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

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)]

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

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the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

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