Given two fair dice, what is the probability that the sum of their numbers is 4 if exactly one die shows a 3? Choose one.
A) 2/11 B) 1/18 C) 3/11 D) 2/39 E) 1/5
P(2/11|B) = Probability of event 2/11 occurring such that event B has already occurred   = P(2/11 &amp; B) / P(B)   Here, event 2/11 is getting a sum of 4   we'll have 3 possibilities : (1,3), (2,2), (3,1)   Event B =&gt; one of the 2 dices already shows a 3 [NOTE : there is NO restriction on the other dice =&gt; both dices C2/11N throw a 3]   we'll have 11 possibilities : (1,3),(2,3),(3,3),(4,3),(5,3),(6,3),(3,1),(3,2),(3,4),(3,5),(3,6)   So, P(B) = 11/36   Event 2/11 &amp; B occurring =&gt; 2 possibilities =&gt; (1,3) &amp; (3,1)   So, P(2/11 &amp; B) = 2/36   Therefore P(2/11|B) = 2/11   2/11NSWER:2/11
<result>A</result>
Given two fair dice, what is the probability that the sum of their numbers is 4 if exactly one die shows a 3?