Question

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Answer 1

Hint : There are 2 different colored balls present in a bag, the number of balls of one color is given and the other can be supposed to be equal to a variable. Probabilities of drawing both the balls can be calculated respectively and then substituted in the given condition so as to find the value of the variable.
Formula to be used:
$P = \dfrac{f}{T}$ where, P is the probability, f is the favorable outcomes and is the total outcomes

** Complete step-by-step answer** :
It is given that a bag contains 5 red balls and some blue balls.
Let the number of blue balls be x
The total number of balls in the bag will be the sum of two i.e. 5 + x
Probability of drawing a red ball P(R) is given as:
$P = \dfrac{f}{T}$
Favorable outcomes (f) = 5 [number of red balls present in the bag]
Total outcomes (T) = 5 + x [total number of balls present in the bag]
$\Rightarrow P\left( R \right) = \dfrac{5}{{5 + x}}$
Probability of drawing a blue ball P(B) is given as:
$P = \dfrac{f}{T}$
Favorable outcomes (f) = x [number of blue balls present in the bag]
Total outcomes (T) = 5 + x [total number of balls present in the bag]
$\Rightarrow P\left( B \right) = \dfrac{x}{{5 + x}}$
Now, according to the question, the probability of drawing a blue ball is double that of a red ball.
$\Rightarrow P\left( B \right) = 2P\left( R \right)$
Substituting the calculated values, we get:
$\Rightarrow \dfrac{x}{{5 + x}} = 2 \times \dfrac{5}{{5 + x}} \\\ \Rightarrow x = 2 \times 5 \\\ \Rightarrow x = 10 \;$
Therefore, the number of blue balls present in the bag are 10.
So, the correct answer is “10”.

Note : The sum of the probabilities of the events is always equal to 1. Using this fact, we can check if the answer obtained is correct or not
We have:
$P\left( R \right) = \dfrac{5}{{5 + x}}$
$P\left( B \right) = \dfrac{x}{{5 + x}}$
The obtained value of x is 10, so:
$P\left( B \right) = \dfrac{{10}}{{5 + 10}} \\\ \Rightarrow P\left( B \right) = \dfrac{{10}}{{15}} \\\ \Rightarrow P\left( B \right) = \dfrac{2}{3} \\\$
Their sum will be given as:
$P\left( B \right) + P\left( R \right) = \dfrac{1}{3} + \dfrac{2}{3} \\\ \Rightarrow P\left( B \right) + P\left( R \right) = \dfrac{3}{3} \\\ \Rightarrow P\left( B \right) + P\left( R \right) = 1 \\\$
Thus, the value of x obtained is correct.