at least 4 counters out of the bag. Since each game is an independent repeat of the same experiment, $W$ follows a binomial distribution.

P(T \geq 4) = P(T = 4) + P(T = 5) = \frac{5}{28} + \frac{1}{21} = \frac{19}{84}

Let $W$ be the number of Lee’s 15 games in which he takes at least 4 counters out of the bag. Then $W \sim B\left( 15,\ \frac{19}{84} \right)$ .

P(W \geq 5) = 1 - P(W \leq 4)

= 1 - 0.762\ldots

= 0.238\ \left( 3\ \text{s.f.} \right)\ \ \text{(by GC)}

$\therefore$ the probability is $0.238$ (3 s.f.).

2017 H2 Maths Paper 2, Question 5

Part (i)

Part (ii)

Part (iii)

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