- In a card game that we typically know, there are 52 cards that make a deck in which there are 4 queens.You have drawn 2 cards: first one, then another. Since it is not explicitly mentioned whether you have replaced the 1st card before drawing the 2nd card, it is appropriate to assume that you have not replaced the 1st card.Now you are looking for scenarios: EITHER 1. (1st card is
**queen**AND 2nd card is**queen)**, OR 2. (1st card is& 2nd card is*not*queen**queen**)Probability of scenario 1 is (4/52)*(3/51) = 12/(52.51)Probability of scenario 2 is (48/52)*(4/51) = 192/(52.51)Total probability = 12/(52.51) + 192/(52.51) = 204/(52.51) = 1/13Regards,S. M. Mamun Ar Rashid

**From:**ahsabhasan <ahsabhasan@...>**To:**math_club@yahoogroups.com**Sent:**Tue, May 31, 2011 6:31:04 PM**Subject:**[math_club] A problem

IF I take two cards from a deck then what is the probability of the second card to be a queen.

- Interestingly, the probability is the same as that of drawing one card (out of the 52) and finding a queen.Murshid

On Wed, Jun 1, 2011 at 4:29 PM, S. M. Mamun Ar Rashid <mamun305@...> wrote:

In a card game that we typically know, there are 52 cards that make a deck in which there are 4 queens.You have drawn 2 cards: first one, then another. Since it is not explicitly mentioned whether you have replaced the 1st card before drawing the 2nd card, it is appropriate to assume that you have not replaced the 1st card.Now you are looking for scenarios: EITHER 1. (1st card is**queen**AND 2nd card is**queen)**, OR 2. (1st card is& 2nd card is*not*queen**queen**)Probability of scenario 1 is (4/52)*(3/51) = 12/(52.51)Probability of scenario 2 is (48/52)*(4/51) = 192/(52.51)Total probability = 12/(52.51) + 192/(52.51) = 204/(52.51) = 1/13Regards,S. M. Mamun Ar Rashid

**From:**ahsabhasan <ahsabhasan@...>**To:**math_club@yahoogroups.com**Sent:**Tue, May 31, 2011 6:31:04 PM**Subject:**[math_club] A problem

IF I take two cards from a deck then what is the probability of the second card to be a queen.