probabilty

[madhur1987]

in many countries around the world couple look to a son to take care of them in their old age. they are inclined to keep having children until they have son.for this problem imagine that govt considers permitting a couple to continue having children until they have a son.assume that birth of amale is a likely as the birth of female a)what is the prob that family would have two children( oldest a girl then a boy) b)what would be the avg no of children per family?

[madhur1987]

a deck of throughly shuffled cards is dealt out of 4 players so that each of them receive 13 cards. what is the prob that each player will receive exactly one ace? ans in d book is 1/4 but i m not getting dis ans

[Sumit]

Sample space=(G,GB,GGB,GGGB......)...where P(G)=1/2, P(GB)=1/4 & P(GGB)=1/8 n so on
This is an GP whose some is=1

Now, A).P(family would have two children)=P(GB)=1/4.

Now Take X=no. of child a family have
So,X 1         2        3       4        5     so on
 P(x) 1/2    1/4    1/8   1/16      1/32 so on

For B) we need to find E(X)=1/2+2/4+3/8+4/16....=2.

[Devika]

Hey, how did you take out the sum of this series? :s

Multiply and then subtract?

[Sinistral]

for these kinds of questions, imagine that you spread all the cards linearly on a table. now the first player will get first set of 13 cards, 2nd will get 2nd set of 13 cards, 3rd will get 3rd set of 13 cards and 4th will get the last remaining set of 13 cards.

now we can choose 1 ace out of 4 in 4C1 ways. first player to have exact one ace implies we can put this chosen ace in the first lot of 13 cards at 13 different places. so total count = 4C1*13

similarly for 2nd player to get exactly one ace we can choose 1 ace out of remaing 3 aces and put it in 13 different places (for the 2nd player) ie 3C1*13

similarly for 3rd player: 2C1 *13

4th player 1C1 *13

multiplying the above 4 wil give the numerator of our probability.

to find denominator: Now if no restriction were there we could have placed those 4 aces anywhere in those 52 cards. ie 52C4* 4!

                                     (4C1*13)( 3C1*13)(2C1 *13)(1C1 *13)
  hence required prob =    ---------------------------------------
                                                 52C4* 4!

[Sinistral]

thats an AGP.

[Devika]

Oh yes! Thanks :)

[madhur1987]

thanx a lot guys:)

[Ayushya Kaul]

Shouldn't E(X)=3?

[duck]

Hey Ayushya.. :)
As Sumit pointed out:
E(X) = 1*1/2+2*(1/2)^2+3*(1/2)^3+........   (I)
Mulitply both the sides by 1/2, you'll get:
1/2E(X)= 1*(1/2)^2+2*(1/2)^3+..............   (II)
Subtracting (II) from (I), you'll get:

1/2E(X) = 1/2+(1/2)^2+(1/2)^3+....
Therefore. E(X) =2.