A fair dice is rolled twice. Find the probability that an odd number will follow an even number.

A fair dice is rolled twice. The probability that an odd number will follow an even number is 

(A) $\dfrac{1}{2}$ 

(B) $\dfrac{1}{6}$ 

(C) $\dfrac{1}{3}$ 

(D) $\dfrac{1}{4}$ 

(GATE 2005) 

Answer: (D) $\dfrac{1}{4}$ 

Explanation: 

Here, the sample space is 

$S = \{(1, 1), \cdots, (1, 6), (2, 1), \cdots, (2, 6), (3, 1), \cdots, \\(3, 6), (4, 1), \cdots, (4, 6), (5, 1), \cdots, (5, 6), (6, 1), \cdots, (6, 6) \}.$ 

So, the number of sample points, $n(S) = 6 \times 6 = 36$. 

Again, favorable event space, 

$E = \{(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), \\~~~~~~~~~~(4, 5), (6, 1), (6, 3), (6, 5)\}.$  

So, the number of event points, $n(E) = 9$. 

Therefore, the required probability is 

$= P(E) = \dfrac{n(E)}{n(S)} = \dfrac{9}{36} = \dfrac{1}{4}$.