Asked by Mily
There are 12 balls in a bag, where n of them are white balls. If 2 balls are chosen at random from the bag without replacement, the probability of choosing at least 1 white balls is 19/33. Find n.
Answers
Answered by
Reiny
number of white --- n
number of non-white -- 12-n
prob of choosing at least 1 white
= 1 - prob(choosing both non-white)
= 1 - (12-n)/12 * (11-n)/11
= 19/33
(12-n)(11-n)/132 = 1 - 19/33 = 14/33
(132 - 23n + n^2)/12 = 14/3
3n^2 - 69n + 396 = 168
3n^2 - 69n +228 = 0
n^2 - 23n + 76 = 0
(n+19)(n-4) = 0
n = 4 or n = -19 , rejecting the negative number of balls
n = 4
check:
prob of picking 2 non-white
= (8/12)(7/11) = 56/132 = 14/33
prob of picking at least 1 white
= 1 - 14/33 = 19/33
number of non-white -- 12-n
prob of choosing at least 1 white
= 1 - prob(choosing both non-white)
= 1 - (12-n)/12 * (11-n)/11
= 19/33
(12-n)(11-n)/132 = 1 - 19/33 = 14/33
(132 - 23n + n^2)/12 = 14/3
3n^2 - 69n + 396 = 168
3n^2 - 69n +228 = 0
n^2 - 23n + 76 = 0
(n+19)(n-4) = 0
n = 4 or n = -19 , rejecting the negative number of balls
n = 4
check:
prob of picking 2 non-white
= (8/12)(7/11) = 56/132 = 14/33
prob of picking at least 1 white
= 1 - 14/33 = 19/33
Answered by
Mily
Oh I see! I got the formula correct but I multiply it wrongly¡Thanks a lot!
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