Martinez39s:
I don't think I would ever try it.

Lol.
Mathematician don't give up hahaha

Martinez39s:
My reasoning is valid for a randomly selected group of 100 students. In fact, throughout this thread, based on my presumption of what your question could possibly mean (as you stated it at first), I have been working with the idea that the group of 100 students were randomly selected.

Let me explain something. When you randomly select a group of 100 students, you have randomly selected a group of 100 birthdays. Each selected birthday has been selected from 365 possible birthdays (dates in a year), and repetition is allowed as you randomly select other birthdays in the group of 100 birthdays. Using the selection of humans, the emboldened statement is tantamount to saying each person from the randomly selected group of hundred has a birthday that is one of the 365 dates (birthdays) in a year, and a person in the group could have a birthday mate in the group (remember that repetition of birthdays is allowed).

When you want to randomly selecting a group of 100 people (birthdays), here are some possible logical questions you can ask if no further information is given or no possible consideration is made (assuming a year contains 365 days):
(1) What is the probability that no two people share the same birthday?

(2) What is the probability that at least two people share the same a birthday Naturalwaves got this, but he applied it to your question which doesn't have anything to do with probability.

(3) What is the number of ways of selecting a group of hundred people such that no two shares the same birthday.

(4) What is the number ways of selecting a group of hundred people such that at least two have the same birthday

This question of yours doesn't make sense: Either you mistakenly left out something or you didn't state the question properly. If this question is complete, and there is nothing more to it, then it's illogical even if you have a unique group of 100 students or a randomly selected group of 100 students.


MrShape, naturalwaves,

From the probability gotten, it is possible to then estimate the number but the problem jackpot has with the solution is that she expects a figure (100-n) and that is not possible given the circumstances of the question.

jackpot:
That's like 1 in 100 cubic trillion trillion chance.

That's why I am saying you should run the experiment. Let's see possibility of getting zero.

I bet that any random sample of 100 that you chose at random will give you about 15-30 persons with birthday mates. That's also surprising considering the fact that there are 365 days in our year.

Hahahaha. Not exactly. I remember how long it took me to find a birthday mate and I had interacted with at least 150 people asking for their birthdays before I saw one. It can be 0 too. Another angle you should look at it from is that the 100 referred to in the question may be a sub group. These sort of questions require a large sample size. The 100 may be a broken-down version from 100,000 because for every 100,000,you will get 3 with an unshared birthday.

Pls lets create a mathematics clinic group chat...... Like online class
Tanx very much

Adedap25:
Pls lets create a mathematics clinic group chat...... Like online class
Tanx very much

my friend, go and pay for the online class fee your school billed you

Your teachers must chop whether COVID-19 or not

naturalwaves:


Hahahaha. Not exactly. I remember how long it took me to find a birthday mate and I had interacted with at least 150 people asking for their birthdays before I saw one. It can be 0 too.

You're seeing it from your perspective (your own birthday matching with others). From your perspective, you'll have to do 150 comparisons of birthdays in your case of trying to see your birthday mate.

The chances of seeing other matching birthday mates increase greatly if you start comparing among all the 150 persons (plus you=151), for then you'll have to do 151Combination2=11325 comparisons

naturalwaves:

Another angle you should look at it from is that the 100 referred to in the question may be a sub group. These sort of questions require a large sample size. The 100 may be a broken-down version from 10000000 because for every 10000000,you will get 3 with an unshared birthday.

I think that with a sound knowledge of statistics especially all these joint probability distribution/density function (PDF), the solution can be readily obtained. It might just be a case of solving one or two multiple integrals as solved in joint PDFs

jackpot:

You're seeing it from your perspective (your own birthday matching with others). From your perspective, you'll have to do 150 comparisons of birthdays in your case of trying to see your birthday mate.

The chances of seeing other matching birthday mates increase greatly if you start comparing among all the 150 persons (plus you=151), for then you'll have to do 151Combination2=11325 comparisons
I think that with a sound knowledge of statistics especially all these joint probability distribution/density function (PDF), the solution can be readily obtained. It might just be a case of solving one or two multiple integrals as solved in joint PDFs

I'm interested in the source of this question o Jackpot.....and I know that whatever is finally solved will definitely not give you the (100-n) you want . It will tend towards 0.00003%.

Please,keep me updated.

Help me to derive a single explicit formula for solving quartic equation.

DeeCherry:


Thank you very much

Hey DeeCherry.. Howdy?

www./1112846675750766/permalink/1129135580788542 please show me some love here, by just clicking on the above link and liking the picture

duchaB:


Hey DeeCherry.. Howdy?

I'm good

DeeCherry:


I'm good

Sorry, I sent you a PM.. Don't know if u have seen it.
I have St important to talk to you about if u don't mind.
Thanks.

duchaB:

Sorry, I sent you a PM.. Don't know if u have seen it. I have St important to talk to you about if u don't mind. Thanks.

Didn't get any

naturalwaves:


I'm interested in the source of this question o Jackpot.....and I know that whatever is finally solved will definitely not give you the (100-n) you want . It will tend towards 0.00003%.

Please,keep me updated.

source: myself really. Question was motivated by the birthday paradox.

