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Twelve balls are distribute among three boxes. The probability that the first box contains three balls is `(110)/9(2/3)^(10)` b. `(110)/9(2/3)^(10)` c. `(^(12)C_3)/(12^3)xx2^9` d. `(^(12)C_3)/(3^(12))`

Text Solution

Verified by Experts

`x+ y = 9`
`P=.^9C_3 (1/3)^3(2/3)^9`
`= (12 xx 11 xx 10)/3 xx 1/3^3 xx (2/3)^9`
`= 220/3^3 xx (2/3)^9`
`= 55/3 xx 2^2/3^2 xx (2/3)^9`
`= 55/3 xx (2/3)^11`
option 1 is correct
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