Probability (Combination??)

well

New Member
#1
1(a). In a group of 15 students, 9 are boys and 6 are girls.
(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​

(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?

(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.​
 
#2
1(a)(i) looks correct to me

1(a)(ii) is a little vague since "exactly 4 girls are selected twice" is not really the same thing as "4 of 5 girls chosen on Sunday is selected again on Monday"

Say the girls are {a, b, c, d, e, f} and Sunday you pick abdef.

On Monday you have to pick 4 out of those 5 women, say bdef. Now what about c, should c be included the remaining 6 spots?

If yes, then it would be:

((5C4) x (10C6)) / (15C10)

(b)Let me restate the question

In a group of 15 children, 6 are boys and 9 are girls. On Sunday, you randomly select 12 children without replacement. What is the probability that you have selected 7 girls?


On Monday, you randomly select 12 children without replacement. What is the probability that you have selected 7 girls?


Once you have those two numbers, just take the product and that is your final answer.
 
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