Probability Ques 5
- Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently, equals
(1998, 2M)
(a) $\frac{1}{2}$
(b) $\frac{7}{15}$
(c) $\frac{2}{15}$
(d) $\frac{1}{3}$
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Answer:
Correct Answer: 5.(b)
Solution: (b) The number of ways of placing 3 black balls without any restriction is ${ }^{10} \mathrm{C}_3$.
Since, we have total $10$ places of putting $10 $ balls in a row. Now, the number of ways in which no two black balls put together is equal to the number of ways of choosing $3$ places marked ‘-’ out of eight places.
$ -W-W-W-W-W-W-W- $
This can be done in ${ }^8 C_3$ ways.
$ \therefore \quad \text { Required probability }=\frac{{ }^8 C_3}{{ }^{10} C_3}=\frac{8 \times 7 \times 6}{10 \times 9 \times 8}=\frac{7}{15} $
BETA
