Question

A bag contains $5$ blue balls and unknown number of red balls, two balls are drawn at random. The probability of both of them are blue is $\frac{5}{14}$, then the number of red balls are

• A. $3$
• B. $2$
• C. $4$
• D. $5$

Answer 1

The correct option is A

$3$

The explanation for the correct option:

Step 1:To find number of red balls:

Number of blue balls = $5$

Let the number of red balls be $x$

Total number of balls = $5+x$

The probability of drawing two blue balls$=\frac{5}{14}$

⇒ $\frac{5}{14}=\frac{{}^5{C}_2}{{}^{5+x}{C}_2}$---------- (1)

Step 2: Finding ${}^{\mathbf{5}}{\mathit{C}}_{\mathbf{2}\mathbf{}}\mathit{a}\mathit{n}\mathit{d}{\mathbf{}}^{\mathbf{5}\mathbf{+}\mathbf{x}}{\mathit{C}}_{\mathbf{2}}$:

We know that, ${}^n{C}_r=\frac{n!}{r!(n-r)!}$

$$\begin{gathered} {}^5{C}_2=\frac{5!}{2!3!} \\ =\frac{5\times 4\times 3!}{2\times 3!} \\ =\frac{5\times 4}{2} \\ =10 \end{gathered}$$

$$\begin{gathered} {}^{5+x}{C}_2=\frac{(5+x)!}{2!(3+x)!} \\ =\frac{(5+x)(4+x)(3+x)!}{2(3+x)!} \\ =\frac{(5+x)(4+x)}{2} \end{gathered}$$

Substitute the values in equation $\left(1\right)$

$$\begin{gathered} \frac{5}{14}=\frac{10\times 2}{(5+x)(4+x)} \\ \Rightarrow 56=(5+x)(4+x) \\ \Rightarrow 56=x^2+9x+20 \\ \Rightarrow 0=x^2+9x-36 \\ \Rightarrow 0=x^2+12x-3x-36 \\ \Rightarrow 0=x(x+12)-3(x+12) \\ \Rightarrow 0=(x+12)(x-3) \end{gathered}$$

On solving this quadratic equation we get,

$x=-12orx=3$

$x$ cannot be negative . So, $x=3$.

Hence, option (A) is the correct answer.