Problem #4: A class of 30 students has 16 boys and 14 girls. How many groups of 8...?

[snake1k1]

A class of 30 students has 16 boys and 14 girls. How many groups of 8 students can be formed where

(a) the eldest boy is included in the group?
(b) the eldest boy is excluded from the group?
(c) the eldest boy is excluded if the eldest girl is included (i.e. they can't both be on the committee)

A group of 10 students is now needed to complete 10 different activities. If 5 girls and 5 boys are chosen to do a different activity each. What is the probability:

(d) the eldest boy is chosen to do an activity?
(e) the eldest boy is not chosen to do an activity given that the eldest girl is not chosen to do an activity?

The class of 30 students is now broken into 2 groups, each containing 15 students.

(f) How many ways can the 2 groups be formed
(g) What is the probability that there are 8 boys and 7 girls in each group?

 

[stapel]

A class of 30 students has 16 boys and 14 girls. How many groups of 8 students can be formed where

(a) the eldest boy is included in the group?
(b) the eldest boy is excluded from the group?
(c) the eldest boy is excluded if the eldest girl is included (i.e. they can't both be on the committee)

A group of 10 students is now needed to complete 10 different activities. If 5 girls and 5 boys are chosen to do a different activity each. What is the probability:

(d) the eldest boy is chosen to do an activity?
(e) the eldest boy is not chosen to do an activity given that the eldest girl is not chosen to do an activity?

The class of 30 students is now broken into 2 groups, each containing 15 students.

(f) How many ways can the 2 groups be formed
(g) What is the probability that there are 8 boys and 7 girls in each group?

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck? For instance, for part (a), after reserving one spot for the specified person, there are seven other slots to be chosen, at random, from the remaining fifteen males and fourteen females (or, more specifically, from the other twenty-nine persons). How many total groups of this sort did you get? And so forth.

Please be complete. Thank you! :wink:

 

[snake1k1]

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck? For instance, for part (a), after reserving one spot for the specified person, there are seven other slots to be chosen, at random, from the remaining fifteen males and fourteen females (or, more specifically, from the other twenty-nine persons). How many total groups of this sort did you get? And so forth.

Please be complete. Thank you! :wink:

Need help with pretty much everything.

 

[stapel]

Need help with pretty much everything.

Since you're completely unable to make any start with anything (even plugging numbers into the customary formula), I would suggest that you consider enrolling in an appropriate sequence of courses at your local college or university, leading up to a course in statistics. In that course, they'll cover all of the material necessary to understanding what is being asked of you, and will provide you with the appropriate formulas, tables, methods, and techniques. Have fun!

 

[snake1k1]

Since you're completely unable to make any start with anything (even plugging numbers into the customary formula), I would suggest that you consider enrolling in an appropriate sequence of courses at your local college or university, leading up to a course in statistics. In that course, they'll cover all of the material necessary to understanding what is being asked of you, and will provide you with the appropriate formulas, tables, methods, and techniques. Have fun!

Thanks for the advice. However, still any help figuring out the solution would be very helpful and much appreciated.

 

[stapel]

Thanks for the advice. However, still any help figuring out the solution would be very helpful and much appreciated.

The "help" was the suggestion that you plug the given values into the standard formulas. But since you haven't yet started a course in statistics, you haven't yet seen those formulas.

Are you asking to links, so you can attempt online self-study? If so, do you know anything about "combinations" and "permutations", or are you needing to start at the very beginning? What online-lessons sites have you tried using so far? Thank you!

 

[Agent Smith]

Kindly check my work and answers:

A class of 30 students has 16 boys and 14 girls. How many groups of 8 students can be formed where

(a) the eldest boy is included in the group?
(b) the eldest boy is excluded from the group?
(c) the eldest boy is excluded if the eldest girl is included (i.e. they can't both be on the committee)

a) If the eldest boy is included, we have only 29 other students to choose from.
Groups of 8 students that can thus be formed = [imath]^{29}C_7[/imath]

b) If the eldest boy is excluded then we have only 29 students to choose from.
Group of 8 students that can thus be formed = [imath]^{29}C_8[/imath]

c) Total number of groups of 8 that can be formed = [imath]^{30}C_8[/imath]
Groups of 8 that have both the eldest girl and the eldest boy = [imath]^{28}C_6[/imath]
Thus, groups of 8 in which both the eldest girl and the eldest boy are not together = [imath]^{30}C_8 - ^{28}C_6[/imath]

Here it's a bit confusing because [imath]^{28}C_7[/imath] also seems like an answer: One "slot" is for the eldest girl ([imath]30 - 1 = 29[/imath]) and the eldest boy is not included ([imath]29 - 1 = 28[/imath]), we have to fill the rest of the [imath]7[/imath] slots with what remains ([imath]29 - 1 = 28[/imath]). I tried doubling this number (to take into account groups with the eldest boy in it), but the answers don't tally.