Any update on the little experiment concerning analyzing the last 100 players?

jackpot:
source: myself really. Question was motivated by the birthday paradox.

Any update on the little experiment concerning analyzing the last 100 players?

Hahahaha. Chai.
I will check out the PDF tomorrow in my free time. I will also try some other experiments.

Hello house kindly help to solve this.

CC: martinez39s, dejt4u, Richiez, mathefaro, naturalwaves, Mrshape

jackpot:
Hello house kindly help to solve this.

CC: martinez39s, dejt4u, Richiez, mathefaro, naturalwaves, Mrshape

I think game theory will apply here

Good day everyone. I need someone who understands Stochastic Programming (farming problem) well. I have a problem with modelling a problem statement and I need insight. I've modeled it actually just need someone to help me check where I'm wrong.

Kindly help me if you know it or know someone who knows it. I'm ready to pay if need be

Please help me to evaluate these questions using logarithms table

only question 23 and 25

ONLINE COURSES 2020 FOR INTERNATIONALS STUDENTS interested students can apply

FUN QUESTION

There are two packs filled with identical cards as follows: Pack 1 contains 4 red cards and 3 yellow cards; Pack 2 contains 3 red cards and 4 yellow cards.

An experiment consists of drawing/picking a card at random from Pack 1, then mixing it up with the cards in Pack 2, shuffling the cards and then picking a card at random from Pack 2.

Find the probability of picking
(1.) a red card
(2.) a yellow card
at the end of the experiment.

Options:
1 (A) 1/2 (B) 5/8 (C) 3/7 (D) 31/56 (E) NOTA

2 (A) 1/2 (B) 3/8 (C) 25/56 (D) 4/7 (E) NOTA


tags: Mrshape naturalwaves dejt4u Richiez mathefaro martinez39s dejt4u

phemolisti:
Please help me to evaluate these questions using logarithms table

only question 23 and 25

I"ll leave you with the no. 25, you should get it now.

jackpot:
FUN QUESTION

There are two packs filled with identical cards as follows: Pack 1 contains 4 red cards and 3 yellow cards; Pack 2 contains 3 red cards and 4 yellow cards.

An experiment consists of drawing/picking a card at random from Pack 1, then mixing it up with the cards in Pack 2, shuffling the cards and then picking a card at random from Pack 2.

Find the probability of picking
(1.) a red card
(2.) a yellow card
at the end of the experiment.

Options:
1 (A) 1/2 (B) 5/8 (C) 3/7 (D) 31/56 (E) NOTA

2 (A) 1/2 (B) 3/8 (C) 25/56 (D) 4/7 (E) NOTA


tags: Mrshape naturalwaves dejt4u Richiez mathefaro martinez39s dejt4u

I've been so busy teaching online. Will attempt this in my free time.

jackpot:
FUN QUESTION

There are two packs filled with identical cards as follows: Pack 1 contains 4 red cards and 3 yellow cards; Pack 2 contains 3 red cards and 4 yellow cards.

An experiment consists of drawing/picking a card at random from Pack 1, then mixing it up with the cards in Pack 2, shuffling the cards and then picking a card at random from Pack 2.

Find the probability of picking
(1.) a red card
(2.) a yellow card
at the end of the experiment.

Options:
1 (A) 1/2 (B) 5/8 (C) 3/7 (D) 31/56 (E) NOTA

2 (A) 1/2 (B) 3/8 (C) 25/56 (D) 4/7 (E) NOTA


tags: Mrshape naturalwaves dejt4u Richiez mathefaro martinez39s dejt4u

My solution

Data

pack 1...4 RED and 3 YELLOW cards
pack 2.....3 RED and 4 yellow cards

1) Probability of picking a red card at the end of the experiment implies
R,R or Y,R = [4/7 x 4/8] + [3/7 x 3/8]
= 2/7 + 9/56
= 25/56

2) Probability of picking a yellow card at the end of the experiment implies
Y,Y or R,Y = [3/7 x 5/8] + [4/7 x 4/8]
=15/56 + 2/7
=31/56

jackpot:
FUN QUESTION

There are two packs filled with identical cards as follows: Pack 1 contains 4 red cards and 3 yellow cards; Pack 2 contains 3 red cards and 4 yellow cards.

An experiment consists of drawing/picking a card at random from Pack 1, then mixing it up with the cards in Pack 2, shuffling the cards and then picking a card at random from Pack 2.

Find the probability of picking
(1.) a red card
(2.) a yellow card
at the end of the experiment.

Options:
1 (A) 1/2 (B) 5/8 (C) 3/7 (D) 31/56 (E) NOTA

2 (A) 1/2 (B) 3/8 (C) 25/56 (D) 4/7 (E) NOTA


tags: Mrshape naturalwaves dejt4u Richiez mathefaro martinez39s dejt4u

SOLUTION
A pair of choices must be made to produce an outcome in this experiment.
(A) If we pick the 1st red ball in the 1st bag, there are 4 possible outcomes of a red ball, and 4 possible outcomes of a yellow ball. This is also true if we pick the 2nd, 3rd, or 4th red ball in the first bag.

