Aptitude - Probability - Discussion

Assume that two independent random samples of size 100 each is taken from a population that has a variance of 36. What is the probability that the difference in the sample means is more than 2?

Please give me the solution.

Please, Solve anyone this question. A committee of 3 is to be selected from a group of 5 girls and 3 boys. What is the probability that two boys are selected?

Is total score nothing but the sum?

Here, I used nCr. But its not working, why?

Can anyone explain me.

Can anyone explain the simplest method to find numarator?

Please, from where E={(1,1),(1,2)...} is came? Please explain.

Why we enter prime numbers sum as 15? Please explain the step.

I am sure the answer is 1/6 for this as this event mentioned has something missing.

E = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),
(5, 2), (5, 6), (6, 1), (6, 5) }

E(correct) = { (1,1), (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) }
(1,1) will have a probability 2/12 rather than 1/12 un-like other pairs.

If you note (2,1) occurs and (1,2) also occurs and hence (1,1) due to dice A and additionally (1,1) due to dice B and hence the probability of this event is doubled.

Prime numbers
2={1,1}
3={1,2};{2,1}
5={1,4};{4,1};{2,3};{3,2}
7={1,6};{6,1};{2,5};{5,2};{3,4};{4,3}
11={5,6};{6,5}
Total 15 possible outcomes.

So, Probability is =15/36=5/12.

The sum of (1, 4) (1, 6) is not prime.