Question

QUESTION 1 (0.5 POINTS) In a large restaurant an average of 3 out of every 5 customers ask for water with their meal. A random sample of 10 customers is selected. What is the probability that less than 3 customers ask for water with their meal? a). 0.012 b). 0.055 c). 0.042 d). 0.011

Answer 1

Answer:

Option A) 0.012

Step-by-step explanation:

We are given the following information:

We treat customers asking for water as a success.

P(customers ask for water) = $\dfrac{3}{5}$ = 0.6

Then the number of customers follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x < 3) = P(x = 0) + P(x = 1) + P(x = 2) \\= \binom{10}{0}(0.6)^0(1-0.6)^{10} +\binom{10}{1}(0.6)^1(1-0.6)^{9} + \binom{10}{2}(0.6)^2(1-0.6)^{8}\\= 0.0001 +0.0015 + 0.0106\\=0.012$

0.012 is the probability that less than 3 customers ask for water with their meal.