Question

The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are then put in a box and mixed thoroughly. A person draws two slips from the box, one after other without replacement. Describe the following events: A = The number on the first slip is larger than the one on the second slip. B=The number on the second slip is greater than 2. C = The sum of the numbers on the two slips is 6 or 7, D = The number on the second slips is twice that on the first slip. Which pair(s) of events is (are) mutually exclusive.

Answer 1

Hint: In logic and probability theory, two events are mutually exclusive or disjoint if they cannot both occur at the same time. A clear example is the set of outcomes of a single coin toss, which can result in either heads or tails, but not both.

Complete step-by-step answer:
Four slips are marked as 1, 2, 3 and 4 which are placed in a box.
Out of these 4 slips, 2 are drawn from it one after the other and without replacement.
Therefore, we will start by taking the sample space, S for the experiment,
S={(1,2),(1,3),(1,4),(2,1),(2,3),(3,1),(3,2),(3,4),(4,1),(4,2),(2,4),(4,3)}
For A = number on the first slip is larger than one on the second slip=
{(2,1),(3,1),(3,2),(4,1),(4,2),(4,3)}
For B = number on the second slip is greater than 2 =
{(1,3),(2,3),(1,4),(2,4),(3,4),(4,3)}
For C = sum of the number on the 2 slips is 6 or 7 =
{(2,4),(3,4),(4,2),(4,3)}
For D = number on the second slip is 2 times the number on first slip =
{(1,2),(2,4)}
Clearly, $\left( {A \cap D} \right) = \phi$
Therefore, A and D are mutually exclusive events.

Note: Make sure to consider the correct sample space in these questions, otherwise the answer will not be correct.