Question

A bag contains $5$ red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball. Find the number of blue balls in the bag.

Answer 1

Let us consider the number of blue balls in the bag as $n$. Then we find out the total number of balls. From there we will find the probability of red and blue balls separately. And using the other information we will find the number of balls in the bag.

Complete step-by-step answer:
It is given that a bag contains $5$ red balls and some blue balls.
Also given that the probability of drawing a blue ball from the bag is thrice that of a red ball.
Let us consider, the number of blue balls in the bag is $n$.
So, total number of balls in the bag is $n + 5$
We know that, if $n$ be the total number of outcomes and $m$ be the outcomes of an event.
So, the probability of any event is $\dfrac{m}{n}$.
Here, the probability of drawing a red ball$= \dfrac{5}{{n + 5}}$
The probability of drawing a blue ball$= \dfrac{n}{{n + 5}}$
According to the problem, the probability of drawing a blue ball from the bag is thrice that of a red ball.
That is $\dfrac{n}{{n + 5}} = 3 \times \dfrac{5}{{n + 5}}$
On simplifying the above equation we get the value of n,
$n = 15$
Hence, the total number of blue balls in the bag is $15$.

Note: We know that, if $n$ be the total number of outcomes and $m$ be the outcomes of an event. So, the probability of any event is $\dfrac{m}{n}$.
From the given problem we can come to a conclusion that total number of balls in the bag is $20$ and hence, the probability of drawing a red ball $= \dfrac{5}{{20}} = \dfrac{1}{4}$
The probability of drawing a blue ball $= \dfrac{{15}}{{20}} = \dfrac{3}{4}$
And the total probability is $\dfrac{3}{4} + \dfrac{1}{4} = 1$.