An urn contains `5` white and `8` black balls. Two successive drawings of three balls at a time are made such that the balls are not replaced before the second draw. Find the probability that the first draw gives `3` white balls and second draw gives `3` black balls.

Explanation

Total number of balls `=5` white +`8` black `= 13` balls

Let,

`A=` Event that the first draw gives `3` white balls

`B=` Event that the first draw gives `3` black balls without replacing


`P(A)=frac{C_3^5}{C_3^13}`

`P(B)=frac{C_3^8}{C_3^10}`


Required probability `=P(A) times P(B)=frac{C_3^5}{C_3^13} times frac{C_3^8}{C_3^10}`

`= frac{5 times 4 times 3}{13 times 12 times 11) times frac{8 times 7 times 6}{10 times 9 times 8}`

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