Seven white balls and three black balls are randomly placed in a … - Vidyadip

MCQ

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
$\frac{1}{2}$
✓
$\frac{7}{15}$
C
$\frac{2}{15}$
D
$\frac{1}{3}$

✓

Answer

Correct option: B.

$\frac{7}{15}$

While placing 7 while balls in a row, total gaps = 8
3 black balls can be placed in 8 gaps
$=\text{C}=\frac{(8\times7\times6)}{(3\times2\times1)}=8\times7=56$
So, the total number of ways of arranging white and black balls such that no two black balls are adjacent = 56 × 3! × 7!
Actual number of arrangement possible with 7 white and 3 black balls = (7 + 3)! = 10!
So, the required Probability
$=\frac{(56\times3!\times7!)}{10!}$
$=\frac{(56\times3!\times7!)}{(10\times9\times8\times7!)}$
$=\frac{(56\times3!)}{(10\times9\times8)}$
$=\frac{(56\times3\times2\times1)}{(10\times9\times8)}$
$=\frac{(7\times3\times2\times1)}{(10\times9)}$
$=\frac{(7\times2)}{(10\times3)}$
$=\frac{7}{(5\times3)}$
$=\frac{7}{15}$

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