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(B) Total number of ways to arrange $7$ white balls and $3$ black balls in a row is given by the combination formula: $\frac{10!}{7!3!} = \frac{10 	i

Vedclass · Accepted Answer

$\frac{7}{15}$

Vedclass · Answer

(B) Total number of ways to arrange $7$ white balls and $3$ black balls in a row is given by the combination formula: $\frac{10!}{7!3!} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$.To ensure no two black balls are adjacent,we first arrange the $7$ white balls in a row: $W \_ W \_ W \_ W \_ W \_ W \_ W$.There are $8$ possible spaces (including the ends) where the $3$ black balls can be placed: $6$ spaces between the white balls and $2$ at the ends.The number of ways to choose $3$ spaces out of these $8$ is given by ${}^8C_3 = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$.The required probability is the ratio of the number of favorable arrangements to the total number of arrangements: $P = \frac{56}{120} = \frac{7}{15}$.

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

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