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In a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?
A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23
A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23
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rajman41
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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truongynhi
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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General Discussion
rajman41
Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?
VeritasPrepKarishma
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?
Jp27
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?
because if we have taken out those 2 cards..then ramining will be 6
6/46?
Marcab
Joined: 03 Feb 2011
Last visit: 22 Jan 2021
Posts: 834
Status:Retaking after 7 years
Location: United States (NY)
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Schools: Simon '15 (M$)
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GPA: 3.75
sanjoo
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
rajman41
40/48 * 39/47
so next is = 8/46 = 4/23
amithyarli
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
_________________
VeritasPrepKarishma
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???
sarathvr
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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truongynhi
Joined: 05 Jun 2015
Last visit: 09 Jan 2025
Posts: 21
Location: Viet Nam
GMAT 1: 740 Q49 V41
GPA: 3.66
GMAT 1: 740 Q49 V41
Posts: 21
chetan2u
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!
truongynhi
Joined: 05 Jun 2015
Last visit: 09 Jan 2025
Posts: 21
Location: Viet Nam
GMAT 1: 740 Q49 V41
GPA: 3.66
GMAT 1: 740 Q49 V41
Posts: 21
chetan2u
Hi,
Thank you! I think I understand this one.
Quote:
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?
Quote:
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!
truongynhi
pl read the highlighted portion as -- and you are not aware what those two cards contain..
It was a typo..
Quote:
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truongynhi
Joined: 05 Jun 2015
Last visit: 09 Jan 2025
Posts: 21
Location: Viet Nam
GMAT 1: 740 Q49 V41
GPA: 3.66
GMAT 1: 740 Q49 V41
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chetan2u
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!
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)
\(= \frac{(40/48)*(39/47)*(8/46)}{(40/48)*(39/47)} = \frac{8}{46}\)
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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).
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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..
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longhaul123
Joined: 03 Jul 2017
Last visit: 11 Nov 2018
Posts: 137
Status:IF YOU CAN DREAM IT, YOU CAN DO IT
Location: India
Concentration: Finance, International Business
Posts: 137
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??
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