introduction to probability Joseph K. Blitzstein , Jessica Hwang A family has tw
ID: 3328365 • Letter: I
Question
introduction to probability Joseph K. Blitzstein , Jessica Hwang A family has two children. that cach child has characteristic C with probability p, independently of cach gender. For example, C could be the characteristic "born in winter as in Show that the probability that both children are girls given that at least on characteristic C is (which is 1/3 if p I et C be a characteristic that a child can have, and assume Example 2.2.7. e is a girl with Example - 1, agree ing with the first part of 2.2.5, and approaches 1/2 from below as p - 0, agreeing with Example 2.2.7.)Explanation / Answer
Let A = Probability that atleast one is a girl with characteristic C
Given, Probability that a child has characteristic C = p
We need to find the probability that both children are girls given that atleast one is a girl with characteristic C
Now, the possible cases for the two children are : {2G, 2B, 1G 1B, 1B 1G}
Let E = Probability that both children are girls = 1/4
Let B = Probability that both children are boys = 1/4
Let K = Probability that one is a girl and one is a boy = 2/4
Using Bayes' Theorem, we can write,
P(E/A) = [P(A/E) * P(E)] / [P(A/E) * P(E) + P(A/B) * P(B) + P(A/K) * P(K) ]
Now, P(A/E) = Probability that atleast one is girl with a characteristic C given both the children are girls
= 1 - Probability that neither girl has a characteristic C given both are girls
= 1 - (1-p)2
= 2p - p2
= p(2-p)
P(A/B) = Probability that atleast one is girl with a characteristic C given both the children are boys
= 0
P(A/K) = Probability that atleast one is girl with a characteristic C given there is one girl and one boy
= p (The only girl has characteristic C)
Therefore, P(E/A) = [p(2-p) * 1/4] / [p(2-p) * 1/4 + 0 + p * 2/4]
= [p(2-p)] / [p(2-p) + 2p]
= (2-p) / (4-p)
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