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Probability and a Bag of Cards

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Probability and a Bag of Cards

by dtweah » Sat May 02, 2009 9:55 am
A bag has cards with 3 questions written on each card. There are 20 cards in the bag, 8 of which are favorable ( you know the answer to all 3 questions on each of the favorable cards). If You answer all three questions you get an A. If cards are drawn at random from the bag, what is the probability you get an A on the second draw?

A. 1/4

B. 1/3

C. 1/5

D. 2/5

E. 1/2
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by DeepakR » Sat May 02, 2009 10:23 am
We have Total 20 cards of which 8 are favourable hence 12 are unfavorable.

1st card is an unfavorable card = P(Unfav)=12C1/20C1=12/20
2nd card is favorable = P(Fav)=8C1/20C1=8/20.

Hence P=(12/20)*(8/20)=6/30=1/5 = C.) is it correct ?

-Deepak
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by dumb.doofus » Sat May 02, 2009 12:40 pm
I think its D i.e. 2/5 Here is how I solved it..

We have to find probability of getting A on the second draw.. this leads us to two conditions
1. Getting A on both draws
Probability of that is = (8/20)*(7/19) --- (1)

2. Getting A only on 2nd draw
Probability of that is = (12/20)*(8/19) --- (2)

So total probability = (1) + (2)
= 152/380
= 2/5
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by Brent@GMATPrepNow » Sat May 02, 2009 2:52 pm
dumb.doofus wrote:I think its D i.e. 2/5 Here is how I solved it..

We have to find probability of getting A on the second draw.. this leads us to two conditions
1. Getting A on both draws
Probability of that is = (8/20)*(7/19) --- (1)

2. Getting A only on 2nd draw
Probability of that is = (12/20)*(8/19) --- (2)

So total probability = (1) + (2)
= 152/380
= 2/5
This strategy is 100% accurate. However, what would we do if the question asked us to find P(getting an A on 13th card). The calculations would get pretty complicated.

Notice that P(getting an A on 2nd card) = P(getting an A on 1st card)
We have 20 cards and 8 of them are "A" cards. So, P(getting an A on 1st card) = 8/20 = 2/5

The point I wish to make is that P(getting an A on 1st card) = P(getting an A on 2nd card) = P(getting an A on 3rd card) etc.

To illustrate my point, imagine the following scenario:
Someone shuffles the deck of cards and selects a card. Now you enter the room.
The person asks you, "What is the probability that this card is an A card?"
Does it matter if the card in question was the first card drawn? No.
The probability will be the same regardless of whether the card is the first drawn or the last drawn.
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by dumb.doofus » Sat May 02, 2009 3:03 pm
Brent Hanneson wrote:
dumb.doofus wrote:I think its D i.e. 2/5 Here is how I solved it..

We have to find probability of getting A on the second draw.. this leads us to two conditions
1. Getting A on both draws
Probability of that is = (8/20)*(7/19) --- (1)

2. Getting A only on 2nd draw
Probability of that is = (12/20)*(8/19) --- (2)

So total probability = (1) + (2)
= 152/380
= 2/5
This strategy is 100% accurate. However, what would we do if the question asked us to find P(getting an A on 13th card). The calculations would get pretty complicated.

Notice that P(getting an A on 2nd card) = P(getting an A on 1st card)
We have 20 cards and 8 of them are "A" cards. So, P(getting an A on 1st card) = 8/20 = 2/5

The point I wish to make is that P(getting an A on 1st card) = P(getting an A on 2nd card) = P(getting an A on 3rd card) etc.

To illustrate my point, imagine the following scenario:
Someone shuffles the deck of cards and selects a card. Now you enter the room.
The person asks you, "What is the probability that this card is an A card?"
Does it matter if the card in question was the first card drawn? No.
The probability will be the same regardless of whether the card is the first drawn or the last drawn.
Brent I have my doubts about that.. if I was replacing the card back in the deck then what you have mentioned would hold good. But if I am not replacing the card back in the deck, which is what I have done in my solution, then the probability will always change..

Taking your example, if I have to find the probability on the 13th card out of the 20 cards..
then as you can see.. if the first 12 were not A cards.. probability of getting an A card on the 13th try is 1, whereas it won't be 1 if in the 12 trys that I had, some times I did get an A card..
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by Brent@GMATPrepNow » Sat May 02, 2009 3:19 pm
Okay, let me ask a slightly different question.
Let's say only one of the 20 cards is an "A card" and we draw each card without replacement.

What is P(1st card drawn is A card)?
We get 1/20

What is P(2nd card drawn is A card)?
We get (19/20)(1/19)=1/20

What is P(3rd card drawn is A card)?
We get (19/20)(18/19)(1/18)=1/20

Notice that numerators and denominators keep cancelling out. If you try the same calculations on the original question, you will keep getting 2/5
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by Brent@GMATPrepNow » Sat May 02, 2009 3:23 pm
Here's a slightly different angle.
We have the original 20 cards and 8 of them are "A cards"
You enter the room and one card has been taken from the deck and is face down on the table.
Someone asks you to determine P(the card is an A card). Do we really need to know whether this card is the 1st, 2nd, 3rd or last card for us to be able to determine the probability?
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by dumb.doofus » Sat May 02, 2009 3:29 pm
Hmmmm.. I am getting your point Brent.. Awesome.. thanks so much.. you are absolutely right..
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by dtweah » Sat May 02, 2009 5:59 pm
dumb.doofus wrote:I think its D i.e. 2/5 Here is how I solved it..

We have to find probability of getting A on the second draw.. this leads us to two conditions
1. Getting A on both draws
Probability of that is = (8/20)*(7/19) --- (1)

2. Getting A only on 2nd draw
Probability of that is = (12/20)*(8/19) --- (2)

So total probability = (1) + (2)
= 152/380
= 2/5
Good Job Dumb. Keep nailing them probability. OA is 2/5.
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by DeepakR » Sat May 02, 2009 6:17 pm
Thanks a lot Dumb. Your answers are very clear and helps a lot. I have a doubt in this question. If nothing is specified do we need to consider the question as without replacement. ?

Thanks Brent that was a great tip.

- Deepak
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by dumb.doofus » Sun May 03, 2009 12:19 am
Thanks buddies..

Well, with or without replacement are kind of very clear from the words that are used in the question..

For ex: in this question, it says, cards are drawn at random from the bag.. essentially states that you are taking out cards from the bag and not keeping them back in..

Some questions very specifically mention with or without replacement.. and some do not.. but the wordings used in a question gives it out..

HTH..
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