Question

One ball is selected at random from a bag containing $12$ balls of which $p$ are white. When further $6$ white balls are added, the probability of selecting a white ball is doubled. Find $p$.

Answer 1

Step 1: Finding the probability of selecting a white ball before adding $6$ white balls

There are $12$ balls in a bag.

Out of which $p$ balls are white.

The number of favorable outcomes is $p$ and the total number of outcomes is $12$.

The formula for probability is,

$$\begin{gathered} Probablity=\frac{Numberoffavorableoutcomes}{Totalnumberofoutcomes} \\ \therefore P\left(white\right)=\frac{p}{12} \end{gathered}$$

Henceforth, the probability of selecting a white ball is $\frac{p}{12}$.

Step 2: Finding the probability of selecting a white ball after adding $6$ white balls

There were $12$ balls in a bag. Further, $6$ white balls are added.

Now, the total number of balls are $12+6=18$.

Out of which $p+6$ are white.

The number of favorable outcomes is $p+6$ and the total number of outcomes is $18$.

The formula for probability is,

$$\begin{gathered} Probablity=\frac{Numberoffavorableoutcomes}{Totalnumberofoutcomes} \\ \therefore P^{\text{'}}\left(white\right)=\frac{p+6}{8} \end{gathered}$$

Henceforth, the probability of selecting a white ball is $\frac{p+6}{8}$.

Step 3: Finding the value of $p$

As per the question, the probability of selecting a white ball is doubled. So,

$$\begin{gathered} \Rightarrow P\text{'}\left(white\right)=2P\left(white\right) \\ \Rightarrow \frac{p+6}{18}=2\times \frac{p}{12} \\ \Rightarrow \frac{p+6}{18}=\frac{p}{6} \\ \Rightarrow 6p+36=18p \\ \Rightarrow 18p-6p=36 \\ \Rightarrow 12p=36 \\ \therefore p=3 \end{gathered}$$

Henceforth, the value of $p$ is $3$.