Re: Elementary probabilty problem
- From: rob@xxxxxxxxxxxxxx (Rob Johnson)
- Date: Mon, 16 Jan 2006 16:20:36 GMT
In article <FLMyf.73659$m05.15591@clgrps12>,
Bipp <bipp@xxxxxxxxxxxxxxxx> wrote:
>There's a deck of 26 cards, each one with a different letter of the
>alphabet on one side. Someone randomly picks 10 cards from the deck
>without showing them to me. I ask him two questions.
>
>First question: "Is there at least one of the three letters A, B, C in
>your hand?"
>Answer: "Yes."
>
>In the following comments, "P(X)" means "probability that letter X is in
>the hand."
>
>At this point, I can infer that P(A) = P(B) = P(C). Intuitively, I would
>say that the probability is 1/3, but I think some elementary
>combinatorial analysis formula would show that the exact value may be
>slightly different. I haven't done this in years and I'm rusty. How do I
>calculate it?
Since A, B, and C can be in the hand at the same time, the fact that
P(A) = P(B) = P(C) does not imply that P(A) = 1/3.
Let Q be the event that one of A, B, or C is in the hand. Because of
the answer to the first question, we are given Q, so what you mean by
P(A) is p(A|Q), the probability that A is in the hand given Q (note
that the lower-case p represents probabilities not assuming Q).
p(A|Q) = p(A&Q)/p(Q) = p(A)/p(Q) since A implies Q.
p(-Q), the probability of Q being false, is the probability that all
three cards are not in the hand. This is C(16,3)/C(26,3) = 14/65
(color A, B, and C white and all 23 other cards black; there are
C(16,3) ways that the 3 cards are in the 16 not in the hand out of
C(26,3) total). Therefore, the p(Q) = 1 - 14/65 = 51/65.
p(A) is simply 10/26, so p(A|Q) = p(A)/p(Q) = (10/26)/(51/65) = 25/51.
Therefore, P(A) = 25/51, which is more than 1/3.
>Second question: "Is the letter C in your hand?"
>
>If the answer is "no", then P(C) becomes 0. From that new information, I
>would infer that P(A) and P(B) become 1/2 roughly. How do I calculate
>the exact probability?
>
>If, instead, the answer is "yes", then P(C) becomes 1. Does this new
>information change P(A) and P(B)? Intuitively, I would guess that P(A)
>and P(B) may slightly decrease at this point. I don't trust my
>intuition. How do I calculate this?
Answer is "no":
P(A) in this case is p(A|-C&Q)
p(-C) is simply 16/26 and p(-Q) is computed above as 14/65. Therefore,
p(-C&Q) = p(-C) - p(-C&-Q) = p(-C) - p(-Q) = 2/5 since -Q implies -C.
p(A&-C) = 16*10/(26*25) = 16/65 since A can be in 10 places and C in
16 out of a total 26*25.
p(A|-C&Q) = p(A&-C&Q)/p(-C&Q) = p(A&-C)/p(-C&Q) since A implies Q.
p(A|-C&Q) = (16/65)/(2/5) = 8/13.
Therefore, P(A) = 8/13, which is more than 1/2.
Answer is "yes":
p(A&C) = 10*9/(26*25) = 9/65 since A can be in 10 places and C in
9 out of a total 26*25. p(C) is simply 10/26.
P(A) in this case is p(A|C&Q) = p(A|C) since C implies Q.
p(A|C) = p(A&C)/P(C) = (9/65)/(10/26) = 9/25.
Therefore, P(A) = 9/25, which is less than 1/2.
Rob Johnson <rob@xxxxxxxxxxxxxx>
take out the trash before replying
.
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- Elementary probabilty problem
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