Question

A bag contains 12 balls out of which x balls are white. \[\left( i \right)\] if one ball is drawn at random, what is the probability that it will be a white ball $\left( {ii} \right)$ if 6 or more white balls are put in the bag and if the probability of drawing a white ball will be twice that of in \[\left( i \right)\] then find x.

Answer 1

Hint: In order to find x, compare the probability of getting a white ball in the two conditions given. After getting the value of x, the probability of getting a white ball could be obtained.

Complete step by step answer:
Formula for probability = $\dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total outcomes}}}}$
In the first situation, when 12 balls are given in the bag and x balls are white.
Total balls = 12
Number of white balls = x
Probability of getting a white ball will be- $\dfrac{x}{{12}}$
In the second case when 6 more white balls have been added in the same bag-
Total balls = 12 + 6= 18
White balls = 6 + x
Probability of getting a white ball - $\dfrac{{x + 6}}{{18}}$
Also it is given that, the probability of getting a white ball in second case is twice as that of getting it in the first case-
$2 \times \dfrac{x}{{12}} = \dfrac{{x + 6}}{{18}}$
Solving for x we get,
$
  \dfrac{x}{6} = \dfrac{{x + 6}}{{18}} \\
  18x = 6x + 36 \\
  12x = 36 \\
  x = 3 \\
 $
So the value of x is 3.
After getting the value of x, we can calculate the probabilities of the two cases.
$
  \left( i \right).\dfrac{x}{{12}} = \dfrac{3}{{12}} = \dfrac{1}{4} \\
  \left( {ii} \right).\dfrac{{x + 6}}{{18}} = \dfrac{{3 + 6}}{{18}} = \dfrac{9}{{18}} = \dfrac{1}{2} \\ $

Note: This is a simple question, not having any particular thing to note. Just to keep in mind that the total number of balls and the number of total white balls gets changed in addition to some more balls, so while finding probability this change should be encountered.