John has 90 unlabelled cans in a box. Of the 90 cans 24 contain
baby corn, 12contain beans, 36 contain peas and 18 contain
mushrooms.
Find the probability that John will have to open more than 2
cans before he finds one which does not contain baby corn.
Hi quailmaster,
Let us have a look at this question step-by-step.
Firstly, let us consider the probability of getting a can of
baby corn on John's first try. That will be 24/90.
Next, let us consider the probability of getting a can of baby
corns on John's second try. That will be 66/90 X 24/89 = 88/445
Thus, the probability of John getting a can of baby corns on his
first two tries will be 24/90 + 88/445 = 124/267
Hence, the probability of John taking more than two tries will
be 1 - 124/267 = 143/267
Cheers.