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probability basic

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probability basic [#permalink] New post 07 Feb 2008, 10:48
Guys,

suppose there are 4 persons - A,B,C,D. Three are to be chosen.
1) What is the probability that B,C and D will be chosen?
2) What is the probability that C will be chosen first, followed by D followed by B?
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Re: probability basic [#permalink] New post 07 Feb 2008, 11:05
Expert's post
A)

p_1=\frac{P^3_3}{P^4_3}=\frac{3!}{4!}=\frac{1}{4}

p_2=\frac{1}{P^4_3}=\frac{1}{4!}=\frac{1}{24}

B)

p_1=\frac{3}{4}*\frac{2}{3}*\frac{1}{2}=\frac{1}{4}

p_2=\frac{1}{4}*\frac{1}{3}*\frac{1}{2}=\frac{1}{24}
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Re: probability basic [#permalink] New post 07 Feb 2008, 11:09
farend wrote:
Guys,

suppose there are 4 persons - A,B,C,D. Three are to be chosen.
1) What is the probability that B,C and D will be chosen?
2) What is the probability that C will be chosen first, followed by D followed by B?


one more way of answering question 1. choosing B, C and D means that A is not chosen. probability of that is 1/4.
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Re: probability basic [#permalink] New post 07 Feb 2008, 11:22
walker wrote:
A)

p_1=\frac{P^3_3}{P^4_3}=\frac{3!}{4!}=\frac{1}{4}

p_2=\frac{1}{P^4_3}=\frac{1}{4!}=\frac{1}{24}


Can we use 3C3 and 4C3 in the steps of A) here as we are just dealing with selection ?
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Re: probability basic [#permalink] New post 07 Feb 2008, 11:46
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farend wrote:
Can we use 3C3 and 4C3 in the steps of A) here as we are just dealing with selection ?


You can but I don't like to do so because in second question you may go a wrong direction.
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Re: probability basic [#permalink] New post 07 Feb 2008, 15:39
walker, why do you have 2 answers for each question ? what am i missing ?
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Re: probability basic [#permalink] New post 07 Feb 2008, 18:09
(A) and (B) are just different approaches (yielding the same results... duh!)

(A) involves counting the number of ways you can "choose" things. You divide the number of desired ways of choosing by the total number of ways possible. This is a bit less intuitive than (B).

(B) involves simply choosing the three that you want one by one, and you consider the probability of choosing each of them one step at a time.

I think I can explain method (A) to a certain degree but I don't know if I can apply it myself. Hope this helps a bit!
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Re: probability basic [#permalink] New post 16 Feb 2008, 00:01
Walker,
I am still having trouble understanding the counting. Let me put up another quest:

-> 2 cards are drawn from a pack of 52 cards.
1) What is the probability that a King and a Queen will be selected ?
2) What is the probability that the first is a King and second is a Queen ?
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Re: probability basic [#permalink] New post 16 Feb 2008, 01:06
This one you would definitely want to use the probabilities, not the "choose" method I think...

1) Select king and queen (order not important):

=(8/52)*(4/52) = (2/13)*(1/13) = 2/169


2) Select king first then queen (order is important)

=(4/52)*(4/52)= (1/13)*(1/13) = 1/169
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Re: probability basic [#permalink] New post 16 Feb 2008, 01:13
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-> 2 cards are drawn from a pack of 52 cards.
1) What is the probability that a King and a Queen will be selected ?
2) What is the probability that the first is a King and second is a Queen ?

Can I assume that there are 4 Kings and 4 Queens in the card batch? :)

1)p=p_{1king}*p_{2queen}+p_{1queen}*p_{2king}=\frac{4}{52}*\frac{4}{51}+\frac{4}{52}*\frac{4}{51}=\frac{32}{52*51}=\frac{8}{13*51}

or

p=p_{(king) or (queen)}*p_{opposite}=\frac{8}{52}*\frac{4}{51}=\frac{32}{52*51}=\frac{8}{13*51}

or

p=\frac{C^4_1*C^4_1*P^2_2}{P^{52}_2}=\frac{4*4*2}{52*51}=\frac{8}{13*51}

or

p=\frac{C^4_1*C^4_1}{C^{52}_2}=\frac{4*4*2}{52*51}=\frac{8}{13*51}

or

p=1-\frac{P^{44}_2+C^8_1*C^3_1+C^8_1*C^{44}_1*P^2_2}{P^{52}_2}=1-\frac{44*43+8*3+8*44*2}{52*51}=1-\frac{473+6+176}{13*51}=\frac{663 -655}{13*51}=\frac{8}{13*51} :)


2) p=p_{1king}*p_{2queen}=\frac{4}{52}*\frac{4}{51}=\frac{16}{52*51}=\frac{4}{13*51}

or

p=\frac{C^4_1*C^4_1}{P^{52}_2}=\frac{4*4}{52*51}=\frac{4}{13*51}

or

p_2=\frac12*p_1=\frac12*\frac{8}{13*51}=\frac{4}{13*51} (symmetry: QK,KQ)
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Re: probability basic   [#permalink] 16 Feb 2008, 01:13
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