Question

A bag contains 12 blue balls and x red balls. If one ball is drawn at random:
(i)What is the probability that it will be a red ball?
(ii)If 8 more red balls are put in the bag and if the probability of drawing a red ball will be twice the probability in part (i) then find x?

Answer 1

Hint: We know that probability of any outcome is equal to the ratio of favorable outcomes by total outcomes. The total outcomes are drawing one ball at random which is equal to selecting 1 ball from $12+x$ balls. In the first part, we have to find the probability that selected ball is a red ball so the favorable outcomes equal to selecting one red ball from x red ball using a combinatorial approach. Now, in part (ii), 8 more red balls are added so the total number of balls is $20+x$. The total outcomes are now equal to selecting a ball from these new numbers of balls and the favorable outcomes are selecting a red ball from $x+8$ number of red balls. Now, apply the condition given in this part and solve the value of x.

Complete step-by-step answer:
We have given a bag containing 12 blue balls and x red balls.
We know that to find the probability of any outcome we have to divide the favorable outcomes by total outcomes.
Total outcomes are equal to the number of possible ways to draw a ball at random from the bag which is equal to selecting a ball from $12+x$ balls. The selection of a ball is done using a combinatorial approach.
${}^{12+x}{{C}_{1}}$
We know that ${}^{n}{{C}_{1}}=n$ so using this relation in the above equation we get,
$12+x$
In the part (i) we are asked to find the probability that the drawn ball is a red ball.
The favorable outcomes for the probability of drawing a red ball are the ways to select a red ball from the number of red balls.
The possible ways to select a red ball from x red balls are:
${}^{x}{{C}_{1}}$
$=x$
Now, the probability that the drawn ball is red ball is equal to:
$\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$
Substituting favorable outcomes as x and the total outcomes as $12+x$ in the above equation we get,
$\dfrac{x}{12+x}$
Hence, the probability that the drawn ball will be a red ball is equal to $\dfrac{x}{12+x}$.
Moving on to part (ii) of the above question, 8 more red balls are added so now the total number of balls in the bag is:
$\begin{align}
  & 12+8+x \\
 & =20+x \\
\end{align}$
As the number of balls in the bag changes to $20+x$ then the total number of outcomes changes to $20+x$.
Now, selecting a red ball from $8+x$ number of red balls we get,
$\begin{align}
  & {}^{8+x}{{C}_{1}} \\
 & =8+x \\
\end{align}$
The favorable outcomes change to $8+x$.
The probability of drawing a red ball is equal to:
$\dfrac{8+x}{20+x}$
It is also given that the probability of drawing a red ball will be twice the probability in part (i) so equating the probability in part (ii) to the multiplication of probability in part (i) we get,
$\begin{align}
  & \dfrac{8+x}{20+x}=2\left( \dfrac{x}{12+x} \right) \\
 & \Rightarrow \dfrac{8+x}{20+x}=\dfrac{2x}{12+x} \\
\end{align}$
On cross multiplication of the above equation we get,
$\begin{align}
  & \left( 8+x \right)\left( 12+x \right)=2x\left( 20+x \right) \\
 & \Rightarrow 96+8x+12x+{{x}^{2}}=40x+2{{x}^{2}} \\
 & \Rightarrow 96+20x+{{x}^{2}}=40x+2{{x}^{2}} \\
\end{align}$
Rearranging the above equation we get,
${{x}^{2}}+20x-96=0$
Solving the above quadratic equation using factorization method we get,
${{x}^{2}}+24x-4x-96=0$
Taking x as common from the first two terms and -4 from the last two terms we get,
$x\left( x+24 \right)-4\left( x+24 \right)=0$
Taking $\left( x+24 \right)$ as common from the above equation we get,
$\left( x+24 \right)\left( x-4 \right)=0$
Equating $\left( x+24 \right)\And \left( x-4 \right)$ to 0 we get,
$\begin{align}
  & x+24=0 \\
 & \Rightarrow x=-24 \\
 & x-4=0 \\
 & \Rightarrow x=4 \\
\end{align}$
As x is the number of balls which cannot be negative so we are rejecting $x=-24$.
Hence, the value of $x=4$.

Note: You can verify that the value of x that you are getting is correct or not by substituting this value of x in the following equation and check whether after substituting this value of x does that equation hold or not.
$\begin{align}
  & \dfrac{8+x}{20+x}=2\left( \dfrac{x}{12+x} \right) \\
 & \Rightarrow \dfrac{8+x}{20+x}=\dfrac{2x}{12+x} \\
\end{align}$
Substituting $x=4$ in the above equation we get,
$\begin{align}
  & \dfrac{8+4}{20+4}=\dfrac{2\left( 4 \right)}{12+4} \\
 & \Rightarrow \dfrac{12}{24}=\dfrac{8}{16} \\
\end{align}$
Cross multiplying the above equation we get,
$\begin{align}
  & 12\left( 16 \right)=8\left( 24 \right) \\
 & \Rightarrow 192=192 \\
\end{align}$
As L.H.S is equal to R.H.S so the value of x that we obtained above are correct.