A group of 10 students is now needed to complete 10 different activities. If 5 girls and 5 boys are chosen to do a different activity each. What is the probability:

(d) the eldest boy is chosen to do an activity?
(e) the eldest boy is not chosen to do an activity given that the eldest girl is not chosen to do an activity?

d) Total number of (5-boy, 5-girl) groups that can be formed = [imath]^{16}C_5 \times ^{14}C_5[/imath]
Number of such groups that have the eldest boy in it = [imath]^{16}C_4 \times ^{14}C_5[/imath]
Pr(the eldest boy is chosen to do an activity) = [imath]\frac{^{16}C_4 \times ^{14}C_5}{^{16}C_5 \times ^{14}C_5}[/imath]

e) Groups in which both the eldest girl and eldest boy are not there = [imath]^{15}C_5 \times ^{13}C_5[/imath]
Pr(both the eldest girl and boy are not given an activity) = [imath]\frac{^{15}C_5 \times ^{13}C_5}{^{16}C_5 \times ^{14}C_5}[/imath]

The class of 30 students is now broken into 2 groups, each containing 15 students.

(f) How many ways can the 2 groups be formed
(g) What is the probability that there are 8 boys and 7 girls in each group?

f) The number of ways groups of 15 can be formed = [imath]^{30}C_{15} \times ^{15}C_{15} = ^{30}C_{15}[/imath]
g) Number of groups with 8 boys and 7 girls that can be formed = [imath]^{16}C_8 \times ^{14}C_7[/imath]

Pr(there are 8 boys and 7 girls in each group) = [imath]\frac{^{16}C_8 \times ^{14}C_7}{^{30}C_{15}}[/imath].

Are all these correct? Kindly help.

 

[Steven G]

They seem to be all correct.

 

[Agent Smith]

@Steven G , sarcasm? Thank you all the same.

 

[Dr.Peterson]

c) Total number of groups of 8 that can be formed = [imath]^{30}C_8[/imath]
Groups of 8 that have both the eldest girl and the eldest boy = [imath]^{28}C_6[/imath]
Thus, groups of 8 in which both the eldest girl and the eldest boy are not together = [imath]^{30}C_8 - ^{28}C_6[/imath]

Here it's a bit confusing because [imath]^{28}C_7[/imath] also seems like an answer: One "slot" is for the eldest girl ([imath]30 - 1 = 29[/imath]) and the eldest boy is not included ([imath]29 - 1 = 28[/imath]), we have to fill the rest of the [imath]7[/imath] slots with what remains ([imath]29 - 1 = 28[/imath]). I tried doubling this number (to take into account groups with the eldest boy in it), but the answers don't tally.

An alternate way similar to your alternate attempt is by addition rather than subtraction: Either the oldest boy is included, but not the oldest girl ([imath]^{28}C_7[/imath]), or the oldest girl is included, but not the oldest boy ([imath]^{28}C_7[/imath]), or neither is included ([imath]^{28}C_8[/imath]). The total is [imath]^{28}C_7+^{28}C_7+^{28}C_8=5,476,185[/imath]. This is the same as your [imath]^{30}C_8 - ^{28}C_6=5,476,185[/imath].

e) Groups in which both the eldest girl and eldest boy are not there = [imath]^{15}C_5 \times ^{13}C_5[/imath]
Pr(both the eldest girl and boy are not given an activity) = [imath]\frac{^{15}C_5 \times ^{13}C_5}{^{16}C_5 \times ^{14}C_5}[/imath]

This doesn't answer the question:

A group of 10 students is now needed to complete 10 different activities. If 5 girls and 5 boys are chosen to do a different activity each. What is the probability:
...
(e) the eldest boy is not chosen to do an activity given that the eldest girl is not chosen to do an activity?

This is asking for a conditional probability. You need to change the denominator.

f) The number of ways groups of 15 can be formed = [imath]^{30}C_{15} \times ^{15}C_{15} = ^{30}C_{15}[/imath]

I think this double-counts: a given group might be the first or the second one chosen in your scheme, and it would be the same two groups. It would be different if the groups were distinguished (e.g. one group to do an activity today, and another for tomorrow).

