# A bag contains 20 balls numbered from 1 to 20. One ball is drawn at random from the bag. What is the probability that the ball drawn is marked with a number which is multiple of 3 or 4?

Last updated date: 18th Mar 2023

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Answer

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Hint – In this question first find out the numbers which are multiple of 3 or 4 from the numbers which are given which are the favorable outcomes, then divide these favorable outcomes to the total given numbers, use this concept to reach the solution of the problem.

Given data

A bag contains 20 balls numbered from 1 to 20.

Therefore total number of balls $ = 20$.

Now we have to find the probability that the ball drawn is marked with a number which is multiple of 3 or 4.

From 1 to 20 multiple of 3 is (3, 6, 9, 12, 15, and 18).

So there are 6 favorable ways.

From 1 to 20 multiple of 4 is (4, 8, 12, 16 and 20).

So there are 5 favorable ways.

Now as we see that 12 is a multiple of both 3 and 4, so it is considered only one time.

Therefore total number of favorable ways $ = 6 + 5 - 1 = 10$

So, the required probability that the ball drawn is marked with a number which is multiple of 3 or 4.

$ = \dfrac{{{\text{Favorable number of outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{{10}}{{20}} = \dfrac{1}{2}$

So, $\dfrac{1}{2}$ is the required probability.

Note – In such types of questions the key concept we have to remember is that always recall the formula of probability which is stated above then first find out the number of favorable ways after that use the probability formula and simplify, we will get the required answer.

Given data

A bag contains 20 balls numbered from 1 to 20.

Therefore total number of balls $ = 20$.

Now we have to find the probability that the ball drawn is marked with a number which is multiple of 3 or 4.

From 1 to 20 multiple of 3 is (3, 6, 9, 12, 15, and 18).

So there are 6 favorable ways.

From 1 to 20 multiple of 4 is (4, 8, 12, 16 and 20).

So there are 5 favorable ways.

Now as we see that 12 is a multiple of both 3 and 4, so it is considered only one time.

Therefore total number of favorable ways $ = 6 + 5 - 1 = 10$

So, the required probability that the ball drawn is marked with a number which is multiple of 3 or 4.

$ = \dfrac{{{\text{Favorable number of outcomes}}}}{{{\text{Total outcomes}}}} = \dfrac{{10}}{{20}} = \dfrac{1}{2}$

So, $\dfrac{1}{2}$ is the required probability.

Note – In such types of questions the key concept we have to remember is that always recall the formula of probability which is stated above then first find out the number of favorable ways after that use the probability formula and simplify, we will get the required answer.

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