Hence,
--- The number of the pairs of choices starting with a red ball that leads to an outcome of a red ball is O_RR = 16.
--- The number of the pairs of choices starting with a red ball that leads to an outcome of a yellow ball is O_RY = 16.

(B) If we pick the 1st yellow ball in the 1st bag, there are 3 possible outcomes of a red ball, and there are 5 possible outcomes of a yellow ball. This is also true if we start with the 2nd or 3rd yellow ball in the first bag.

Hence,
--- The number of the pairs of choices starting with a yellow ball that leads to an outcome of a red ball is O_YR = 9.
--- The number of the pairs of choices starting with a yellow ball that leads to an outcome of a yellow ball is O_YY = 15.

(C) From (A) & (B),
--- it is clear that the number of the pairs of choices leading to an outcome a red ball is O_RR + O_YR = 16 + 9 = 25.
--- It is also clear that the number of the pairs of choices leading to an outcome of a yellow ball is O_RY + O_YY = 16 + 15 = 31.
--- it then follows that the number of the pairs of choices resulting into an outcome of the experiment is
O_RR + O_YR + O_RR + O_YR = 56

The probability of picking a red ball is 25/56.
The probability of picking a yellow ball is 31/56.

The ancestors have spoken and they have spoken well. I've been overwhelmed by online classes these days. Never thought online classes could be as demanding as this. Well-done guys. May the mathematical force continue to be with us all

mathefaro:
The ancestors have spoken and they have spoken well. I've been overwhelmed by online classes these days. Never thought online classes could be as demanding as this. Well-done guys. May the mathematical force continue to be with us all

Pls I need help with this
Cost price of 12 Oranges is equal to the selling price of 9 Oranges and the discount on 10 oranges is equal to the profit on 5 oranges.What is the percentage point difference between the profit percentage and discount percentage?
A. 20
B. 22.22
C.16.66
D. 15

Aybalance:
Pls I need help with this
Cost price of 12 Oranges is equal to the selling price of 9 Oranges and the discount on 10 oranges is equal to the profit on 5 oranges.What is the percentage point difference between the profit percentage and discount percentage?
A. 20
B. 22.22
C.16.66
D. 15

SOLUTION
Let the cost price of 1 orange be = x
Let the selling price of 1 orange = y

From 1st statement,
12x = 9y
x/y = 9/12
x:y = 9:12
x:y = 3:4
This implies that the ratio of cost price to selling price is 3:4
Thus, ratio of selling price to cost price (y:x) is 4:3
This will require a bit of thinking hereon.

For every 3 things you buy, you sell at a value of 4. Profit is (4-3= 1).This means that your profit is actually 1/3 of your c.p
Thus, % profit = 1/3 x 100 = 33.333%

PART 2- Discount on 10 oranges
Let the marked price be = z
marked price of 10 oranges = 10z
s.p of 10 oranges = 10y (Recall that s.p =y)
Thus, Discount = 10z-10y

PART 3- Profit on 5 oranges
Profit on 5 oranges will be s.p of 5 oranges - c.p of 5 oranges.
Thus, profit on 5 oranges = 5y-5x

FROM QUESTION,
Discount on 10 oranges = profit on 5 oranges
Thus,
10z-10y = 5y- 5x
Recall that, x= (9/12)y from equation 1
10z-10y=5y-5(9/12)y
10z= 5y-(45y/12) + 10y
120z= 135y
8z=9y
y:z = 9:8
y-z = 9-8 = 1
This implies that discount value is 1/9 of the marked price.
Thus, discount % = (1/9 x 100) = 11.111%

Finally, % point difference between the profit % and discount % = 33.333% - 11.111% = 22.222%. Thus, the correct answer is option B

the correct answer is option B [/b]
[/quote]Thanks natural waves.How come it is an objective question.Am asking because of how long it will take to do all these

Aybalance:
the correct answer is option B [/b]
Thanks natural waves.How come it is an objective question.Am asking because of how long it will take to do all these

You are welcome.
Where did you get the question from? It depends
I know what they do in some places is to memorise the process and then start solving any question of this form in a very short time.
Do you know that because you know the method now, you can solve any question on this under 1 minute if it follows the same pettern?
Just look for your sp:cp ratio and get your %
do same for discount:sp and get your % and you are good to go.
If one is just seeing such for the first time, it will take a while o.
Do you know that I spent at least 15 minutes on this question trying to look for the best approach that will work?

naturalwaves:

You are welcome.
Where did you get the question from? It depends
I know what they do in some places is to memorise the process and then start solving any question of this form in a very short time.
Do you know that because you know the method now, you can solve any question on this under 1 minute if it follows the same pettern?
Just look for your sp:cp ratio and get your %
do same for discount:sp and get your % and you are good to go.
If one is just seeing such for the first time, it will take a while o.
Do you know that I spent at least 15 minutes on this question trying to look for the best approach that will work?

It is a GMAT test past question.15 minutes....Lol that's almost the time for the whole test....seems the only green flag is to have crossed the question before.
I also noticed in your solution that y:z=9:8 is z:y=9:8 and z-y=1 as there is a discount