A) \[\frac{1}{2}\]
B) \[\frac{7}{15}\]
C) \[\frac{2}{15}\]
D) \[\frac{1}{3}\]
Correct Answer: B
Solution :
The number of ways to arrange 7 white an 3 black balls in a row \[=\frac{10\,!}{7\,\,!\,.\,3\,\,!}=\frac{10.9.8}{1.2.3}=120\] Numbers of blank places between 7 balls are 6. There is 1 place before first ball and 1 place after last ball. Hence total number of places are 8. Hence 3 black balls are arranged on these 8 places so that no two black balls are together in number of ways \[={}^{8}{{C}_{3}}=\frac{8\times 7\times 6}{1\times 2\times 3}=56\] So required probability \[=\frac{56}{120}=\frac{7}{15}.\]
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