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09 Oct 2003, 21:06
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A box contains black chips and red chips . A person draws two chips without replacement. If the probability of selecting black chip and a red chip is 15/56 and the probability of drawing a black chip on the first draw is 3/4, what is the probability of drawing a red chip on the second draw, if you know that the first chip drawn was black?
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09 Oct 2003, 22:46
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araspai
A box contains black chips and red chips . A person draws two chips without replacement. If the probability of selecting black chip and a red chip is 15/56 and the probability of drawing a black chip on the first draw is 3/4, what is the probability of drawing a red chip on the second draw, if you know that the first chip drawn was black?
P(Red on Second Draw/First Black) = P( Red on second and first black) /P(First Black)
= 15/56 /( 3/4) = 15/56 * 4/3 = 5/14
answer 5/14
thanks
praetorian
14 Oct 2003, 01:07
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the stem does not specifically says that 15/56 is the probability of taking a black one first and a red one second.
So, we have to conclude that P(BR)+P(RB)=15/56
P(BR)=P(RB)=15/112
15/112=3/4*X
X=15*4/112*3=5/28
So, we have to conclude that P(BR)+P(RB)=15/56
P(BR)=P(RB)=15/112
15/112=3/4*X
X=15*4/112*3=5/28
