Expert: Jack Cheng - 11/12/2006

Question
Hi, I have the following question.  I have most sections of the question:
Mrs. Smith has 3 children who are part of a group of 8 children who go on a field trip. School officials later announce that 2 of the children in the group have been exposed to hepatitis, but they provide no information as to which 2 children were exposed.
What is the probability that both of those exposed to hepatitis were children of Mrs. Smith?
What is the probability exactly one of the exposed children was a Smith?
What is the probability that at least one of the children exposed was a Smith?
What is the probability that neither exposed child was a Smith?
One of Mrs. Smith's children in the group is named Amy. What is the probability Amy was exposed?
thank you,
J

Answer
You wrote:

Hi, I have the following question.  I have most sections of the question:

Mrs. Smith has 3 children who are part of a group of 8 children who go on a field trip. School officials later announce that 2 of the children in the group have been exposed to hepatitis, but they provide no information as to which 2 children were exposed.

What is the probability that both of those exposed to hepatitis were children of Mrs. Smith?

What is the probability exactly one of the exposed children was a Smith?

What is the probability that at least one of the children exposed was a Smith?

What is the probability that neither exposed child was a Smith?

One of Mrs. Smith's children in the group is named Amy. What is the probability Amy was exposed?

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Hi Jennifer,

To be frank with you, I am a little shaky with probabilities, but here's my three cents:

For both children exposed to be a Smith, the probability would be 3/8 (probability of one Smith exposed) times 3/7 (probability of one Smith exposed in the second selection; the second selection has one less person  because the first selection took one away), which is 9/56.

For exactly one of the children exposed to be Smith, the probability would be 3/8 (probability of one getting exposed) times 5/7 (probability that someone else gets exposed; in the second selection, there are only 7 people, and only two of them are Smiths - the first selection took one away, so five of them are not Smiths), or 15/56. If you reverse the problem, 5/8 (probability of someone else getting exposed in the first selection) times 3/7 (probability of a Smith exposed in the second selection), you still get the same answer.

Now, for the probability of at least one children exposed, I am not sure if it's right, but you add 3/8 (probability of a Smith exposed in the first selection) and 3/7 (probability of a Smith exposed in the second selection), which is 45/56 (that seems a little high though).

For the probability of neither children exposed being a Smith, it would be 5/8 (probability of people not a Smith getting exposed in first selection) times 4/7 (probability of people in the second selection not being a Smith; seven people left, and three are Smiths, so 4 are not), which is 20/56.

The probability of Amy getting exposed would be - again, I am not absolutely sure - 1/8 + 1/7 or 15/56.

Again, I am not completely sure that my answers are right. I would try at another expert here. Paul K. was Math professor at a college, so he should give you a better answer than I can.

Sorry if I am not of any help,
Jack