(i). On Sunday, you randomly select 12 students without replacement, what is the probabilty that you have selected 5 girls?

I guess:

P(8 girls) = ((6C5) x (9C7)) / (15C12) = 0.4747

P(8 girls) = ((6C5) x (9C7)) / (15C12) = 0.4747

(ii) Continued from part (i): On Monday, with the same group of 15 students, you randomly select 10 students(again without replacement). That is the probability that exactly 4 girls are selected twice (That is, 4 of 5 girls chosen on Sunday is selected again on Monday)?

I guess:

P(exactly 4 girls are selected twice) = ((5C4) x (9C6)) / (15C10) = 0.13986

I am not sure about this, am I right?

P(exactly 4 girls are selected twice) = ((5C4) x (9C6)) / (15C10) = 0.13986

I am not sure about this, am I right?

(b) In a group of 15 children, 6 are boys and 9 are girls. On Sunday, you randomly select 12 children without replacement. On Monday, you select 10 children(the same group of children as on Sunday) without replacement. What is the probabilty that exactly 7 girls are selected twice (selected on both Sunday and Monday)?

I really do not how to do this question.