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Probability — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsProbability1 Mark
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 $= 56 \times 3! \times 7!$
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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