A bag contains n n\,n red counters and 8 blue counters.
Two counters are removed from the bag at random, one after the other, without replacement.
The probability that both counters are red is 15\dfrac{1}{5}51.
Show that n2−5n−14=0n^{2} - 5n - 14 = 0n2−5n−14=0.
Hence find the number of red counters in the bag, explaining why you reject the other root.
Given that the two counters removed are the same colour, find the probability that they are both blue.