Question

A bag contains \[6\]red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red ball. Find the number of blue balls in the bag.

Answer 1

Hint:In this question we have to find the number of blue balls in the bag. They give the relation of probability of drawing a blue ball and red ball from the bag and also give the number of red balls.Let us consider the number of blue balls in the bag. Then we find out the total number of balls. From there we will find the probability of red and blue balls separately.Then,equating both the probabilities we will get the number of blue balls in the bag.


Complete step-by-step answer:
It is given that; a bag contains \[6\]red balls and some blue balls.
The probability of drawing a blue ball from the bag is twice 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 \[n + 6\]
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\[ = \dfrac{6}{{n + 6}}\]
The probability of drawing a blue ball\[ = \dfrac{n}{{n + 6}}\]
According to the problem, the probability of drawing a blue ball from the bag is twice that of a red ball.
So, we have,
\[\dfrac{n}{{n + 6}} = 3 \times \dfrac{6}{{n + 6}}\]
Simplifying we get,
\[n = 18\]
Hence, the total number of blue balls in the bag is \[18\].

Note:We know that, if \[n\] be the total number of outcomes and \[m\] be the favourable outcomes of an event. So, the probability of any event is \[\dfrac{m}{n}\]. A probability distribution is a function that describes the likelihood of obtaining the possible values that a random variable can assume. In other words,the values of the variable vary based on the underlying probability distribution.