Question

A bag contains 2 blue, 4 yellow, 3 red balls, what is the probability of getting a ball at random which is either blue or yellow?
(a) $\dfrac{6}{9}$
(b) $\dfrac{4}{9}$
(c) $\dfrac{2}{9}$
(d) $\dfrac{1}{9}$

Answer 1

Hint: To solve this problem we need to know what is the probability of an event happening, it is generally defined as the ratio of number of ways of happening the event cases to the total number of cases. Now in the question we have to calculate that the selected ball at random from the bag will be either blue or yellow, for this we will count the total number of favourable outcomes i.e. blue and yellow balls and divide it by total number of cases i.e. total number of balls.

Complete step by step answer:
We are given a bag which contains 2 blue, 4 yellow and 3 red balls.
And we have to find the probability that if we select a ball at random from the bag then it is either blue or yellow.
But first we need to know what is probability of an event happening,
It is given by,
$\operatorname{Probability}\,of\,an\,event\,happening=\dfrac{Number\,of\,ways\,it\,can\,happen}{Total\,number\,of\,ways}$
Now in the given problem favourable case will be the case in which we select either blue or yellow ball from the bag and the total number of cases will be equal to the number of total balls in the bag.
So we get favourable cases as,
Number of yellow balls + Number of blue balls = 2 + 4 = 6
And total number of cases as,
Number of red balls + Number of yellow balls + Number of blue balls = 3 + 2 + 4 = 9
Now we will calculate probability as,
$\operatorname{probability}=\dfrac{number\,of\,favourable\,cases}{total\,number\,of\,cases}=\dfrac{6}{9}$

So, the correct answer is “Option A”.

Note: To solve this problem you first need to know what are the favourable cases according to the question and what are unfavourable cases, otherwise you may mix up unfavourable cases with the favourable ones and end up with the wrong answer. And remember the definition of the probability given in the solution for future references.