truongynhi wrote:
VeritasPrepKarishma wrote:
Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).
I have given the answer to this question here:
in-a-certain-baord-game-a-stack-of-48-cards-8-of-which-136256.html#p1130713Quote:
First question: I solved this question using conditional probability. Is my solution correct?
Probability (at least 2 non-share cards) = Probability (2 non-share AND 1 share cards) + Probability (3 non-share cards)
\(= \frac{40*39*8}{48*47*46} + \frac{40*39*38}{48*47*46}\)
\(=\frac{40*39*46}{48*47*46}\)
Desired probability = Probability(2 non-share AND 1 share cards)/Probability(at least 2 non-share cards)
\(=\frac{40*39*8}{40*47*46}\)
\(=\frac{8}{46}=\frac{4}{23}\)
You can do it using conditional probability.
Your calculations are correct. The reason you have done them is not convincing. When you calculate Probability (at least 2 non-share cards), you are saying that you are calculating the probability of selecting 2 of the 3 cards as non share. So first card could be share card and other two non share or second could be share card and other two non share etc. But you already know that first two are non share. So this is not correct. The arrangement matters. First two should be non share cards.
So what you instead need to do is find the probability of picking first two non share cards.
The probability of that is (40/48)*(39/47)
You got the same thing by adding the two probabilities. Note that you considered both cases: third is share or third is non share and added them. That is equivalent to a probability of 1 for the third card. Hence it is the same as (40/48)*(39/47)*1.
Desired Probability = P(First two are non share and third card is share)/P(First two are non share)
\(= \frac{(40/48)*(39/47)*(8/46)}{(40/48)*(39/47)} = \frac{8}{46}\)
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