A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number

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Given:

A die is thrown once.

To do:

We have to find the probability of getting

(i) a prime number

(ii) a number lying between 2 and 6

(iii) an odd number

Solution:

When a die is thrown, the total possible outcomes are 1, 2, 3, 4, 5 and 6.

This implies,

The total number of possible outcomes $n=6$.

(i) Prime numbers between 1 and 6 (including 1 and 6) are 2, 3 and 5.

Total number of favourable outcomes $=3$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability of getting a prime number $=\frac{3}{6}$

$=\frac{1}{2}$

The probability of getting a prime number is $\frac{1}{2}$.

(ii) Numbers lying between 2 and 6 are 3, 4 and 5.

Total number of favourable outcomes $=3$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability of getting a number lying between 2 and 6 $=\frac{3}{6}$

$=\frac{1}{2}$

The probability of getting a number lying between 2 and 6 is $\frac{1}{2}$.

(iii) Odd numbers from 1 to 6 are 1, 3 and 5.

Total number of favourable outcomes $=3$.

We know that,

Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

Therefore,

Probability of getting an odd number $=\frac{3}{6}$

$=\frac{1}{2}$

The probability of getting an odd number is $\frac{1}{2}$.

Updated on 10-Oct-2022 13:26:57