There are n sweets in a bag. 6 of the sweets are orange. The rest of the sweets are yellow. Hannah takes at random a sweet from the bag. She eats the sweet. Hannah then takes at random another sweet from the bag. She eats the sweet. The probability that Hannah eats two orange sweets is 1/3 (a) Show that n^2 - n - 90 = 0 (b) Solve n^2 - n - 90 = 0 to find the value of n
+33661 Probability that the first sweet is orange = 6/n
Probability that, if the first was orange, the second is also orange = 5/(n-1) beacuse there are only 5 orange sweets left and there are only n-1 sweets left in total.
Hence overall probability that her two sweets are orange = (6/n)*(5/(n-1)) = 30/(n(n-1))
We are told this equals 1/3, so 30/(n(n-1)) = 1/3
Multiply both sides by 3n(n-1): 90 = n(n-1) or 90 = n^2 - n
Subtract 90 from both sides: n^2 - n - 90 = 0
This factors as (n-10)(n+9) = 0
Since we can't have a negative number of sweets, we must have n = 10 sweets
