An urn contains 6 yellow balls and x black ba | vedclass

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(C) Total number of balls = $6 + x$. Number of ways to draw $2$ balls from $6 + x$ balls is given by $\binom{6+x}{2} = \frac{(6+x)(5+x)}{2}$. Number o

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Total number of balls = $6 + x$. Number of ways to draw $2$ balls from $6 + x$ balls is given by $\binom{6+x}{2} = \frac{(6+x)(5+x)}{2}$. Number of ways to draw $2$ yellow balls from $6$ yellow balls is $\binom{6}{2} = \frac{6 	imes 5}{2} = 15$. The probability of drawing $2$ yellow balls is $\frac{15}{\frac{(6+x)(5+x)}{2}} = \frac{30}{(6+x)(5+x)}$. Given that the probability is $\frac{5}{26}$,we have $\frac{30}{(6+x)(5+x)} = \frac{5}{26}$. Simplifying,we get $(6+x)(5+x) = \frac{30 	imes 26}{5} = 6 	imes 26 = 156$. Expanding the equation: $x^2 + 11x + 30 = 156$,which leads to $x^2 + 11x - 126 = 0$. Factoring the quadratic equation: $(x + 18)(x - 7) = 0$. Since $x$ must be positive,$x = 7$.

An urn contains $6$ yellow balls and $x$ black balls. When two balls are drawn at random,the probability that both are yellow is $\frac{5}{26}$,then the value of $x$ is

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