Question

A bag contains \[15\] balls of which x are blue and the remaining are red. If the number of red balls is increased by \[5,\] the probability of drawing the red ball doubles. Find
(i)P (red ball)
(ii)P (blue ball)
(iii)P (blue ball if \[5\] red balls are actually added).

Answer 1

Hint: To solve this question, we will first obtain the probability of red and blue balls in the terms of x, by applying probability formula, then using the condition given, i.e., the number of red balls is increased by \[5,\] the probability of drawing the red ball doubles, we will obtain the value of x and hence we will get our required answers of all the three parts.

Complete step-by-step answer:
We have been given a bag which contains \[15\] balls, out of which x are blue balls and the remaining are red. It is given that the number of red balls is increased by \[5,\] then the probability of drawing the red ball doubles. We need to find the probability of red ball, blue ball and probability of blue ball if \[5\] red balls are actually added.
Total number of outcomes of balls \[ = {\text{ }}15\]
And number of favourable outcomes of red balls \[ = {\text{ }}15{\text{ }} - {\text{ }}x\]
We know that, probability \[ = \dfrac{{{\text{favourable outcomes}}}}{{{\text{total outcomes}}}}\]
On putting the values in the above formula, we get
Probability of getting red ball $ = \dfrac{{15 - x}}{{15}}$
Number of favourable outcomes of blue balls \[ = {\text{ }}x\]
And, probability of getting blue ball $ = \dfrac{x}{{15}}$
Probability of getting red ball if \[5\] red balls are actually added $ = \dfrac{{20 - x}}{{20}}$
Now, according to the question, If the number of red balls is increased by \[5,\] the probability of drawing the red ball doubles.
$
 \Rightarrow (\dfrac{{15 - x}}{{15}}) \times 2 = \dfrac{{20 - x}}{{20}} \\
\dfrac{{30 - 2x}}{{15}} = \dfrac{{20 - x}}{{20}} \\
600 - 40x = 300 - 15x \\
300 = 25x \\
x = \dfrac{{300}}{{25}} \\
x = 12 \\
$
So, out of \[15\] balls, there are \[12\] blue balls and \[3\] red balls.
(i) P (red ball) $ = \dfrac{3}{{15}} = \dfrac{1}{5}$
(ii) P (blue ball) $ = \dfrac{{12}}{{15}}$
(iii) P (red balls if \[20\] balls are there) $ = 2(\dfrac{1}{5})$
Total probability \[ = {\text{ }}1\]
P (blue ball if \[5\] red balls are actually added) $ = 1 - \dfrac{2}{5}$
$ = \dfrac{3}{5}$
Thus, the probability of red ball, blue ball and probability of blue ball if \[5\] red balls are actually added is $\dfrac{1}{5},\dfrac{{12}}{{15}}$ and $\dfrac{3}{5}$ respectively.

Note: Students should notice, there in the solutions we have solved two times the probability of red and blue balls, one in the terms of x and second after getting the value of x, as they didn’t ask us the values in the terms of x, so we are supposed to solve the values after evaluating x also.