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 A1 for a correct second term and award A1 for a correct third term.
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 (A1) for \(a = \frac{1}{3}\) and \(r = \frac{2}{9}\).
\( = \frac{3}{7}\;\;\;( = 0.429)\) 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]