If three caps are picked at random , what is the proabability that two are red and one is green?

Options

A. `(9)/(22)`
B. `(6)/(19)`
C. `(1)/(6)`
D. `(3)/(22)`

Answer

D

Explanation

Total number of caps = 12
`n(S) =""^(12)C_(3)=(12!)/(3!.9!)=(12xx11xx10xx9!)/(3xx2xx1xx9!)=220`
`n(E_(1))=` Out of 4 red caps ,number of ways to pick 2 caps `=""^(4)C_(2)`
`=(4!)/(2!.2!)=(4xx3xx2xx1)/(2xx1xx2xx1)=6`
`n(E_(2))`= out of 5 green caps , number of ways to pick one cap `="^(5)C_(1)=5`
`P(E) =(n(E_(1))xxn(E_(2)))/(n(S))=(6xx5)/(220)=(3)/(22)`