A single ball is taken from the bag. Let \(X\) denote the value shown on the ball.

Find \({\text{E}}(X)\).

the total of the three numbers is 5;

the median of the three numbers is 1.

Ten balls are taken from the bag. Find the probability that less than four of the balls are numbered 2.

Find the least number of balls that must be taken from the bag for the probability of taking out at least one ball numbered 2 to be greater than 0.95.

Another bag also contains balls numbered 1 , 2 or 3.

Eight balls are to be taken from this bag at random. It is calculated that the expected number of balls numbered 1 is 4.8 , and the variance of the number of balls numbered 2 is 1.5.

Find the least possible number of balls numbered 3 in this bag.

\({\text{E}}(X) = 1 \times \frac{1}{6} + 2 \times \frac{2}{6} + 3 \times \frac{3}{6} = \frac{{14}}{6}{\text{ }}\left( { = \frac{7}{3} = 2.33} \right)\)     (M1)A1

[2 marks]

\(3 \times {\text{P}}(113) + 3 \times {\text{P}}(122)\)     (M1)

\(3 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} + 3 \times \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} = \frac{7}{{72}}{\text{ }}( = 0.0972)\)     A1

Note:     Award M1 for attempt to find at least four of the cases.

[3 marks]

recognising 111 as a possibility \(\left( {{\text{implied by }}\frac{1}{{216}}} \right)\)     (M1)

recognising 112 and 113 as possibilities \(\left( {{\text{implied by }}\frac{2}{{216}}{\text{ and }}\frac{3}{{216}}} \right)\)     (M1)

seeing the three arrangements of 112 and 113     (M1)

\({\text{P}}(111) + 3 \times {\text{P}}(112) + 3 \times {\text{P}}(113)\)

\( = \frac{1}{{216}} + \frac{6}{{216}} + \frac{9}{{216}} = \frac{{16}}{{216}}{\text{ }}\left( { = \frac{2}{{27}} = 0.0741} \right)\)     A1

[3 marks]

let the number of twos be \(X,{\text{ }}X \sim B\left( {10,{\text{ }}\frac{1}{3}} \right)\)     (M1)

\({\text{P}}(X < 4) = {\text{P}}(X \leqslant 3) = 0.559\)     (M1)A1

[3 marks]

let \(n\) be the number of balls drawn

\({\text{P}}(X \geqslant 1) = 1 - {\text{P}}(X = 0)\)     M1

\( = 1 - {\left( {\frac{2}{3}} \right)^n} > 0.95\)     M1

\({\left( {\frac{2}{3}} \right)^n} > 0.05\)

\(n = 8\)     A1

[3 marks]

\(8{p_1} = 4.8 \Rightarrow {p_1} = \frac{3}{5}\)     (M1)A1

\(8{p_2}(1 - {p_2}) = 1.5\)     (M1)

\(p_2^2 - {p_2} - 0.1875 = 0\)     (M1)

\({p_2} = \frac{1}{4}{\text{ }}\left( {{\text{or }}\frac{3}{4}} \right)\)     A1

reject \(\frac{3}{4}\) as it gives a total greater than one

\({\text{P}}(1{\text{ or }}2) = \frac{{17}}{{20}}{\text{ or P}}(3) = \frac{3}{{20}}\)     (A1)

recognising LCM as 20 so min total number is 20     (M1)

the least possible number of 3’s is 3     A1

[8 marks]

Part (a) was generally well done, although many candidates lost their way after that.

Candidates had difficulty recognising all the different cases in part (b).

Candidates had difficulty recognising all the different cases in part (b).

Parts (c) and (d) should have been more standard questions, but many were unable to tackle them.

Parts (c) and (d) should have been more standard questions, but many were unable to tackle them.

Part (e) was poorly answered in general.