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Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

$\begin {array} {1 1} (A)\;\large\frac{1}{2} & \quad (B)\;\large\frac{7}{15} \\ (C)\;\large\frac{2}{15} & \quad (D)\;\large\frac{1}{15} \end {array}$


2 Answers

Total number of ways in which all the balls can be placed is
$ \large\frac{10!}{7!\: 3!} $$= 120$
If no two black balls are placed adjacently, the possible number of ways is
$ ^8C_3 = 56$
Hence the required chance is = $ \large\frac{56}{120}$
$ = \large\frac{7}{15}$
Ans : (B)
answered Jan 8, 2014 by thanvigandhi_1
edited Mar 25, 2014 by thanvigandhi_1

PROBABILITY = no. of possible outcomes/ttotal number of outcomes

so here probabllity= 8c3/10c3=56/120=7/15

answered Jul 9 by phaniajit2005

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