A box contains 6 red counters and some blue counters, and no other counters. When one counter is drawn at random, the probability that it is blue is \(\frac{3}{5}\). How many blue counters are there in the box?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 9
A counter is either red or blue, so P(red) = 1 − P(blue), and P(red) is the number of red counters divided by the total.
P(red) \(\displaystyle =1-\frac{3}{5}=\frac{2}{5}\)
Let the total number of counters be \(t\). Then \(\displaystyle \frac{6}{t}=\frac{2}{5}\), so \(\displaystyle t=\frac{6\times5}{2}=15\)
Blue counters \(=15-6=9\)