Question

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 girkl and 2 boys. Select the correct answer from above options

Answer 1

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`.