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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).

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)

Second question: I cannot tell the difference between the problem under discussion and this problem: A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?

In the latter problem, Bunuel and other experts confirmed that the probability of getting a certain ball (black ball) will not change for any successive drawing. This means the probability of getting a black ball remains 5/8 no matter what.

I thought the same line of thinking would apply here. On the third drawing, the probability of getting a share remains 8/48. I know it is dead wrong, but I cannot understand the difference between the two questions. Could you please clarify?

Thank you very much!

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).
_________________

But what happens when you have picked two cards andand you are not aware what those two cards contain.. you had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X.. Our probability will continue to be out of all 48..

So, in this case, the probability that the third card is of type X will equal \(\frac{8}{48}\). Is that right?

Thank you very much! You've been very helpful!

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..
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Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]

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16 Jul 2017, 07:36

there are total of 48 cards out of which 8 are stock cards so the probability of picking a card from the stock card is 8/48 =S and the other cards are 40/48=N therefore, NNS=5/6*4/5*1/6=1/9 where am i going wrong??

there are total of 48 cards out of which 8 are stock cards so the probability of picking a card from the stock card is 8/48 =S and the other cards are 40/48=N therefore, NNS=5/6*4/5*1/6=1/9 where am i going wrong??

The probability of picking a Non-stock card, Non-stock card and a Stock card is different from this question.

Here, you are given that the first two are non-stock cards so you do not need to account for the probability of those being non-stock. It is already known that they are.

Also, the probability of selecting NNS would be (40/48)*(39/47)*(8/46)
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Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]

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25 Jul 2017, 08:51

total possibilities 48 after elimnating 2 now 46 now what is the probability from these 46 on 3rd pick the card can be of stock that is 8/46=4/23 Option:E

Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]

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19 Oct 2017, 15:32

there is a total of 48 cards, 2 got picked and are not shares of stock, 48-2=46, 46 cards are remaining that may be picked and represent shares of stock. Therefore, the probability of the third one will be share of stock is 8/46=4/23