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35052 00875 pb0g3p3 girls and total of 3 childrenp3

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Unformatted text preview: 0 0 0 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 ⎞ྏ children)= 0.3 * ⎜ྎ ⎟ྏ⎜ྎ ⎟ྏ ⎜ྎ ⎟ྏ =0.1125 ⎝ྎ 2⎠ྏ⎝ྎ 2 ⎠ྏ ⎝ྎ 2 ⎠ྏ We can answer many questions with this table. € (a) Find out the joint probability P(B<2, G<2) (b) Find the marginal distribution for the number of girls, marginal expectations and marginal variance of G. 3 6. Suppose that 3 balls are chosen without replacement from an urn consisting of 5 white and 8 red balls. Let X i equal 1 if the ith ball selected is white, and let it equal 0 otherwise. Give the joint probability mass function of (a) X 1 , X 2 ; (b) X 1 , X 2 , X 3 ; 4...
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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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