Answer:
see below
Explanation:
The probability a member is a girl is independent of the probability that a member plays sports if the fraction of members that are girls does not depend on whether the member plays sports or not.
That is, the fraction of members that are girls will be the same for sports players and non-sports players.
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Examples
(a) Suppose we have a membership that consists of ...
1 girl and 1 boy who play sports (as many girls as boys), and 1 girl who does not play sports.
We want to see if p(g) and p(s) are independent, so we compute p(g) and p(g|s).
p(g) = 2/(2+1) = 2/3
p(g|s) = p(g&s)/p(s) = (1/3)/(2/3) = 1/2 . . . . not independent
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(b) Suppose we have a membership that consists of ...
1 girl and 1 boy who play sports, and 2 girls and 2 boys who don't play sports (girls are half the members who play, and half the members who don't play).
p(g) = 3/6 = 1/2
p(g|s) = p(g&s)/p(s) = (2/6)/(4/6) = 2/4 = 1/2 . . . . independent
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(c) Suppose we have a membership that consists of ...
2 girls and 1 boy who play sports, and 2 boys and 1 girl who do not play sports (fraction of girl players = fraction of boy non-players)
p(g) = 3/6 = 1/2
p(g|s) = p(g&s)/p(s) = (2/6)/(3/6) = 2/3 . . . . not independent
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(d) Suppose we have a membership that consists of ...
1 girl and 1 boy who play sports, and 1 girl who does not play sports (as many girls play as do not)
p(g) = 2/3
p(g|s) = p(g&s)/p(s) = (1/6)/(2/6) = 1/2 . . . . not independent
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These examples match our selection of the 2nd choice as being the one that demonstrates independence.