IB DP Math AA: Topic: SL 4.5: The probability of an event A HL Paper 2


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]



\(\left( {\begin{array}{*{20}{c}}
  {10} \\
\end{array}} \right) = 210\)     (M1)A1

[2 marks]


\(2 \times \left( {\begin{array}{*{20}{c}}
  8 \\
\end{array}} \right) = 112\)     (M1)A1A1

 Note: Accept \(210 – 28 – 70 = 112\)


[3 marks]


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

[2 marks]


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]



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



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]

a.the total of the three numbers is 5;[3]
b.i.the median of the three numbers is 1.[3]

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]



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

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