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 G1, G2, G3 denote events for selecting a girl and B1, B2, B3 denote events for selecting a boy from 1st, 2nd and 3rd groups respectively.
Then P(G1) = `3/4`, P(G2) = `2/4`, P(G3) = `1/4`
P(B1) = `1/4`, P(B2) = `2/4`, P(B3) = `3/4`
Where G1, G2, G3, B1, B2, and B3 are mutually exclusive events.
Let E be the event that 1 girl and 2 boys are selected
∴ E = (G1 ∩ B2 ∩ B3) ∪ (B1 ∩ G2 ∩ B3) ∪ (B1 ∩ B2 ∩ G3)
∴ P(E) = P(G1 ∩ B2 ∩ B3) + P(B1 ∩ G2 ∩ B3) + P(B1 ∩ B2 ∩ G3)
= P(G1)·P(B2)·P(B3) + P(B1)·P(G2)· P(B3) + P(B1)· P(B2)·P(G3)
`=(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`
