Question

At Foodland supermarket 45%of the customers pay by credit card. Find the probability that in a randomly selected sample of fifteen customers (a)exactly four pay by credit card. (b) more than ten pay by cash.

Answer 1

$n=15 \\ p = 0.45 \\ q = 1 -p = 1 -0.45 = 0.55$

(a) According to Bernoulli's formula:

$P(X=4) = C_{15}^4p^4q^{15-4} = \frac{15!}{4! \times 11!} \times 0.45^4 \times 0.55^{11} = 0.07798$

(b)

$P(X > 10) = 1 -[P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)+ P(X = 6)+ P(X = 7) + P(X = 8) + P(X = 9)] \\ P(X=0) = \frac{15!}{0! \times 15!} \times 0.45^0 \times 0.55^{15} = 0.00012 \\ P(X=1) = \frac{15!}{1! \times 14!} \times 0.45^1 \times 0.55^{14} = 0.00156 \\ P(X=2) = \frac{15!}{2! \times 13!} \times 0.45^2 \times 0.55^{13} = 0.00896 \\ P(X = 3) = \frac{15!}{3! \times 12!} \times 0.45^3 \times 0.55^{12} = 0.03176 \\ P(X = 4) = 0.07798 \\ P(X=5) = \frac{15!}{5! \times 10!} \times 0.45^5 \times 0.55^{10} = 0.14036 \\ P(X=6) = \frac{15!}{6! \times 9!} \times 0.45^6 \times 0.55^9 = 0.19140 \\ P(X=7) = \frac{15!}{7! \times 8!} \times 0.45^7 \times 0.55^8 = 0.20134 \\ P(X=8) = \frac{15!}{8! \times 7!} \times 0.45^8 \times 0.55^7 = 0.16473 \\ P(X=9) = \frac{15!}{9! \times 6!} \times 0.45^9 \times 0.55^6 = 0.10483 \\ P(X > 10) = 1 -( 0.00012 + 0.00156 + 0.00896 + 0.03176 + 0.07798 + 0.14036 + 0.19140 + 0.20134 + 0.16473 + 0.10483 ) \\ = 1 -0.92304 \\ = 0.07696$