22M.2.AHL.TZ2.8

pestleMathematicsAAHLPaper 222M· sl-4-6-combined-mutually-exclusive-conditional-independence-prob-diagramssource ↗

Rachel and Sophia are competing in a javelin-throwing competition.

The distances, $R$ metres, thrown by Rachel can be modelled by a normal distribution with mean $56.5$ and standard deviation $3$ .

The distances, $S$ metres, thrown by Sophia can be modelled by a normal distribution with mean $57.5$ and standard deviation $1.8$ .

In the first round of competition, each competitor must have five throws. To qualify for the next round of competition, a competitor must record at least one throw of $60$ metres or greater in the first round.

Find the probability that only one of Rachel or Sophia qualifies for the next round of competition.

Markscheme / solution

Rachel: $R ~ N (56.5, 3^{2})$

$P (R \geq 60) = 0.1216 \dots$ (A1)

Sophia: $S ~ N (57.5, 1 . 8^{2})$

$P (S \geq 60) = 0.0824 \dots$ (A1)

recognises binomial distribution with $n = 5$ (M1)

let $N_{R}$ represent the number of Rachel’s throws that are longer than $60$ metres

$N_{R} ~ B (5, 0.1216 \dots)$

either $P (N_{R} \geq 1) = 0.4772 \dots$ or $P (N_{R} = 0) = 0.5227 \dots$ (A1)

let $N_{S}$ represent the number of Sophia’s throws that are longer than $60$ metres

$N_{S} ~ B (5, 0.0824 \dots)$

either $P (N_{S} \geq 1) = 0.3495 \dots$ or $P (N_{S} = 0) = 0.6504 \dots$ (A1)

EITHER

uses $P (N_{R} \geq 1) P (N_{S} = 0) + P (N_{S} \geq 1) P (N_{R} = 0)$ (M1)

$P (one of Rachel or Sophia qualify) = (0.4772 \dots \times 0.6504) + (0.3495 \dots \times 0.5227 \dots)$

uses $P (N_{R} \geq 1) + P (N_{S} \geq 1) - 2 \times P (N_{R} \geq 1) \times P (N_{S} \geq 1)$ (M1)

$P (one of Rachel or Sophia qualify) = 0.4772 \dots + 0.3495 \dots - 2 \times 0.4772 \dots \times 0.3495 \dots$

THEN

$= 0.4931 \dots$

$= 0.493$ A1

Note: M marks are not dependent on the previous A marks.

[7 marks]

Examiners’ report

In general, well answered with many candidates getting full marks. On the other end of the spectrum, some candidates could only get the first two marks for normal distribution without recognising the binomial distribution and often just stopping there. Some candidates lost the last two marks for not being able to combine the probabilities correctly.