Example 6: HSC Mathematics Extension 1 paper, 2002. Q4

Question 4

(a)

Lyndal hits the target on average 2 out of every 3 shots in archery competitions.During a competition she has 10 shots at the target.

(i)

What is the probability that Lyndal hits the target exactly 9 times? Leave your answer in unsimplified form. 1 mark

(ii)

what is the probability that Lyndal hits the target fewer than 9 times? Leave your answer in unsimplified form. 2 marks

ANSWER

(a)

(i)

$$P(X=9) = {{10} \choose 9} \left[ \frac{2}{3} \right]^9 \left[ \frac{1}{3} \right]^1$$

(ii)

$$\begin{eqnarray*} P(X < 9) & = & 1 - P(X=9) - P(X=10) \\ & = & 1 - {{10} \choo... ...rac{2}{3} \right]^{10} = 1 - 6 \left[ \frac{2}{3} \right]^{10} \end{eqnarray*}$$

EXAMINERS' REMARKS, 2002

Question 4

(a)

(i)

This was usually well done. The most common error was for candidates to produce the answer $\left(\frac{2}{3}\right)^9\left(\frac{1}{3}\right)$, rather than the correct ${{10} \choose 9} \left(\frac{2}{3}\right)^9\left(\frac{1}{3}\right)$. Perhaps this was due to thinking of a sequence of 9 successes followed by 1 failure without realizing that there are 10 different ways of obtaining 9 successes and one failure.

(ii)

The majority of candidates understood the concept of complementary events. errors with parentheses and signs were common, for example, many candidates wrote $P[<9] = 1 - (P[9] - P[10])$ or $P[<9] = 1 - P[9] + P[10]$. candidates also appeared to misinterpret `fewer than 9 times' as `9 or fewer times' and so gave their answer as $P[9] = 1 - P[10]$.