#20 
by Calilasseia » Jun 06, 2015 2:17 am
Step by step:
[1] If there are n total items to select from randomly, and k items (0<=k<=n) are of a given type, then the probability of selecting one of those k items from that collection of n total items is k/n.
[2] After the first selection takes place above, if the selected item is one of the k items of the specified type, then one is left with a total n-1 items, of which k-1 are of the given type.
[3] If a second selection now takes place, then the probability of selecting one of the k-1 items of the given type from the remaining n-1 items is (k-1)/(n-1).
[4] Since these are independent selection events, the total probability of selecting two of the k items of the given type from the available n items at the start, is therefore (k/n)×[(k-1)/(n-1)] = [k(k-1)]/[n(n-1)]
[5] If that total probability is specified as P, then we have:
P = [k(k-1)]/[n(n-1)]
Therefore n(n-1) = [k(k-1)]/P
[6] We are given that k=6, and P=1/3. This gives us:
n(n-1) = (6×5)/(1/3) = 90
Therefore the resulting equation is n2-n-90 = 0.
Factorising this equation gives us (n-10)(n+9) =0. This has solutions n=-9 and n=10.
Since the total number of items n has to be greater than 0, the desired solution is n=10. There are 10 sweets in total in the selection set, of which 6 are orange, 4 are yellow. As a check, the probability of the first selection of an orange sweet is 6/10, and the probability of the second selection of an orange sweet is 5/9. This gives us a probability for the two selections of (6×5)/(10×9) = 30/90 = 1/3.
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