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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
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 $\frac{1}{2}$.
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