saha wrote:
I think it helps to invert the problem. So whats the probability that you don't have any quarters after N coin withdrawals.
That would be F(N)=15/21*14/20.. (N terms).
The probability that you have at least one quarter is (1-F(N)).
F(1) = 15/21
F(2) = (15/21) (14/20) = 0.5
Since 1-F(2) = 0.5, the answer is 2.
old_dream_1976 wrote:
it is 2
[6c2 + (6c1 x 15c1)]/21C2
but the more important question is should this be a trial and error
p(1), p(2) ........ approach ?
Can anyone suggest a more direct method ?
i guess the above approaches both work. the answer should be 2 as well. remember the question says "at least 1 qt".
prob of getting "at least 1 qt" in drawing 1 qt = 6/21 = 2/7 i.e less than 50%. so not correct.
prob of getting "at least 1 qt" in 2 coins = [(prob of getting getting 1 qt and 1 dime or nickel) + (prob of getting getting 2 qts)]/(total ways of geeting 2 coins)
prob = [(6 x 15) + (6c2)]/21c2 = (90+15)/210 = 50%
the first one is also coreect way...
thanx guys...