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)]
![={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](https://tex.z-dn.net/?f=%3D%7B5%5Cchoose%201%7D%280.09%29%5E%7B1%7D%281-0.09%29%5E%7B5-1%7D%5B%7B5%5Cchoose%203%7D%280.50%29%5E%7B3%7D%281-0.50%29%5E%7B5-3%7D%2B%7B5%5Cchoose%204%7D%280.50%29%5E%7B4%7D%281-0.50%29%5E%7B5-4%7D%5C%5C-%7B5%5Cchoose%205%7D%280.50%29%5E%7B5%7D%281-0.50%29%5E%7B5-5%7D%5D%5C%5C%3D0.3086%5B0.3125%2B0.1563-0.0313%5D%5C%5C%3D0.3086%5Ctimes%200.4375%5C%5C%3D0.1350)
Thus, the probability that exactly 1 student is using Internet Explorer and at least 3 students are using Chrome is 0.1350.