Question

In a large resturanat an average of 2 out of every 5 customers ask for chilies with their meals. A random sample of 15 customers is selected. Find the probability that i. Exactly 7 customers ask for chilies with their meal ii. Less than 3 customers ask for chilies with their meals. 

Answer 1

$p = \frac{2}{5} = 0.4 \\ q = 1 - 0.4 = 0.6$

X~Binomial(15,0.4)

$P(X) = C^{15}_x(0.4)^x(0.6)^{15-x}$

i. Exactly 7 customers ask for chilies with their meal

$P(X=7) = C^{15}_7 \times (0.4)^7 \times (0.6)^{15-7} = 0.1770$

ii. Less than 3 customers ask for chilies with their meals.

$P(X<3) = P(X=0) +P(X+1) +P(X=2) \\ P(X=0) = C^{15}_0 \times (0.4)^0 \times (0.6)^{15-0} = 0.00047 \\ P(X=1) = C^{15}_1 \times (0.4)^1 \times (0.6)^{15-1} = 0.00470 \\ P(X=2) = C^{15}_2 \times (0.4)^2 \times (0.6)^{15-2} = 0.02194 \\ P(X<3) = 0.00047 + 0.00470 + 0.02194 = 0.02711$