probability basic

Question

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?

walker · Answer

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}

maratikus · Answer

one more way of answering question 1. choosing B, C and D means that A is not chosen. probability of that is 1/4.

farend · Answer

Can we use 3C3 and 4C3 in the steps of A) here as we are just dealing with selection ?

walker · Answer

You can but I don't like to do so because in second question you may go a wrong direction.

pmenon · Answer

walker, why do you have 2 answers for each question ? what am i missing ?

aimecon · Answer

(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!

farend · Answer

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 ?

jankynoname · Answer

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

walker · Answer

-> 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)

probability basic

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