Aptitude - Probability - Discussion

If we draw this in a cartesian plan we can easily find the answer. Take d axis and Y axis. Number from 1 to 6 in each axis. The find the sum of the numbers using plan.

If two numbers are selected at random, one after the other with replacement from the set A = (5, 6, 7, 8, 9). Find the probability of selecting at least one prime number.

Can anyone help me to solve this problem.

Hey @Prasad,

Here is the clarification of your doubt. You have to choose those number of pairs whose sum is a prime number. If you choose (1, 3), (2, 2) & (3, 1) then their sum will be 4 which is not a prime number.

Thanks!

Could any one of you explain me in simplistic processes?

Why 4 kings are taken?

The total no of unique combinations on throw of 2 dice is 36:

i.e. {(1,1),(1,2),(2,1),(1,3),(3,1),(2,2),(1,4),(4,1),(2,3),(3,2),(1,5),(5,1),(2,4),(4,2),(3,3),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(2,6),(6,2),(3,5),(5,3),(4,4),(3,6),(6,3),(4,5),(5,4),(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)}.

Out of these pairs, how many give us a prime number sum = 15.

i.e. {(1,1),(1,2),(2,1),(1,4),(4,1),(2,3),(3,2),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3),(5,6),(6,5)}.

So, probability = 15/36 = 5/12.

No detail about the Dice's aren't they Identical (0,0).

How we got 15 point please anyone can explain?

Don't you think we should not count (1,2) and (2,1) as 2 it must counted as 1 only.

We have prime numbers between 2 To 12 as 2,3,5,7,11.

There is 1 Way to get sum as 2, 2 ways to get sum 3, 4 To get 5, 6 Ways to get 7 And 2 Ways to get 11. So total of 15, therefore prob = 15/36.