Question

b) At a supermarket 80% of customers pay by credit card. Find the probability that a randomly selected sample of ten customers,
(i) Exactly two pay by credit card
(ii)Less than three pay by credit card

Answer 1

Answer on Question #54000 – Math – Statistics and Probability

Question

At a supermarket 80% of customers pay by credit card. Find the probability that a randomly selected sample of ten customers,

i) Exactly two pay by credit card

ii) Less than three pay by credit card.

Solution

Let $\xi$ be a random variable of customers who pay by credit card. Obviously $\xi$ has a binomial distribution with the following parameters: $n = 10$ independent attempts; $p = 0.8$ is a probability of success in one attempt.

i) $P\{\xi = 2\} = C_n^2 \cdot p^2 (1 - p)^8 = \frac{10!}{2! \cdot 8!} \cdot 0.8^2 \cdot (0.2)^8 \approx 0.00007$.

ii) $P\{\xi < 3\} = P\{\xi = 0\} + P\{\xi = 1\} + P\{\xi = 2\} = \sum_{k=0}^{2} C_{10}^{k} p^{k} (1 - p)^{10 - k} =$

$$\begin{array}{l} = \sum_{k=0}^{2} C_{10}^{k} (0.8)^{k} (0.2)^{10-k} = \frac{10!}{0! \cdot 10!} \cdot 0.8^{0} \cdot (0.2)^{10} + \frac{10!}{1! \cdot 9!} \cdot 0.8^{1} \cdot (0.2)^{9} + \frac{10!}{2! \cdot 8!} \cdot 0.8^{2} \cdot (0.2)^{8} \approx \\ \approx 0.0000779. \end{array}$$

Answer: i) 0.00007; ii) 0.0000779.

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