There are 5 red, 4 black and 3 white in a bag. Three balls are drawn one by one without replacement find the probability that all three balls are red.

Answer

N/a

Explanation

Let A,B and C are the events of drawing first second and third red balls respectively.
`:.` Required probability `P(AnnBnnC)`
`=P(A).P(B//A).P(C//AnnB)`
Now `P(A)=(.^(5)C_(1))/(.^(12)C_(1))=5/12`
After drawing first red ball, remaining balls red `=4` black `=4` white `=3`
Total balls `=11`
`:.P(B//A)=(.^(4)C_(1))/(.^(11)C_(1))=4/11`
After drawing second red ball, remaining balls red `=3`, black `=4`, white` =3`
Total balls `=10`
`:. P(C//AnnB)=(.^(3)C_(1))/(.^(10)C_(1))=3/10`
`P(AnnBnnC)=P(A).P(B//A).P(C//AnnB)`
`=5/12xx4/11xx3/10=1/22`