A) \[\frac{7}{15}\]
B) \[\frac{1}{3}\]
C) \[\frac{2}{15}\]
D) \[\frac{4}{15}\]
Correct Answer: A
Solution :
[a] Total no. of balls \[=7+3=10.\] Without any restriction, no. of ways of placing 3 black balls \[{{=}^{10}}{{C}_{3}}.\] Now, no. of ways in which no two black balls put together by arranging this type \[-W-W-W-W-W-W-W\] \[{{=}^{8}}{{C}_{3}}.\left[ \begin{align} & -position\to black\text{ }ball \\ & Wpositions\to white\text{ }ball \\ \end{align} \right]\] Hence, the required probability \[=\frac{^{8}{{C}_{3}}}{^{10}{{C}_{3}}}=\frac{8\times 7\times 6}{10\times 9\times 8}=\frac{7\times 2}{10\times 3}=\frac{7}{15}\] Hence, option [a] is correct.
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