Find the probability he has his third success of hitting the target on his sixth throw.

(i)     Find the expected number of throws required for Eric to hit the target three times.

(ii)     Write down his expected profit or loss if he plays until he wins the \($10\).

If he has just \($8\), find the probability he will lose all his money before he hits the target three times.

METHOD 1

let \(X\) be the number of throws until Eric hits the target three times

\(X \sim {\text{NB(3, 0.2)}}\)     (M1)

\({\text{P}}(X = 6) = \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right){0.8^3} \times {0.2^3}\)     (A1)

\( = 0.04096\;\;\;\left( { = \frac{{128}}{{3125}}} \right)\;\;\;\)(exact)     A1

METHOD 2

let \(X\) be the number of hits in five throws

\(X\) is \({\text{B}}(5,{\text{ }}0.2)\)     (M1)

\({\text{P}}(X = 2) = \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right){0.2^2} \times {0.8^3}\;\;\;(0.2048)\)     (A1)

\(P\)(3rd hit on 6th throw) \( = \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right){0.2^2} \times {0.8^3} \times 0.2 = 0.04096\left( { = \frac{{128}}{{3125}}} \right)\;\;\;\)(exact)     A1

[3 marks]

(i)     \({\text{expected number of throws}} = \frac{3}{{0.2}} = 15\)     (M1)A1

(ii)     \({\text{profit}} = (10 - 15) =  - \$ 5{\text{ or loss}} = \$ 5\)     A1

[3 marks]

METHOD 1

let \(Y\) be the number of times the target is hit in \(8\) throws

\(Y \sim {\text{B}}(8,{\text{ }}0.2)\)     (M1)

\({\text{P}}(Y \le 2)\)     (M1)

\( = 0.797\)     A1

METHOD 2

let the \({3^{{\text{rd}}}}\) hit occur on the \({Y^{{\text{th}}}}\) throw

\(Y{\text{ is NB}}(3,{\text{ }}0.2)\)     (M1)

\({\text{P}}(Y > 8) = 1 - {\text{P}}(Y \le 8)\)     (M1)

\( = 0.797\)     A1

[3 marks]

Total [9 marks]

Part (a) was well answered, using the negative binomial distribution \(NB(3,{\text{ }}0.2)\), by many candidates. Some candidates began by using the binomial distribution \(B(5,{\text{ }}0.2)\) which is a valid method as long as it is followed by multiplying by 0.2 but this final step was not always carried out successfully.

Part (b) was well answered by the majority of candidates.

In (c), candidates who used the binomial distribution \(B(8,{\text{ }}0.2)\) were generally successful. Candidates who used the negative binomial distribution \(Y \approx NB(3,{\text{ }}0.2)\) to evaluate \(P(Y > 8)\) were usually unsuccessful because of the large amount of computation involved.