Question
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
Solution
S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (1, H), (1, T), (2, H), (2, T), (4, H), (4, T), (5, H), (5, T)}
The outcomes of S are not equally likely. First 12 outcomes are equally likely and are such that the sum of their probabilities is 
So each of the first 12 outcomes has a probability equal to 
Remaining eight outcomes are equally likely and are such that the sum of their probabilities is 
So, each of these has a probability equal to 
Let E: 'the coin shows a tail'
and F: 'at least one die shows up a 3',
and 
Required probability = 
