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Question

(A) Total number of balls = $8 + x$. Number of ways to choose $3$ balls out of $8 + x$ is given by $\binom{8+x}{3}$. Number of ways to choose $3$ red

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Total number of balls = $8 + x$. Number of ways to choose $3$ balls out of $8 + x$ is given by $\binom{8+x}{3}$. Number of ways to choose $3$ red balls out of $8$ is $\binom{8}{3}$. The probability is given by: $P = \frac{\binom{8}{3}}{\binom{8+x}{3}} = \frac{7}{15}$. Calculating $\binom{8}{3} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$. So,$\frac{56}{\binom{8+x}{3}} = \frac{7}{15}$. $\binom{8+x}{3} = \frac{56 	imes 15}{7} = 8 	imes 15 = 120$. We know that $\binom{n}{3} = \frac{n(n-1)(n-2)}{6} = 120$,so $n(n-1)(n-2) = 720$. Since $10 	imes 9 	imes 8 = 720$,we have $n = 10$. Since $n = 8 + x$,then $10 = 8 + x$,which gives $x = 2$.

$A$ box contains $8$ red and $x$ number of green balls. $3$ balls are drawn at random. If the probability that all $3$ balls are red is $\frac{7}{15}$,then the number of green balls $x$ is:

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