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Intermediate Probability

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Intermediate Probability

by Aman verma » Mon Feb 22, 2010 6:38 am
Q: A and B pick up a card at random from a well shuffled pack of cards , one after the other , replacing it every time till one of them gets a heart . If A begins the game, then the probability that the game ends with B is

a) 3/7

b)4/7

c)3/4

d)1/4

e)2/7

Ans[spoiler]a) 3/7[/spoiler]
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by harsh.champ » Mon Feb 22, 2010 6:46 am
Aman verma wrote:Q: A and B pick up a card at random from a well shuffled pack of cards , one after the other , replacing it every time till one of them gets a heart . If A begins the game, then the probability that the game ends with B is

a) 3/7

b)4/7

c)3/4

d)1/4

e)2/7

Its imp. to get your thought process correct.Here it goes for me:-
At first glance I note that-"Now,here I will get an infinite series."
If in the 2nd chance ,the game ends :- P = 39/52[A chooese any of the other 39 cards) x 13/52(B chooses one of the 13 hearts) = 3/4 x 1/4 =3/16
If in the 4th chance,the game ends:-P = (39/52)^3 x 13/52
If in the 6th chance,the game ends:-P = (39/52)^5 x 13/52

So,we get the series as (39/52 x 13/52)*[ 1 + (39/52)^2 + (39/52)^4 +...]
=(3/4 x 1/4)*[ 1 + (3/4)^2 + (3/4)^4 +...]
Sum of an infinite series = 1st term/(1 - common ratio)
Hence the term in [ ........ ] would be 1/( 1 - 9/16)=16/7
[spoiler]Hene the answer is 3/16 x 16/7 =3/7 Hence A is the answer.[/spoiler]
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by gmatwriter » Mon Feb 22, 2010 12:20 pm
Hi,

Sorry can you explain how 9/16 is the common ratio? Shouldn't it be 3/16?

harsh.champ wrote:
Aman verma wrote:Q: A and B pick up a card at random from a well shuffled pack of cards , one after the other , replacing it every time till one of them gets a heart . If A begins the game, then the probability that the game ends with B is

a) 3/7

b)4/7

c)3/4

d)1/4

e)2/7

Its imp. to get your thought process correct.Here it goes for me:-
At first glance I note that-"Now,here I will get an infinite series."
If in the 2nd chance ,the game ends :- P = 39/52[A chooese any of the other 39 cards) x 13/52(B chooses one of the 13 hearts) = 3/4 x 1/4 =3/16
If in the 4th chance,the game ends:-P = (39/52)^3 x 13/52
If in the 6th chance,the game ends:-P = (39/52)^5 x 13/52

So,we get the series as (39/52 x 13/52)*[ 1 + (39/52)^2 + (39/52)^4 +...]
=(3/4 x 1/4)*[ 1 + (3/4)^2 + (3/4)^4 +...]
Sum of an infinite series = 1st term/(1 - common ratio)
Hence the term in [ ........ ] would be 1/( 1 - 9/16)=16/7
[spoiler]Hene the answer is 3/16 x 16/7 =3/7 Hence A is the answer.[/spoiler]
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by arzanr » Mon Feb 22, 2010 7:43 pm
If in the 2nd chance ,the game ends :- P = 39/52[A chooese any of the other 39 cards) x 13/52(B chooses one of the 13 hearts) = 3/4 x 1/4 =3/16
How do you get (39/52)*(13/52)?

Shouldn't it be (39/52)*(13/51) because when B chooses one of the 13 hearts, A has already picked one and therefore, there are only 51 cards left.

Similarly when analyzing longer sequences I'm not sure if you can raise (39/52) n times since after every turn, the fraction would change as follows:

1. 39/52 38/51
2. 37/50 13/49

and so on...
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