A bag contains 5 blue and an unknown number x | vedclass

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(C) Total number of balls = $5 + x$. Number of ways to choose $2$ balls from $(5 + x)$ balls is $^{5+x}C_2 = \frac{(5+x)(4+x)}{2}$. Number of ways to

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$3$

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(C) Total number of balls = $5 + x$. Number of ways to choose $2$ balls from $(5 + x)$ balls is $^{5+x}C_2 = \frac{(5+x)(4+x)}{2}$. Number of ways to choose $2$ blue balls from $5$ blue balls is $^5C_2 = \frac{5 	imes 4}{2} = 10$. The probability of drawing $2$ blue balls is given by $P = \frac{^5C_2}{^{5+x}C_2} = \frac{10}{\frac{(5+x)(4+x)}{2}} = \frac{20}{(5+x)(4+x)}$. Given $P = \frac{5}{14}$,we have $\frac{20}{(5+x)(4+x)} = \frac{5}{14}$. $\Rightarrow (5+x)(4+x) = \frac{20 	imes 14}{5} = 56$. $\Rightarrow x^2 + 9x + 20 = 56$. $\Rightarrow x^2 + 9x - 36 = 0$. $\Rightarrow (x+12)(x-3) = 0$. Since $x$ must be positive,$x = 3$.

$A$ bag contains $5$ blue and an unknown number $x$ of red balls. Two balls are drawn at random. If the probability of both of them being blue is $\frac{5}{14}$,then the value of $x$ is equal to

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