Question

(6) Suppose there is a special trait amongst humans which occurs with proba- bility p. Suppose a couple have two chil (a) Find the probability that they have two girls, given that at least one of dren their children is a girl with that special trait with a y'' number 123-45-6789" (b) Compute this probability if the special trait is "born on a day ending (c) Approximate this probability if the special trait is "has social security

Answer 1

(a)

It is a question of conditional probability ,

P(G) = 1/2, (probability of being a girl, assumption here is there can be only 2 outcomes either girl or boy)

P(B) =1/2 ( probability of being a boy)

given information :

p - special traits among human

probability of a girl with special traits P(Gs) = probability of special trait * probability being a girl = p * 1/2

probability of a boy with special traits P(Bs) = probability of special trait * probability being a boy =  p * 1/2

Question is probability of being both girls, given the information that at-least one of two children is a girl with special ability.

P( Both girl / one of the children is girl with special ability) = P(Both girl  $\cap$ one of the children is girl with special ability) / P( one of the children is girl with special ability)

P(one of the children is girl with special ability) = P(Gs)P(B) + P(Gs)P(Bs) + P(Gs)P(G) + P(Gs)P(Gs) = p/2*1/2+p/2*p/2+p/2*1/2 + p/2*p/2 = (2p + 2p*p)/4 = 2p(1+p)/4 = p(1+p)/2

P(Both girl  $\cap$ one of the children is girl with special ability) = P(Gs)P(G) + P(Gs)P(Gs) = p/2*1/2+p/2*p/2 = p(1+p)/4

P(Both girl  $\cap$ one of the children is girl with special ability) / P( one of the children is girl with special ability) = (p(1+p)/4) /

(p(1+p)/2) = 1/2 = .5