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24 Apr 2016, 20:27
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truongynhi
I have given the answer to this question here: in-a-certain-baord-game-a-stack-of-48-cards-8-of-which-136256.html#p1130713
Quote:
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)
\(= \frac{(40/48)*(39/47)*(8/46)}{(40/48)*(39/47)} = \frac{8}{46}\)
24 Apr 2016, 20:43
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truongynhi
This balls question is exactly the same as the modified question I asked and answered here: in-a-certain-baord-game-a-stack-of-48-cards-8-of-which-136256.html#p1130713
The modified question was this:
"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)."
The probability for successive drawing do not change in both these questions.
But our original question has conditional probability. It is different. We already know the first two picks. We have more information now. We know that two non share cards are out of the pool. Hence the probability of getting a share card increases to 8/46 from 8/48 (the probability if we don't know what the first two picks are).
24 Apr 2016, 21:05
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truongynhi
Hi truongynhi,
you are absolutely correct..
the probability of any card let it be 10th or 3rd or 48th will remain same as we are not aware what has happened prior to that..
