Problem 3. Suppose that 15 percent of the families in a certain community have n

Question

Problem 3. Suppose that 15 percent of the families in a certain community have no children, 20 percent have 1 child, 35 percent have 2 children, and 30 percent have 3. Suppose further that in each family each child is equally likely (independently) to be a boy or a girl. If a family is chosen at random from this community, then writing B, the number of boys, and G, the number of girls, in this family, determine a) the joint probability mass function of B and G; b) the marginal probability mass functions of B and G c) whether B and G independent?

Explanation / Answer

here P(B=0;G=0)=P(no children family)*P(no boy and no girl|no children family)+P(1 children family)*P(1 boy and no girl|1 children family)+P(2 children family)*P(no boy and no girl|2 children family)+P(3 children family)*P(no boy and no girl|3 children family)

=0.15*1+0.2*0+0.35*0+0.3*0=0.15

similarly for other combinations :

a) joint probability mass function of B and G is as follows:

b)

marginal probability of B:

marginal probability of G:

c)

here asP(G=0;B=0) =0.15 ; while P(G=-0)*P(B=0)=0.3750*0.3750=0.140625

both are not equal therefore condition of independence P(A n B)=P(A)*P(B) is not satisfied

hence B and G are not independent.,

B G 0 1 2 3 Total 0 0.1500 0.1000 0.0875 0.0375 0.3750 1 0.1000 0.1750 0.1125 0.0000 0.3875 2 0.0875 0.1125 0.0000 0.0000 0.2000 3 0.0375 0.0000 0.0000 0.0000 0.0375 Total 0.3750 0.3875 0.2000 0.0375 1.0000

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