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Two Probability questions

  • #1
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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?
 

Answers and Replies

  • #2
118
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1) The numerator is the number of ways you can get one person of each ethnicity if you pick 4 people. How many ways can you pick four people so that one person is Chicano, one is Asian, one is African American, and one is Caucasian? There are 5 choices for a Chicano. For each Chicano you can pick one of 2 Asians, so there are 5x2 ways to pick one Chicano and one Asian. Extend this to picking African Americans and Caucasians and you get 5x2x3x3.

2) I think that's right.
 

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