Six customers wait in a queue in a supermarket. A customer can choose to pay with cash or a credit card. Assume that whether or not a customer pays with a credit card is independent of any other customers’ methods of payment.

It is known that 60% of customers choose to pay with a credit card.

(a)     Find the probability that:

          (i)     the first three customers pay with a credit card and the next three pay with cash;

          (ii)     exactly three of the six customers pay with a credit card.

There are n customers waiting in another queue in the same supermarket. The probability that at least one customer pays with cash is greater than 0.995.

(b)     Find the minimum value of n.

(a)     (i)     \({0.6^3} \times {0.4^3}\)     (M1)

 

Note:     Award (M1) for use of the product of probabilities.

 

          \( = 0.0138\)     A1

 

          (ii)     binomial distribution \(X:{\text{B(6, 0.6)}}\)     (M1)

 

Note:     Award (M1) for recognizing the binomial distribution.

 

          \({\text{P}}(X = 3) = \) \(^6{C_3}{(0.6)^3}{(0.4)^3}\)

          \( = 0.276\)     A1

 

Note:     Award (M1)A1 for \(^6{C_3} \times 0.0138 = 0.276\).

 

[4 marks]

 

(b)     \(Y:{\text{B(}}n,{\text{ 0.4)}}\)

\({\text{P}}(Y \geqslant 1) > 0.995\)

\(1 - {\text{P}}(Y = 0) > 0.995\)

\({\text{P}}(Y = 0) < 0.005\)     (M1)

 

Note:     Award (M1) for any of the last three lines. Accept equalities.

 

\({0.6^n} < 0.005\)     (M1)

 

Note:     Award (M1) for attempting to solve \({0.6^n} < 0.005\) using any method, eg, logs, graphically, use of solver. Accept an equality.

 

\(n > 10.4\)

\(\therefore n = 11\)     A1

[3 marks]

 

Total [7 marks]

[N/A]