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# What is the probability of getting two cards belonging to

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Senior Manager
Joined: 09 Aug 2005
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What is the probability of getting two cards belonging to [#permalink]  25 Jun 2006, 06:08
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What is the probability of getting two cards belonging to two different suits when two cards are drawn at random from a pack of well shuffled cards comprising 52 cards?
a)1/13
b)13/17
c)13/51
d)26/51
e)7/13

how do you figure out that the two cards are drawn one after the other or simultaneously ?

the answers differ based on the assumption one makes.
VP
Joined: 25 Nov 2004
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Re: PS: probability interpretation [#permalink]  25 Jun 2006, 06:27
depends on whether the first is replaced or not. lests suppose the first card is drawan and was not replaced. then now the

P (x =getting two cards from the same suit) = 13/52 x 12/51 = 3/17
P (two cards belonging to two different suits) = 1 - 3/17 = 14/17. but its not the answer.

it is done differently.

P (of getting first card from any suit) = 52/52 = 1
P (of getting second card from any suits other than the first suit) = (51-12) / (51) = 39/51=13/17

So it should be B.
Manager
Joined: 01 Jun 2006
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the prob must be 1-4*13C2/52C2=13/17
Yup. B is the correct answer.
Senior Manager
Joined: 07 Jul 2005
Posts: 406
Location: Sunnyvale, CA
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(B)
Choose first card in 1 way.
2nd card in 39/51 = 13/17 ways.
Director
Joined: 28 Dec 2005
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Re: PS: probability interpretation [#permalink]  25 Jun 2006, 20:03
old_dream_1976 wrote:
What is the probability of getting two cards belonging to two different suits when two cards are drawn at random from a pack of well shuffled cards comprising 52 cards?
a)1/13
b)13/17
c)13/51
d)26/51
e)7/13

how do you figure out that the two cards are drawn one after the other or simultaneously ?

the answers differ based on the assumption one makes.

Lets solve it first for the case where the card is replaced:

For any one of the 4 types, P of getting the same type = 1/4 * 1/4 = 1/16.
since there are 4 card types, p = 4 * 1/16 = 1/4. Therefore, P of not getting the same type (which is what we want) is 1 -1/4 = 3/4.

Lets now solve it for the case where the card is not replaced.

P of the same card type (for each card type)= 1/4 * 12/51 = 1/17
Since there are 4 card types, P = 4 * 1/17 = 4/17
Thus, P that this is not the case = 1 -14/17 = 3/17, hence B
Re: PS: probability interpretation   [#permalink] 25 Jun 2006, 20:03
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