Three groups of children contain 3 girls and 1 boy; 2 girls and 2 boys;1 girl and 3 boys respectively. One child is selected at random from each group.Find the chance that the three selected comprise one girkland 2 boys.
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Let `G_(1), G_(2), G_(3)` be the events of selecting a girl from the first, second and third group respectively, and let `B_(1), B_(2), B_(3)` be the events of selecting a boy from the first, second and third group respectively. Then,
`P(G_(1))=3/4, P(G_(2))=2/4=1/2, P(G_(3))=1/4`.
`P(B_(1))=1/4, P(B_(2))=2/4=1/2` and `P(B_(3))=3/4`
`:.` P(selecting 1 girl and 2 boys)
`=P[(G_(1)B_(2)B_(3)) or (B_(1)G_(2)B_(3)) or (B_(1)B_(2)G_(3))]`
`=P(G_(1)B_(2)B_(3))+P(B_(1)G_(2)B_(3))+P(B_(1)B_(2)G_(3))`
`={P(G_(1))xxP(B_(2))xxP(B_(3))}+{P(B_(1))xxP(G_(2))xxP(B_(3))}+{P(B_(1))xxP(B_(2))xxP(G_(3))}`
`=(3/4xx1/2xx3/4)+(1/4xx1/2xx3/4)+(1/4xx1/2xx1/4)=(9/32+3/32+1/32)=13/32`.
Hence, the chances of selecting 1 girl and 2 boys are `13/32`.