Question

In a student body, 50% use Google Chrome, 9% use Internet Explorer, 10% Firefox, 5% Mozilla, and the rest use Safari. In a group of 5 students, what is the probability exactly one student is using Internet Explorer and at least 3 students are using Chrome

Answer 1

Answer:

The probability that exactly 1 student is using Internet Explorer and at least 3 students are using Chrome is 0.1350.

Step-by-step explanation:

Denote the events as follows:

C = a student uses Google chrome

E = a student uses Internet explorer

F = a student uses Firefox

M = a student uses Mozilla

S = a student uses Safari

Given:

P (C) = 0.50

P (E) = 0.09

P (F) = 0.10

P (M) = 0.05

P (S) = 0.26

A sample of n = 5 students is selected.

The probability that exactly 1 student is using Internet Explorer and at least 3 students are using Chrome is:

P (E = 1 ∩ C ≥ 3) = P (E = 1 ∩ C = 3) + P (E = 1 ∩ C = 4) - P (E = 1 ∩ C = 5)

The probability distribution of a student using any of the browser is Binomial.

Compute the probability that exactly 1 student is using Internet Explorer and at least 3 students are using Chrome as follows:

P (E = 1 ∩ C ≥ 3) = P (E = 1 ∩ C = 3) + P (E = 1 ∩ C = 4) - P (E = 1 ∩ C = 5)

                          = P (E = 1) [P (C = 3) + P (C = 4) - P (C = 5)]

       $={5\choose 1}(0.09)^{1}(1-0.09)^{5-1}[{5\choose 3}(0.50)^{3}(1-0.50)^{5-3}+{5\choose 4}(0.50)^{4}(1-0.50)^{5-4}\\-{5\choose 5}(0.50)^{5}(1-0.50)^{5-5}]\\=0.3086[0.3125+0.1563-0.0313]\\=0.3086\times 0.4375\\=0.1350$

Thus, the probability that exactly 1 student is using Internet Explorer and at least 3 students are using Chrome is 0.1350.