When a die is thrown, equally likely events are getting 

Correct option is (A) even number and odd number

When a die is thrown, total outcomes are S = {1, 2, 3, 4, 5, 6}.

(A) Even numbers are 2, 4 & 6 and odd numbers are 1, 3 & 5.

P(E) = Probability of getting an even number \(=\frac36=\frac12\)

P(0) = Probability of getting an odd number \(=\frac36=\frac12\)

\(\because\) P(E) = P(0)

\(\therefore\) Events of getting even number and odd number are equally likely.

(B) Prime numbers are 2, 3 & 5 and composite numbers are 4 & 6.

\(P(E_1)\) = Probability of getting a prime number \(=\frac36=\frac12\)

\(P(E_2)\) = Probability of getting a composite number \(=\frac26=\frac13\)

\(\because\) \(P(E_1)\neq P(E_2)\)

\(\therefore\) Events of getting a prime number and a composite number are not equally likely.

(C) \(E_3\) = Multiple of 3 among {1, 2, 3, 4, 5, 6} = {3, 6}

\(E_4\) = Multiple of 2 among {1, 2, 3, 4, 5, 6} = {2, 4, 6}

\(P(E_3)\) \(=\frac26=\frac13\) & \(P(E_4)\) \(=\frac36=\frac12\)

Hence, event of getting multiple of 3 & multiple of 2 are not equally likely.

(D) \(E_5\) = Numbers less than 3 in sample space = {1, 2}

\(E_6\) = Numbers greater than 3 in sample space = {4, 5, 6}

\(\therefore\) \(P(E_5)\) \(=\frac26=\frac13\) & \(P(E_6)\) \(=\frac36=\frac12\)

Hence, both events are not equally likely.

A) even number and odd number