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35052 00875 pb0g3p3 girls and total of 3 childrenp3

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

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