Question

Problem 7. (Graded, 10 points) Assume that identical twins are always of the same sex, equally likely boys or girls. Assume that for fraternal twins, the firstborn is equally likely to be a boy or a girl, and so is the secondborn, independently of the first. Assume that proportion p of twins are identical, and q = 1-p of twins are fraternal. Find formulae in terms of p for the following probabilities: a) P(both boys) b) P(firstborn boy and secondborn girl) c) P(secondborn girlfirstborn boy) d) P(secondborn girlfirstborn girl)

Answer 1

a)

Our sample space is the space of twins, which is partitioned by the sets I = {Identical twins} and F = {Fraternal twins}. Let BB = {both boys}.

Then P(BB) = P(BB ∩ I) + P(BB ∩ F)

= P(BB|i)P(I) + P(BB|F)P(F) = 0.5p + 0.25(1 − p)

$=\frac{1+p}{4}$

b)

Let BG =firstborn boy, a secondborn girl.

Since identical twins are always of the same gender,

P(BG ∩ I) = 0.

We have P(BG) = P(BG ∩ I) + P(BG ∩ F)

= 0 + P(BG|F)P(F)

= 0.25(1 − p) = (1 − p)/4

c)

We need to first calculate P(firstborn boy).

Let 1B=firstborn boy.

We have P(1B) = P(1B ∩ I) + P(1B ∩ F)

= P(1B|I)P(I) + P(1B|F)P(F)

= 0.5p + 0.5(1 − p) = 1/2

Let 2G = secondborn girl.

We have P(2G ∩ 1B) = P(2G ∩ 1B|F)P(F) + P(2G ∩ 1B|I)P(I)

= 0.25(1 − p) + 0p = 0.25(1 − p)

and therefore,

P(2G|1B) = P(2G ∩ 1B) P(1B)

= 0.25(1 − p) 0.5 = (1 − p)/2

d)

P(2G|1G) = P(2G ∩ 1G)/P(2G)

P(2G) = P(2G|I)P(I) + P(2G|F)P(F)

= 0.5p + 0.5(1 − p)

= 0.5

P(2G ∩ 1G) = P(2G ∩ 1G|I)P(I) + P(2G ∩ 1G|F)P(F)

= 0.5p + 0.25(1 − p)

= 0.25 + 0.25p

⇒ P(2G|1G) = 0.25 + 0.25p 0.5

= $\boldsymbol{\frac{1+p}{2}}$