Find \(k\).

(i)     A day is chosen at random. Write down the probability that Machine A has no breakdowns.

(ii)     Five days are chosen at random. Find the probability that Machine A has no breakdowns on exactly four of these days.

Find \({\text{E}}(B)\).

(i)     Find the probability that there are exactly two breakdowns on Tuesday.

(ii)     Given that there are exactly two breakdowns on Tuesday, find the probability that both breakdowns are of Machine A.

evidence of summing to 1     (M1)

eg\(\,\,\,\,\,\)\(0.55 + 0.3 + 0.1 + k = 1\)

\(k = 0.05{\text{ (exact)}}\)    A1     N2

[2 marks]

(i)     0.55     A1     N1

(ii)     recognizing binomial probability     (M1)

eg\(\,\,\,\,\,\)\(X:{\text{ }}B(n,{\text{ }}p),{\text{ }}\left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right),{\text{ }}{(0.55)^4}(1 - 0.55),{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){p^r}{q^{n - r}}\)

\(P(X = 4) = 0.205889\)

\(P(X = 4) = 0.206\)    A1     N2

[3 marks]

correct substitution into formula for \({\text{E}}(X)\)     (A1)

eg\(\,\,\,\,\,\)\(0.2 + (2 \times 0.08) + (3 \times 0.02)\)

\({\text{E}}(B) = 0.42{\text{ (exact)}}\)    A1     N2

[2 marks]

(i)     valid attempt to find one possible way of having 2 breakdowns     (M1)

eg\(\,\,\,\,\,\)\(2A,{\text{ }}2B,{\text{ }}1A\) and \(1B\), tree diagram

one correct calculation for 1 way (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(0.1 \times 0.7,{\text{ }}0.55 \times 0.08,{\text{ }}0.3 \times 0.2\)

recognizing there are 3 ways of having 2 breakdowns     (M1)

eg\(\,\,\,\,\,\)A twice or B twice or one breakdown each

correct working     (A1)

eg\(\,\,\,\,\,\)\((0.1 \times 0.7) + (0.55 \times 0.08) + (0.3 \times 0.2)\)

\({\text{P(2 breakdowns)}} = 0.174{\text{ (exact)}}\)     A1     N3

(ii)     recognizing conditional probability     (M1)

eg\(\,\,\,\,\,\)\({\text{P}}(A|B),{\text{ P}}(2A|{\text{2breakdowns}})\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\frac{{0.1 \times 0.7}}{{0.174}}\)

\({\text{P}}(A = 2|{\text{two breakdowns}}) = 0.402298\)

\({\text{P}}(A = 2|{\text{two breakdowns}}) = 0.402\)     A1     N2

[8 marks]

Candidates generally found parts (a), (b)(i) and (c) of this question the most straightforward and those who recognised the binomial distribution in (b)(ii) were usually able to obtain the required solution using their GDCs. Part (d)(i) proved to be more problematic with many candidates identifying one possible way of having two breakdowns (usually 1A and 1B), but not recognising three ways of having two breakdowns. Furthermore, many were not able to successfully calculate the probability of two breakdowns on one machine (and none on the other). The conditional probability in (d)(ii) was generally recognised though and those who showed their working in full were able to score follow through marks in this part.

Candidates generally found parts (a), (b)(i) and (c) of this question the most straightforward and those who recognised the binomial distribution in (b)(ii) were usually able to obtain the required solution using their GDCs. Part (d)(i) proved to be more problematic with many candidates identifying one possible way of having two breakdowns (usually 1A and 1B), but not recognising three ways of having two breakdowns. Furthermore, many were not able to successfully calculate the probability of two breakdowns on one machine (and none on the other). The conditional probability in (d)(ii) was generally recognised though and those who showed their working in full were able to score follow through marks in this part.

Candidates generally found parts (a), (b)(i) and (c) of this question the most straightforward and those who recognised the binomial distribution in (b)(ii) were usually able to obtain the required solution using their GDCs. Part (d)(i) proved to be more problematic with many candidates identifying one possible way of having two breakdowns (usually 1A and 1B), but not recognising three ways of having two breakdowns. Furthermore, many were not able to successfully calculate the probability of two breakdowns on one machine (and none on the other). The conditional probability in (d)(ii) was generally recognised though and those who showed their working in full were able to score follow through marks in this part.

Candidates generally found parts (a), (b)(i) and (c) of this question the most straightforward and those who recognised the binomial distribution in (b)(ii) were usually able to obtain the required solution using their GDCs. Part (d)(i) proved to be more problematic with many candidates identifying one possible way of having two breakdowns (usually 1A and 1B), but not recognising three ways of having two breakdowns. Furthermore, many were not able to successfully calculate the probability of two breakdowns on one machine (and none on the other). The conditional probability in (d)(ii) was generally recognised though and those who showed their working in full were able to score follow through marks in this part.