Answer & Explanation
We have 15 balls out of which 9 balls have not been previously used, so the remaining 6 balls are already used.
Let E1 be the first event when the 3 balls are randomly chosen and played with.
Now when choosing 3 balls, the combination can be choosing all 3 unused, 2 unused and 1 used, 1 unused and 2 used & all 3 used.
Here n is the no of unused balls. For eg:- when n is 1, we can choose 1 unused ball from 9 unused in (9c1) ways and remaining 2 from 6 used balls in (6c2) ways .
Now let E2 be event of picking 3 unused balls after E1.
Now total probability
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