Question

A die is thrown once. What is the probability of getting an even number?
$(a){\text{ }}\dfrac{2}{5} \\\ (b){\text{ }}\dfrac{1}{2} \\\ (c){\text{ }}\dfrac{1}{3} \\\ (d){\text{ }}\dfrac{2}{3} \\\$

Answer 1

Hint – This question uses the basic definition of probability which is the favorable number of outcomes divided by the total possible outcome. The favorable cases will all be the even numbers that can appear while throwing a dice and total possible cases are the all 6 numbers on a dice.

Complete step-by-step answer:

As we know that the die has a total six numbers.
So the total number of outcomes = 6.

The set of numbers = (1, 2, 3, 4, 5 and 6).

So, we know that even numbers are those which are divisible by 2.

So in these numbers the set of even numbers are (2, 4 and 6).

So the favorable number of outcomes = 3.

Now as we know that the probability (P) is the ratio of favorable number of outcomes to the total number of outcomes.

$\Rightarrow P = \dfrac{{{\text{Favorable number of outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{3}{6} = \dfrac{1}{2} = 0.5$

So the required probability of getting an even number is $\dfrac{1}{2}$.

Hence option (B) is correct.

Note – Natural numbers can be broadly categorized into two categories one is even and other is odd. All numbers which leave a remainder 0 that are completely divisible by 2 can be termed as an even number whereas all numbers which are not eventually completely divisible by 2 are odd numbers.