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Cards - Probability

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by Vemuri » Thu Mar 05, 2009 6:02 am
Two cards are drawn from a well-shuffled pack of cards. Given that both the cards are numbered cards, what is the probability that the cards have

(i) Same color on them?
A) 17/35
B) 17/70
C) 34/35
D) 1/35
E) 18/35

(ii) same number & different colors on them?
A) 1/70
B) 1/35
C) 1/26
D) 3/35
E) 2/35
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by Feep » Thu Mar 05, 2009 1:15 pm
We know that a standard deck of cards has 52 cards in it, but we must remove all non-numbered cards (J, Q, K, A) for the purposes of this problem. There are four suits of each of those cards, so a total of sixteen we must remove, leaving a deck of 36 cards.

In the first problem, we draw the first card, which of course can be any card, leaving 35 cards. Of those 35, only 17 remain that are the same color as the first one you drew. So, 17/35 is the correct answer.

For the second, you draw the first card, which can be any card. Of those remaining, only three are the same number (the other three suits), and two of those three are of a different color than the one you drew. So, the answer is 2/35.
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by Vemuri » Thu Mar 05, 2009 7:41 pm
You are cool !!! Both the answers are right.

Is there any other approach method?
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by cramya » Thu Mar 05, 2009 8:40 pm
Another approach:

Same color:

P(BB) + P(RR)

= 18/36*17/35+18/36*17/35

= 1/2(2 * 17/35)

= 17/35
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by Vemuri » Sat Mar 07, 2009 3:16 am
cramya wrote:Another approach:

Same color:

P(BB) + P(RR)

= 18/36*17/35+18/36*17/35

= 1/2(2 * 17/35)

= 17/35
Thank you. This is a simple & clear approach.
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by devesh99 » Mon Mar 09, 2009 6:12 pm
Hi

Cramya .. can you answer the second part of the question??

Feep .. What did u take in the denominator. I understood the logic for numerator i.e. 9*2 but Denominator should be 36*35 rite??

Thanks
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by devesh99 » Mon Mar 09, 2009 6:13 pm
Hi

Cramya .. can you answer the second part of the question??

Feep .. What did u take in the denominator. I understood the logic for numerator i.e. 9*2 but Denominator should be 36*35 rite??

Thanks
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