Combinatorics and Conditional Probabilities: council of 5 students is to be chosen...

Sirius

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A school council of 5 students is to be chosen from a group of 8 male students and 4 female students.
(a) Given that the first student chosen for the council is male, find the probability that there are exactly 2 female students on the council.
(b) Given that the first student chosen is male, find the probability that the last chosen is female.
Could anyone please talk through a worked solution to parts (a) and (b), in terms of combinatorics, since I have conflicting answers for each question and would like to know what the 'accepted' answer for the problem would be.
 
A school council of 5 students is to be chosen from a group of 8 male students and 4 female students.
(a) Given that the first student chosen for the council is male, find the probability that there are exactly 2 female students on the council.
(b) Given that the first student chosen is male, find the probability that the last chosen is female.
Could anyone please talk through a worked solution to parts (a) and (b), in terms of combinatorics, since I have conflicting answers for each question and would like to know what the 'accepted' answer for the problem would be.
What are your thoughts?

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A school council of 5 students is to be chosen from a group of 8 male students and 4 female students.
(a) Given that the first student chosen for the council is male, find the probability that there are exactly 2 female students on the council.
(b) Given that the first student chosen is male, find the probability that the last chosen is female.
Could anyone please talk through a worked solution to parts (a) and (b), in terms of combinatorics, since I have conflicting answers for each question and would like to know what the 'accepted' answer for the problem would be.
I am glad that you tried this problem even though you got conflicting answer. Please show us how you arrived at your multiple answers and I am sure someone will explain where you went wrong and give a hint on how to fix it.
 
A school council of 5 students is to be chosen from a group of 8 male students and 4 female students.
(a) Given that the first student chosen for the council is male, find the probability that there are exactly 2 female students on the council.
HINT. If one male is already chosen then there are \(\displaystyle \dbinom{7}{2}\dbinom{4}{2} \) more ways to choose that committee having exactly two females.
 
A school council of 5 students is to be chosen from a group of 8 male students and 4 female students.
(a) Given that the first student chosen for the council is male, find the probability that there are exactly 2 female students on the council.
Choose any one of the 8 males to be on the council. That leaves 7 male students and 4 female students from which to choose the remaining four members.
The probability the second student is chosen is male is 7/11. That leaves 6 male students and 4 female students. The probability the third student chosen is male is 6/10. That leaves 5 male students and 4 female students. The probability the fourth student chosen is female is 4/9. That leaves 5 male students and 3 female students. The probability the last student chosen is female is 3/8. The probability the five students chosen (including the first who must be male) are MMMFF in that order is (7/11)(6/10)(4/9)(3/8). Doing exactly the same analysis with a different order, say, "MMFMF" gives (7/11)(6/10)(4/9)(3/8). Those fractions have exactly the same denominators and the same numerators but in a different order- their product is exactly the same. It should be easy to see that this is true for any order of "MMFF". And there are \(\displaystyle \frac{4!}{2!2!}= \frac{4(3)(2)}{2(2)}= 6 different orders.
(They are MMFF, MFMF. MFFM. FMMF, FMFM, and FFMM.) The probability there are 3 male and 2 female students, given that the first chosen is male, is 6(7/11)(6/10)(4/9)(3/8).

(b) Given that the first student chosen is male, find the probability that the last chosen is female.
Could anyone please talk through a worked solution to parts (a) and (b), in terms of combinatorics, since I have conflicting answers for each question and would like to know what the 'accepted' answer for the problem would be.
Given that the first student chosen is male and the last student chosen is female, there are \(\displaystyle 2^3= 8\) choices for the gender of the other three:
MMM, MMF, MFM, MFF, FMM, FMF, FFM, and FFF. We can divide those into four classes- all male (MMM), two male and one female (MMF, MFM, and FMM), two females and one male (MFF, FMF, and FFM) and all female (FFF).
(1) The probability of "MMM" is (7/11)(6/10)(5/9). In that case, there are 3 male and 4 female students left so here the probability the last student chosen is female is 4/7. The probability the first student chosen is male, the next three male, and the last female is (4/7)(7/11)(6/10)(5/9).
(2) The probability of "MMF, MFM, or FMM" is 3(7/11)(6/10)(4/9). In that case there are 4 male and 3 female students left so here the probability the last student chosen is female is 3/7. The probability the first student chosen is male, the next three two males and one female, and the last female is 3(3/7)(7/11)(6/10)(4/9).
(3) The probability of "MFF, FMF, or FFM" is 3(7/11)(6/10)(4/9). In that case, there are 6 male and 2 female students left so here the probability the last student chosen is female is 2/8. The probability the first student chosen is male, the next three one male and two females, and the last female is 3(2/8)(7/11)(6/10)(4/9).
(4) The probability of "FFF" is (4/11)(3/10)(2/9). In that case, there are 7 male and one female left so the probability the last student chosen if female is 1/8. The probability the first student chosen is male and the remaining four female is (4/11)(3/10)(2/9)(1/8).

Putting those together, given that the first student chosen is male, the probability the last student chosen is female is
(4/7)(7/11)(6/10)(5/9)+ 3(3/7)(7/11)(6/10)(4/9)+ 3(2/8)(7/11)(6/10)(4/9)+ (4/11)(3/10)(2/9)(1/8).\)
 
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