## Question

A team of 6 players is to be selected from 10 volleyball players, of whom 8 are boys and 2 are girls.

a.In how many ways can the team be selected?[2]

b.In how many of these selections is exactly one girl in the team?[3]

c.If the selection of the team is made at random, find the probability that exactly one girl is in the team.[2]

**▶️Answer/Explanation**

## Markscheme

\(\left( {\begin{array}{*{20}{c}}

{10} \\

6

\end{array}} \right) = 210\) **(M1)A1****[2 marks]**

\(2 \times \left( {\begin{array}{*{20}{c}}

8 \\

5

\end{array}} \right) = 112\) *(M1)A1A1*

**Note: **Accept \(210 – 28 – 70 = 112\)

*[3 marks]*

\(\frac{{112}}{{210}}\,\,\left( { = \frac{8}{{15}} = 0.533} \right)\) *(M1)A1*

*[2 marks]*

## Question

Ava and Barry play a game with a bag containing one green marble and two red marbles. Each player in turn randomly selects a marble from the bag, notes its colour and replaces it. Ava wins the game if she selects a green marble. Barry wins the game if he selects a red marble. Ava starts the game.

a.Find the probability that Ava wins on her first turn.[1]

b.Find the probability that Barry wins on his first turn.[2]

c.Find the probability that Ava wins in one of her first three turns.[4]

d.Find the probability that Ava eventually wins.[4]

**▶️Answer/Explanation**

## Markscheme

\({\text{P(Ava wins on her first turn)}} = \frac{1}{3}\) *A1*

*[1 mark]*

\({\text{P(Barry wins on his first turn)}} = {\left( {\frac{2}{3}} \right)^2}\) *(M1)*

\( = \frac{4}{9}\;\;\;( = 0.444)\) *A1*

*[2 marks]*

\(P\)(Ava wins in one of her first three turns)

\( = \frac{1}{3} + \left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right)\frac{1}{3} + \left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right)\frac{1}{3}\) *M1A1A1*

**Note: **Award ** M1 **for adding probabilities, award

**for a correct second term and award**

*A1***for a correct third term.**

*A1*Accept a correctly labelled tree diagram, awarding marks as above.

\( = \frac{{103}}{{243}}\;\;\;( = 0.424)\) *A1*

*[4 marks]*

\({\text{P(Ava eventually wins)}} = \frac{1}{3} + \left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right)\frac{1}{3} + \left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right)\left( {\frac{2}{3}} \right)\left( {\frac{1}{3}} \right)\frac{1}{3} + \ldots \) *(A1)*

using \({S_\infty } = \frac{a}{{1 – r}}\) with \(a = \frac{1}{3}\) and \(r = \frac{2}{9}\) *(M1)(A1)*

**Note: **Award ** (M1) **for using \({S_\infty } = \frac{a}{{1 – r}}\) and award

**for \(a = \frac{1}{3}\) and \(r = \frac{2}{9}\).**

*(A1)*\( = \frac{3}{7}\;\;\;( = 0.429)\) *A1*

*[4 marks]*

*Total [11 marks]*

## Question

Six balls numbered 1, 2, 2, 3, 3, 3 are placed in a bag. Balls are taken one at a time from the bag at random and the number noted. Throughout the question a ball is always replaced before the next ball is taken.

Three balls are taken from the bag. Find the probability that

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

Find \({\text{E}}(X)\).[2]

b.ii.Ten balls are taken from the bag. Find the probability that less than four of the balls are numbered 2.[3]

c.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.[3]

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

e.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.[8]

**▶️Answer/Explanation**

## Markscheme

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