Question

Number of X is randomly selected from the set of odd numbers and Y is randomly selected from the set of even numbers of the set { 1,2,3,4,5,6,7 }. Let Z = X + Y, then What is P ( Z = 10 ) equal to?
(a) 1
(b) -1
(c) 0
(d) None of the above

Answer 1

Hint: To solve this kind of problem, we are to find the sets from which X and Y can be chosen at the beginning. X is said to be the set of odd numbers and Y is the set of even numbers. So, the sum of an odd and even number is always odd. From here the problem can be easily solved.

Complete step by step solution:
According to the question, we have a given set, { 1,2,3,4,5,6,7 } and also Z = X + Y. Where X is randomly selected from the set of odd numbers and Y is randomly selected from the set of even numbers of the set { 1,2,3,4,5,6,7 }.
Then, X belongs to { 1,3,5,7} and Y belongs to {2,4,6}.
So, in any case, Z will be odd and must also be greater than 1.
As, the sum of an odd and even number would always be an odd number.
But, it is given that, Z = 10.
And we also can see, 10 is an even number.
So, Z = 10 can not take place.
Thus, it can be concluded, P ( Z = 10 ) = 0

So, the correct answer is “Option C”.

Note: Here we have concluded that the solution of the probability problem is 0. This gives us a scenario that the event can not take place in any place of the solution set. And again if we get the solution as 1, the event should take place in every case of the sample space.