Hey there!

Note that a lot of the introduction to the question is pretty irrelevant. All you need to know from the intro is that there are 5 classrooms, each with two aircons, and the information given in the table. Please use the hints to solve the questions yourself before opening the spoilers for the answers

a) If every air con is turned off, that means all five classrooms have two air cons off. First, what is the chance that one classroom has both air cons off, then what is the chance that all five have them all off?

b) If two classrooms have no air cons off, and there is at most one classroom with one off, we have two cases:

- Two rooms with two air cons off, one room with one air con off, two rooms with no air cons off

- Three rooms with two air cons off, two rooms with no air cons off

Consider why this is an exhaustive list, and check that they both satisfy the criteria in the question. From here, you can calculate the probabilities of each case individually in a similar manner to a), then sum them up.

Spoiler

\(0.45^2\times 0.3\times 0.25^2 + 0.45^2\times 0.25^3\)

c) The 'at least' part of the question should be tipping you off to use the complement in some way, where possible and here, we can do exactly that. The probability that at least one classroom has no air cons off is the complement of no classroom has no air cons off. This is a rather more difficult question, but have a go nevertheless.

Hint 1

What is one arrangement of every classroom having at least one air con on?

Hint 2

Have you considered every way in which you can have every classroom having at least one air con on?

Answer

\(1 - (5\times 0.25\times 0.3^4)\)