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1) A total of 46% of the voters in a certain city classify themselves as Independents, where as 30% classify themselves as Liberals and 24% say they are Conservatives. In s recent local election, 35% of the Independents, 62% of the Liberals and 58% of the Conservatives voted. A voter is chosen at random. Given that this person voted in the local election, the probability that he or she is an Independent is _________ ans  0.331

2) A family has children with probability P_i where P1=0.01,P2=0.25,P3=0.35,P4=0.3. A child from this family is randomly chosen. Given that this child is the eldest child in the family, the conditional probability that the family has 4 children _______.ans  0.18

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Given ,

P(Independent)   =  0.46   =   r (say)

P(Liberal)   =   0.30       =    s

P(Conservative)   =  0.24   =  t

P(Vote | Independent)  =  0.35   =   u

P(Vote | Liberal)   =  0.62         =   v

P(Vote | Conservative)  =  0.58   =  w

So

P(Independent | Vote)  =   ru / ( ru + sv + tw )

=   (0.46 * 0.35) / [(0.46 * 0.35) + (0.3 * 0.62) + (0.24 * 0.58)]

=    0.331

Hence the answer to the first question is 0.331

The second question has a bit lengthy calculation actually..

selected

II)P1=0.01,P2=0.25,P3=0.35,P4=0.3.

P(E|P1)=1,P(E|P2)=0.5 , P(E|P3)=0.33,P(E|P4)=0.25.

$P(P4|E) = \frac{P(E|P4) * P(P4))}{ P(E|P3) * P(P3) +P(E|P2) * P(P2)+P(E|P1) * P(P1)+P(E|P4) * P(P4)}$

= 0.075 / 0.3255= 0.230

please explain the 2nd Q ..i have difficulty in understanding the Questn
P1 =>family has 1 child,P2=>family has 2 child same for P3,P4

P(E|P1) = eldest from family P1 = 1 ( only one child no comparision)

P(E|P2) = eldest from family P2 = 0.5 (  two child ,one is eldest and other young )

P(E|P3) = eldest from family P3 = 0.33 (  three child ,one is eldest and two  young )

P(E|P4) = eldest from family P4  = 0.25 (  four child ,one is eldest and three young )