In a certain baord game, a stack of 48 cards, 8 of which

Yes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.

The probability of picking a stock card keeps increasing as we keep pulling out the non stock cards.
Think: You have 5 cards and 1 of them is the Queen. What is the probability that you will pull out a queen? It's 1/5. What happens after you pull out a non queen? The probability of pulling out a queen now becomes 1/4. It keeps increasing till you reach the last card. If you still haven't pulled out a queen, the last one must be the queen and the probability becomes 1.

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).
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Hi,
you have given a Q above wherein there are 3 blacks and 5 white, and we are to find the probability that 4th is black..
here you are not aware what has happened in the first three DRAWS, so picking of all 8 will remain PROBABILITY..

But what happens when you have picked two cards and you are told they do not contain the card we are lookin for..
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..

If you are told first two are not type X...
so now you have ONLY 46 cards left, Because in the TWO picked, there is no probability involved since we know what those cards are..
Probability is ONLY there where we do not know the out come..

another extension of above Q..
you have picked 40 cards and none of them are type X..
what is the probability that 41st card will be X..
we now do not care of what we already know. they have moved from Probability to REALITY..
now we know 48-40=8 cards are left and none of type X has been picked..
so now all these 8 will be type X..
prob = 8/8 = 1.. that is it will be type X for sure..
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Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?

My answer is E.
We can substract 2 from 48 and find the probability for the third card.

What exactly is your doubt.

Hello Karishma - I guess we can list them out?

Pick Share : Pick share : pick share =6/46
Pick share : No pick Share : pick share = 7/46
No pick Share : pick : pick = 7/46
No : No : pick = 8/46

add them all to get the overall probability?

You can list it out though you don't need to. The probability will stay 8/48. It doesn't matter whether it is the first pick, the second or the third or a later pick. I have discussed this here: a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html?hilit=probability%20same

When you list it out, you will obviously get the same answer.

Share, Share, Share = (8/48) * (7/47) * (6/46)
Share, No Share, Share = (8/48) * (40/47) * (7/46)
No Share, Share, Share = (40/48) * (8/47) * (7/46)
No Share, No Share, Share = (40/48) * (39/47) * (8/46)
When you add all these up, you get 8/48.
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Will we not change 8 to 6??

because if we have taken out those 2 cards..then ramining will be 6
6/46?
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Read the question carefully. These 2 cards dont represent the stocks. Hence the 8 cards that represented stocks are still there, i.e. those 8 cards are still intact. But total number of cards has reduced from 48 to 46. Hence, probability can be found easily.
Hope that helps
-s
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i have a big trouble ..i think it should be done in this way. we select 2 from the 40 , in the first beginning. then select 1 from the 8 , in total we have 3 from 48.
so what I got was C1 8C2 40/C3 48

40/48 * 39/47
so next is = 8/46 = 4/23

well done! I got it, meanwhile, I found my problem. what I did do not follow the first 2 cards and the third. what i did means just select cards randomly , two are PPPPPP, one is PPPPP...
thank you very much

Hi,
there are 48 cards , out of which 8 are shares of stack..
the first two are not share of stack..
so the remaining cards now are 48-2=46, again out of which there are 8 shares of stock..
the prob that next is one of these 8 is 8/46=4/23
E
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So when do you use combinatorics? I thought the solution would work out as:

Probability of third card being a stock card = 40/48* 39/47*8/46

When to use combinatorics???

Hi,
the first two not being a stock card is no more a probability but a fact..
there is some action carried out and you have to work further to it..
out of 48 cards, you have already picked up two cards.. so the present case is that you are left with 46 cards and the prob will depend on these cards now..
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Hi Karishma, can you please give the answer to the above query?

I have 2 other questions.

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)
\(=\frac{40*39*8}{40*47*46}\)
\(=\frac{8}{46}=\frac{4}{23}\)


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!

Hi Chetan2u,

A silly question: How can we differentiate between a fact and a probability?

I am very confused maybe because I am not familiar with the wording. I thought the first two cards constitute a probability. Could you please give some wording examples for both 'a fact' and 'a probability' types of question? Could you please help me understand the difference between the two?

Always appreciate your help!

Hi,
Thank you! I think I understand this one.



I am a bit confused here. If I am told that the two cards do not contain the card I'm looking for, then how come the probability continues from 48? Doen't it start from the remaining 46 cards?



Doesn't this contradict with your above statement? You said we start from 48 after the first two cards. But here you point out that we calculate on the remaining 46 cards. Is it a typo or something? Please clarify.

Thank you very much!

pl read the highlighted portion as -- and you are not aware what those two cards contain..
It was a typo..


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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!