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# probability basic

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Manager
Joined: 27 Jul 2007
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07 Feb 2008, 11:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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CEO
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07 Feb 2008, 12:05
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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Director
Joined: 01 Jan 2008
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07 Feb 2008, 12:09
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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Manager
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07 Feb 2008, 12:22
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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CEO
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07 Feb 2008, 12:46
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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SVP
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07 Feb 2008, 16:39
walker, why do you have 2 answers for each question ? what am i missing ?

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Intern
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07 Feb 2008, 19:09
(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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Manager
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16 Feb 2008, 01:01
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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Manager
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16 Feb 2008, 02:06
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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CEO
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16 Feb 2008, 02:13
-> 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, 02:13
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