Question

A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find the value of x.

Answer 1

Hint: Consider the first case in which the bag contains 12 balls and apply the formula for probability which is the ratio of the number of favorable outcomes to the total number of outcomes. Here, the number of favorable outcomes means the number of black balls. Now, consider the second case in which 6 more black balls are added and find the probability using the same formula. Use the given condition to form a linear equation in x and solve for its value to get the answer.

Complete step-by-step solution:
Here, we have been provided with a box containing 12 black balls in which x balls are black. Two conditions are provided to us and we have to find the value of x. So, let us check the conditions one by one.
(1) Here, we are asked to find the probability of drawing a black ball from the bag containing 12 balls in which x balls are black.
Now, we know that the probability of an event to occur is the ratio of the number of favorable outcomes to the total number of outcomes. So, we have,
Total number of outcomes = Total number of balls
\[\Rightarrow \text{Total number of balls}=n\left( S \right)\]
\[\Rightarrow \text{Total number of balls}=12\]
Favorable number of outcomes = Number of black balls
\[\Rightarrow \text{Favourable number of outcomes}=n\left( F \right)\]
\[\Rightarrow \text{Favourable number of outcomes}=x\]
Therefore, the probability of drawing a black ball, denoted by P (E) is given as
\[\Rightarrow P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}\]
\[\Rightarrow P\left( E \right)=\dfrac{x}{12}\]
Therefore, the probability that the randomly drawn ball is black will be \[\dfrac{x}{12}.\]
(2) In the second condition, it is said that 6 more black balls are added in the box and then the probability of drawing a black ball becomes twice the value it was before. So, let us denote the total number of outcomes as n’(S), number of favourable outcomes as n’(E) and the probability of drawing the black ball as P’(E) for this new case. So, we have,
Total number of outcomes = Total number of balls
\[\Rightarrow \text{Total number of outcomes}={{n}^{'}}\left( S \right)\]
\[\Rightarrow \text{Total number of outcomes}=12+6\]
\[\Rightarrow \text{Total number of outcomes}=18\]
Favourable number of outcomes
\[\Rightarrow \text{Favourable number of outcomes}={{n}^{'}}\left( E \right)\]
\[\Rightarrow \text{Favourable number of outcomes}=x+6\]
Therefore, applying the formula to find the probability of drawing the black ball for this new case, we get,
\[\Rightarrow {{P}^{'}}\left( E \right)=\dfrac{{{n}^{'}}\left( E \right)}{{{n}^{'}}\left( S \right)}\]
\[\Rightarrow {{P}^{'}}\left( E \right)=\dfrac{x+6}{18}\]
Now, according to the given condition in the question, we have,
\[\Rightarrow {{P}^{'}}\left( E \right)=2\times P\left( E \right)\]
\[\Rightarrow \dfrac{x+6}{18}=2\times \dfrac{x}{12}\]
\[\Rightarrow \dfrac{x+6}{3}=x\]
By cross multiplication we get,
\[\Rightarrow x+6=3x\]
\[\Rightarrow 2x=6\]
\[\Rightarrow x=3\]
Hence, the value of x is 3.

Note: One may note that we cannot find the value of ‘x’ by just considering the first condition because we were not provided with sufficient conditions there. We needed the second condition to form an equation in x to determine its value. You must remember the formula of the probability of an event to occur to solve the above question. Do not change the numerator and denominator in the formula of probability because the probability of any event to occur is always less than 1 or equal to 1. Note that in the second condition when 6 more black balls were added then the total number of balls will also increase and that is why it is considered as n’(S) = 18 in the formula.