session7

35052 00875 pb0g3p3 girls and total of 3 childrenp3

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ave 1, 35 percent have 2, and 30 percent have 3; and 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 B, the number of boys, and G, the number of girls, in this family will have the joint probability mass function shown below. Table 6.2 in Ross, p. 261 P(B=i, G=j) G=0 G=1 G=2 G=3 B=0 0.15 0.1 0.0875 0.0375 B=1 0.1 0.175 0.1125 0 B=2 0.0875 0.1125 0 0 B=3 0.0375 0 0 0 3 P(B=0,G=0)= P(no children) =0.15 P(B=0, G=1) = P(1 girl and total of 1 child)= P(one child)P(one girl |one child)=(0.2)(0.5) = 0.1 P(B=0, G=2)= P(2 girls and total of two children)=P(2 children)P(2 girls |2 children)=(0.35)(0.52) =0.0875 P(B=0,G=3)=P(3 girls and total of 3 children)=P(3 children)P(3 girls |3 children)=(0.3)(0.53)=0.0375 P(B=1,G=1)=P(1 boy, 1 girl and 2 children)=P(2 children)P(boy and a girl | 2 children)= (0.35)(2/4)=0.175 (remember BB, GG, BG, GB are the possibilities ) P(B=2,G=1)= P(2 boys and 1 girls and 3 children)=P(3 children)P(2 boys and 1 girl | 3 ⎛ྎ 3⎞ྏ⎛ྎ 1 ⎞ྏ 2 ⎛ྎ 1 ⎞ྏ 0.3 * ⎜ྎ ⎟ྏ⎜ྎ ⎟ྏ ⎜ྎ ⎟ྏ =0.1125 children)= ⎝ྎ 2⎠ྏ⎝ྎ 2 ⎠ྏ ⎝ྎ 2 ⎠ྏ Table 6.2 in Ross, p. 261 P(B=i, G=j) G=0 G=1 G=2 G=3 € B=0 0.15 0.1 0.0875 0.0375 B=1 0.1 0.175 0.1125 0 B=2 0.0875 0.1125 0 0 B=3 0.0375 0 0 0 We can answer many quest...
View Full Document

This note was uploaded on 03/04/2014 for the course STAT 100A taught by Professor Wu during the Winter '10 term at UCLA.

Ask a homework question - tutors are online