Question
Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals
Answer: Option B
:
B
The number of ways to arrange 7 white an 3 black balls in a row
10!7!.3!=10.9.81.2.3=120
Numbers of blank places between 7 balls are 6. There is 1 place before first ball and 1 place after last ball. Hence total number of places are 8.
Hence 3 black balls are arranged on these 8 places so that no two black balls are together in number of ways.
=8C3=8×7×61×2×3=56
So required probability = 56120=715
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:
B
The number of ways to arrange 7 white an 3 black balls in a row
10!7!.3!=10.9.81.2.3=120
Numbers of blank places between 7 balls are 6. There is 1 place before first ball and 1 place after last ball. Hence total number of places are 8.
Hence 3 black balls are arranged on these 8 places so that no two black balls are together in number of ways.
=8C3=8×7×61×2×3=56
So required probability = 56120=715
Was this answer helpful ?
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