A die is thrown once. The probability of getting a prime number is

Question:

A die is  thrown once. The probability of getting a prime number is 

(a) $\frac{2}{3}$

(b) $\frac{1}{3}$

(C) $\frac{1}{2}$

 

(d) $\frac{1}{6}$

 

Solution:

(C) $\frac{1}{2}$

Explanation:
In a single throw of a die, the possible outcomes are:
1, 2, 3, 4, 5, 6
Total number of possible outcomes = 6
Let E be the event of getting a prime number.
Then, the favourable outcomes are 2, 3 and 5.                                        
Number of favourable outcomes = 3

$\therefore$ Probability of getting a prime number $=P(E)=\frac{\text { Number of favorable outcomes }}{\text { Total number of possible outcomes }}=\frac{3}{6}=\frac{1}{2}$

 

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