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

• A. $\frac{1}{2}$
• B. $\frac{7}{15}$
• C. $\frac{2}{15}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B

$\frac{7}{15}$

Explanation for the correct answer:

Finding the probability that no two black balls are placed adjacently

Given that there are seven white balls and three black balls are randomly placed in a row

The number of ways of placing three black balls at ten places is: ${}^{10}{C}_3$

There are $6$ blank spaces between $7$ white balls. And also there is one space before the first ball and another space after the last ball. Hence, the total number of blank spaces is $8$.

This can be done in ${}^8{C}_3$ ways.

Therefore, The probability that no two black balls are placed adjacently

$=\frac{{}^8{C}_3}{{}^{10}{C}_3}$

$$\begin{gathered} =\frac{{\displaystyle \frac{8\times 7\times 6}{3\times 2\times 1}}}{{\displaystyle \frac{10\times 9\times 8}{3\times 2\times 1}}} \\ =\frac{56}{120} \\ =\frac{7}{15} \end{gathered}$$

Therefore, the correct answer is option (B).