Cards - Probability

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.

You are cool !!! Both the answers are right.

Is there any other approach method?

Another approach:

Same color:

P(BB) + P(RR)

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

= 1/2(2 * 17/35)

= 17/35

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.

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

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