Seven white balls and three black balls are randomly placed in a row. The prob

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Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

Options:

A .

\frac{1}{2}

B .

\frac{7}{15}

C .

\frac{2}{15}

D .

\frac{1}{3}

Answer: Option B
:
B

The number of ways to arrange 7 white an 3 black balls in a row
$\frac{10!}{7! .3!} = \frac{10.9.8}{1.2.3} = 120$
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.
$=^{8} C_{3} = \frac{8 \times 7 \times 6}{1 \times 2 \times 3} = 56$
So required probability = $\frac{56}{120} = \frac{7}{15}$

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