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:
<em>C</em> = a student uses Google chrome
<em>E</em> = a student uses Internet explorer
<em>F</em> = a student uses Firefox
<em>M</em> = a student uses Mozilla
<em>S</em> = 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 <em>n</em> = 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)]
Thus, the probability that exactly 1 student is using Internet Explorer and at least 3 students are using Chrome is 0.1350.
