Three groups of children contain respectively 3 girls and 1 boy; 2 girls and 2 boys; 1 girl and 2 boys. One child is selected at random from each group. Show that the probability that the three selected consist of 1 girl and 2 boys is $3 over 8.$

Solution

Group I Group II Group III
Boys 1 2 2
Girls 3 2 1
P (1 girl and two boys) = P(GBB) + P(BGB) + P(BBG)
= P(G) P(B) P(B)+ P(B) P(G) P(B) + P(B) P(B) P(G)
$equals space 3 over 4 cross times 2 over 4 cross times 2 over 3 plus 1 fourth cross times 2 over 4 plus 2 over 3 cross times 1 fourth cross times 2 over 4 cross times 1 third space equals space 12 over 48 plus 4 over 48 plus 2 over 48 space equals space 18 over 48 space equals space 3 over 8$

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Probability

Three groups of children contain respectively 3 girls and 1 boy; 2 girls and 2 boys; 1 girl and 2 boys. One child is selected at random from each group. Show that the probability that the three selected consist of 1 girl and 2 boys is $3 over 8.$

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