Question

A box contains 4 pears and 7 Oranges. Three fruited are taken out at random and eaten. Find the probability that (1) 2 pears and 1 orange are eaten (2) The third fruit is eaten is an orange (3) The first fruit eaten was a pears. Given that the third fruit eaten is an orange

Answer 1

ppo
4/11 * 3/10 * 7/9 = a

pop
4/11 * 7/10 * 3/9 = b


opp
7/11 * 4/10 * 3/9 = c

add a+b+c
for the answer to (1)
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first is p = 4/11
second is p = 3/10
third is o = 7/9
answer to (2) is 4/11*3/10*7/9

for ( 30), Bayes
try
http://en.wikipedia.org/wiki/Bayes'_theorem

if the third one is an orange
then we really have 10 total
4 pears and 6 oranges to chose from for our first pick
4/10