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.

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.