There's a school in India where only 60% of students have internet access. What is the probability of choosing eight students randomly, five of whom have internet access? (Info: Each student's internet access (or lack of it) is independent from all others

Imagine a bag of balls. Each ball represents a student, and those with A on them are students with internet access, while balls without an A are those without internet access. We know from the question, when we pick a ball from the bag at random (ie. when we pick a student from random) the probability of A (student with internet) is 60% or 0.6. That means the probability of a student without internet is 1 - 0.6 = 0.4. Let's call this ball A' (pronounced "not A")

We're looking to find the probability of choosing five students with internet and three without by random. That is the same as choosing 5A and 3A' . Let's break this down:

The probability of choosing 1A is 0.6 (from the question). The probability of choosing 2A is 0.6 x 0.6. Why? Because we assume choosing balls (students) from the bag are independent events. That basically means, whether I have internet does not depend on whether you, or anyone else, has internet. Each student's probability of having internet is independent of all other students' internet access. 

Alright so we know choosing balls (students) are independent events which means the probability of choosing 5A is 0.6 x 0.6 x 0.6 x 0.6 x 0.6 = (0.6)^5. Similarly, we know the probability of chosing 3A' is 0.4 x 0.4 x 0.4 = (0.4)^3

So probability of choosig 5A and 3A' is (0.6)^5 x (0.4)^3 right? Wrong! This is where many students will lose marks. Let's say we have our eight students lined up and numbered 1,2,3,4,5,6,7,8. 

We know five have internet and three don't, but the five might be 1,2,3,4,5 and the three is 6,7 and 8. Or maybe the five is 2,4,5,6,7 and the three is 1,3 ad 8. Each of these combinations happen with probability: (0.6)^5 x (0.4)^3, but we must count all the combinations: recall from previous knowledge 8C5 combinations of 5 from 8, and the other three must not have internet. So the final answer is: 8C5 x (0.6)^5 x (0.4)^3 = 0.2787 (4 d.p.) (calculator work).