 

[Agent Smith]

@Dr.Peterson
Thanks. Is the following correct then?
B = The eldest boy is not chosen
G = The eldest girl is not chosen

[imath]P(B|G) = \frac{P(B \wedge G)}{P(G)}[/imath]

[imath]P(B \wedge G) = \frac{ ^{15}C_5 \times ^{13}C_5}{ ^{16}C_5 \times ^{14}C_5}[/imath]

[imath]P(G) = \frac{^{16}C_5 \times ^{13}C_5}{^{16}C_5 \times ^{14}C_5}[/imath]

So [imath]P(B|G) = \left(\frac{ ^{15}C_5 \times ^{13}C_5}{ ^{16}C_5 \times ^{14}C_5} \right) \div \left( \frac{^{16}C_5 \times ^{13}C_5}{^{16}C_5 \times ^{14}C_5} \right) = \frac{^{15}C_5 \times ^{13}C_5}{^{16}C_5 \times ^{13}C_5}[/imath]

---

I think I'm counting the total number of groups of 15 and I suppose it should be just [imath] ^{30}C_{15}[/imath]
Since the number of groups with 8 boys and 7 girls = [imath]^{16}C_8 \times ^{14}C_7[/imath], the probability that a group has 8 boys and 7 girls = [imath]\frac{^{16}C_8 \times ^{14}C_7}{^{30}C_{15}}[/imath]

Good day.

 

[Dr.Peterson]

View attachment 38877
@Dr.Peterson
Thanks. Is the following correct then?
B = The eldest boy is not chosen
G = The eldest girl is not chosen

[imath]P(B|G) = \frac{P(B \wedge G)}{P(G)}[/imath]

[imath]P(B \wedge G) = \frac{ ^{15}C_5 \times ^{13}C_5}{ ^{16}C_5 \times ^{14}C_5}[/imath]

[imath]P(G) = \frac{^{16}C_5 \times ^{13}C_5}{^{16}C_5 \times ^{14}C_5}[/imath]

So [imath]P(B|G) = \left(\frac{ ^{15}C_5 \times ^{13}C_5}{ ^{16}C_5 \times ^{14}C_5} \right) \div \left( \frac{^{16}C_5 \times ^{13}C_5}{^{16}C_5 \times ^{14}C_5} \right) = \frac{^{15}C_5 \times ^{13}C_5}{^{16}C_5 \times ^{13}C_5}[/imath]

---

That looks reasonable.

You could also use the number of ways to choose without either eldest, over the total number of ways to choose without the eldest girl, which gives the same result: [imath]\frac{^{15}C_5 \times ^{13}C_5}{^{16}C_5 \times ^{13}C_5}[/imath]

But you could also recognize that the choice of girls is independent of the choice of boys, so P(B | G) should be equal to P(B). And, in fact, your answer simplifies easily to [imath]\frac{^{15}C_5}{^{16}C_5}[/imath].

View attachment 38879
I think I'm counting the total number of groups of 15 and I suppose it should be just [imath] ^{30}C_{15}[/imath]
Since the number of groups with 8 boys and 7 girls = [imath]^{16}C_8 \times ^{14}C_7[/imath], the probability that a group has 8 boys and 7 girls = [imath]\frac{^{16}C_8 \times ^{14}C_7}{^{30}C_{15}}[/imath]

Good day.

Do you not see what I am saying about double-counting?

If I select a certain group of 15 and put them on my left, or if I select everyone but those 15 to put on my left, and put the others on my right, I have the same two groups; they are just in different places. So I've counted the same pair of groups twice.

This is a subtle distinction; but combinatorics is all about subtle distinctions!

But there's a reason I started with "I think"! One can read the question, about the number of ways to form two groups, as allowing us to think of my two cases as different "ways" (distinguishing the two groups). I don't know what was intended.

Fortunately, the probability (g) will be the same whichever way we interpret (f). Your answer there is correct.

 

[Agent Smith]

@Dr.Peterson gracias. The [imath]2[/imath] groups (of 15) threw me off. We could take one group as A and the other group as B. If now one group of 15 is F1 and the other group of 15 is F2, we could have (A, F1), (B, F2) and (A, F2), (B, F1). Is this a good reason to "double count"?

Of [imath]^{30}C_{15}[/imath] groups of 15, since these are themselves being grouped into 2, we have [imath]\frac{1}{2} \times ^{30}C_{15}[/imath] in each group.


I'm not sure if we could say that [imath]\frac{1}{2} \times ^{16}C_8 \times ^{14}C_7[/imath] will be the distribution over the 2 groups of groups of 15 consisting of 8 boys and 7 girls. Hence I can't find a good numerator for [imath] ^{30}C_{15}[/imath]