Three groups of children contain respectively 3 girls and 1 boy, 2 girls and 2 boys and 1 girl and 3 boys. One child is selected at random from each group. What is the chance that three selected consists of 1 girl and 2 boys?

#### Solution

Group-I |
Group-II |
Group-III |
|||

Girls | Boys | Girls | Boys | Girls | Boys |

3 | 1 | 2 | 2 | 1 | 3 |

Let G_{1}, G_{2}, G_{3} denote events for selecting a girl and B_{1}, B_{2}, B_{3} denote events for selecting a boy from 1^{st}, 2^{nd} and 3^{rd} groups respectively.

Then P(G_{1}) = `3/4`, P(G_{2}) = `2/4`, P(G_{3}) = `1/4`

P(B_{1}) = `1/4`, P(B_{2}) = `2/4`, P(B_{3}) = `3/4`

Where G_{1}, G_{2}, G_{3}, B_{1}, B_{2,} and B_{3} are mutually exclusive events.

Let E be the event that 1 girl and 2 boys are selected

∴ E = (G_{1} ∩ B_{2} ∩ B_{3}) ∪ (B_{1} ∩ G_{2} ∩ B_{3}) ∪ (B_{1} ∩ B_{2} ∩ G_{3})

∴ P(E) = P(G_{1} ∩ B_{2} ∩ B_{3}) + P(B_{1} ∩ G_{2} ∩ B_{3}) + P(B_{1} ∩ B_{2} ∩ G_{3})

= P(G_{1})·P(B_{2})·P(B_{3}) + P(B_{1})·P(G_{2})· P(B_{3}) + P(B_{1})· P(B_{2})·P(G_{3})

`=(3/4 xx 2/4 xx 3/4) + (1/4 xx 2/4 xx 3/4) + (1/4 xx 2/4 xx 1/4)`

= `(18 + 6 + 2)/64`

= `26/64`

= `13/32`