Question

6. (a).

In a particular town 10% of the families have no children, 30% have one child, 20% have
two children, 40% have three children, and 0% have four. Let T represent the total
number of children, and G the number of girls, in a family chosen at random from this
town. Assuming that children are equally likely to be boys or girls, find the distribution
of G. Display your answer in a table and sketch the histogram.

(b). Find E(T | G=1) = conditional expectation of number of children T, given 1 girl.

(c). Find the sum over k= 0, , 2, 3 of

E (T | G=k) P( G= k).

HINT: The hard way is to compute both factors of all 4 terms and do the arithmetic. The easy way is to use the R.A.C.E.

Answer 1

a) Let T : Number of child

The probability mass function of T is

T	0	1	2	3
P(T=t)	0.10	0.30	0.20	0.40

G : Number of girls in family

G take values 0,1,2 and 3

p =P ( A child is a girl) = 0.5

Since Conditional distribution of G given T=t is binomial with parameter n = 3 and p = 0.5

G / T=t ~ Bin ( n=t, p = 0.5)

Hence Joint probability distribution of G and T can be written as

The joint probability table of G and T is

		T				Total
		0	1	2	3
G	0	0.1	0.15	0.05	0.05	0.35
	1	0	0.15	0.1	0.15	0.4
	2	0	0	0.05	0.15	0.2
	3	0	0	0	0.05	0.05
Total		0.1	0.3	0.2	0.4	1

Hence probability distribution of G is

G	0	1	2	3
P(G=g)	0.35	0.40	0.20	0.05

To draw histogram, we use R

By using R

> class=seq(-0.5,3.5,1)
> g=0:3
> f=c(35,40,20,5)
> y1=rep(g,f)
> t=0:3
> f1=c(10,30,20,40)
> y2=rep(t,f1)
> par(mfrow=c(1,2))
> hist(y1,breaks=class,main="Number of girls")
> hist(y2,breaks=class,main="Total Number of Childs")

Histogram

b) The conditional distribution T given G =1 is

T	0	1	2	3
P(T/G=1)	0	0.375	0.25	0.375
T* P(T/G=1)	0	0.375	0.50	1.125

E( T / G=1) = 0 + 0.375 + 0.50 + 1.125 = 2

c) Conditional distribution of T given G =0 is

T	0	1	2	3
P(T/G=0)	2/7	3/7	1/7	1/7
T* P(T/G=1)	0	3/7	2/7	3/7

E(T /G=0) = 9/7 = 1.2857

Conditional distribution of T given G=2

T	0	1	2	3
P(T/G=1)	0	0	0.25	0.75
T* P(T/G=1)	0	0	0.50	2.25

E(T / G=2) = 2.75

Conditional distribution of T given G =3

T	0	1	2	3
P(T/G=1)	0	0	0	1
T* P(T/G=1)	0	0	0	3

E(T / G=3) = 3

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