- #1
BrownianMan
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1. A committee consists of five Chicanos, two Asians, three African Americans, and two Caucasians. A subcommittee of four is chosen at random. What is the probability that all the ethnic groups are represented on the subcommittee?
I know the answer for this is (5x3x2x2)/(12 choose 4), but I'm not sure exactly why. I understand why the denominator is 12 choose 4), but I can't figure out why the numerator is (5x3x2x2).
2. Urn A has three red balls and two white balls, and urn B has two red balls and five white balls. A fair coin is tossed; if it lands heads up, a ball is drawn from urn A, and otherwise, a ball is drawn from urn B. What is the probability that a red ball is drawn? If a red ball is drawn, what is the probability that the coin landed heads up?
Ok, so for this I got,
P(red ball is drawn) = P(heads)P(red ball drawn from urn A) + P(tails)P(red ball drawn from urn B) = (1/2)(3/5) + (1/2)(2/7) = 31/70
then,
P(heads given red ball drawn) = P(heads and red ball drawn)/P(red ball drawn) = (1/2)(3/5)/(31/70) = 21/31
Is this right?
I know the answer for this is (5x3x2x2)/(12 choose 4), but I'm not sure exactly why. I understand why the denominator is 12 choose 4), but I can't figure out why the numerator is (5x3x2x2).
2. Urn A has three red balls and two white balls, and urn B has two red balls and five white balls. A fair coin is tossed; if it lands heads up, a ball is drawn from urn A, and otherwise, a ball is drawn from urn B. What is the probability that a red ball is drawn? If a red ball is drawn, what is the probability that the coin landed heads up?
Ok, so for this I got,
P(red ball is drawn) = P(heads)P(red ball drawn from urn A) + P(tails)P(red ball drawn from urn B) = (1/2)(3/5) + (1/2)(2/7) = 31/70
then,
P(heads given red ball drawn) = P(heads and red ball drawn)/P(red ball drawn) = (1/2)(3/5)/(31/70) = 21/31
Is this right?