Question

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

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

Answer 1

Answer: (B)

$\frac{7}{15}$

Total ways of arranging balls $=\frac{10!}{7!3!}$

Arranging such that no two black balls are together $1^{st}$ arrange $7$ white balls in $\frac{7!}{7!}=1$ ways .

Now, we have $8$ places when we can keep those balls.

$\therefore$ no. of ways $=1.8C_3=\frac{8!}{3!5!}$

$\therefore$ P (no. black ball together) $=\frac{\frac{8!}{3!5!}}{\frac{10!}{7!3!}}$

$=\frac{8!}{3!5!}\times \frac{7!\times 3!}{10!}$

$=\frac{8!\times 7\times 6\times 5!}{5!\times 10\times 9\times 8!}$

$=\frac{42}{90}$

$=\frac{7}{15}$