Mobile phone batteries are produced by two machines. Machine A produces 60% of the daily output and machine B produces 40%. It is found by testing that on average 2% of batteries produced by machine A are faulty and 1% of batteries produced by machine B are faulty.

(i)     Draw a tree diagram clearly showing the respective probabilities.

(ii)     A battery is selected at random. Find the probability that it is faulty.

(iii)     A battery is selected at random and found to be faulty. Find the probability that it was produced by machine A.

In a pack of seven transistors, three are found to be defective. Three transistors are selected from the pack at random without replacement. The discrete random variable X represents the number of defective transistors selected.

(i)     Find \({\text{P}}(X = 2)\).

(ii)     Copy and complete the following table:

(iii)     Determine \({\text{E}}(X)\).

(i)
     A1A1

Note:     Award A1 for a correctly labelled tree diagram and A1 for correct probabilities.

(ii)     \({\text{P}}(F) = 0.6 \times 0.02 + 0.4 \times 0.01\)     (M1)

\( = 0.016\)     A1

(iii)     \({\text{P}}(A|F) = \frac{{{\text{P}}(A \cap F)}}{{{\text{P}}(F)}}\)

\( = \frac{{0.6 \times 0.02}}{{0.016}}{\text{ }}\left( { = \frac{{0.012}}{{0.016}}} \right)\)     M1

\( = 0.75\)     A1

[6 marks]

(i)     METHOD 1

\({\text{P}}(X = 2) = \frac{{^3{C_2}{ \times ^4}{C_1}}}{{^7{C_3}}}\)     (M1)

\( = \frac{{12}}{{35}}\)     A1

METHOD 2

\(\frac{3}{7} \times \frac{2}{6} \times \frac{4}{5} \times 3\)     (M1)

\( = \frac{{12}}{{35}}\)     A1

(ii)          A2

Note:     Award A1 if \(\frac{4}{{35}},{\text{ }}\frac{{18}}{{35}}\) or \(\frac{1}{{35}}\) is obtained.

(iii)     \({\text{E}}(X) = \sum {x{\text{P}}(X = x)} \)

\({\text{E}}(X) = 0 \times \frac{4}{{35}} + 1 \times \frac{{18}}{{35}} + 2 \times \frac{{12}}{{35}} + 3 \times \frac{1}{{35}}\)     M1

\( = \frac{{45}}{{35}} = \left( {\frac{9}{7}} \right)\)     A1

[6 marks]