Respuesta :
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)]
[tex]={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[/tex]
Thus, the probability that exactly 1 student is using Internet Explorer and at least 3 students are using Chrome is 0.1350.
