Question

The number of customers in line at a supermarket express checkout counter is a random variable with the following probability distribution. x 0 1 2 3 4 5 P(x) 0.10 0.25 0.30 0.20 0.10 0.05

Find P(at least 3)

Answer 1

P(at least 3) = 0.35

Explanation:

We are given the following probability distribution in the question:

x: 0 1 2 3 4 5

P(x): 0.10 0.25 0.30 0.20 0.10 0.05

where x is the number of customers in line at a supermarket

Verification for a discrete probability distribution:

`\displaystyle\sum P(x_i) = 1\\\\\displaystyle\sum P(x_i) = 0.10 +0.25 +0.30 +0.20 +0.10 +0.05 = 1`

Hence, it is a discrete probability distribution.

We have to evaluate P(at least 3).

`P(x\geq 3)\\=1 - P(x<3)\\=1- (P(x=0) + P(x=1) + P(x=2))\\=1 - (0.10 +0.25 +0.30)\\=0.35`

Thus,

P(at least 3) = 0.35

0.35 is the probability that there are atleast 3 customers in line at a supermarket.