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

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

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.

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.

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.

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:
when a random experiment gave me 85 without birthday mates (and 15 with birthday mates) out of 100

If anything, the two are not close. Random experiments often give clue to magnitude of the answer.

Try and run the experiment with a section of the players data. It helps to appreciate results when they come.

But you may use a consecutive 100 players for the experiment to have semblance of randomness

A random sample can also give you 0 people without birthday mates in a size of 100.

jackpot:
my question is different from the birthday paradox. Birthday paradox has it's origin in the probability of at least a member of a group of size k (an integer>2) having a birthday mate.

But birthday paradox helps a bit to understand the question.

If the probability that a member of a group having a birthday mate in a group of 100 is 99.99997%, one can infer that on average those that have birthday mate(s) in a random group of 100 are a significant number x.

I'm looking for (100-x) being number of those that don't have birthday mates.

I think I get what you are trying to say now

Modified :
....... but you must realize that what you're asking is impossible to determine without towing the probability lane. The best we can do is to calculate the probability of an unshared birthday and then tweak it. Take for instance if the probability of an unshared birthday is 1/5. It means that for every 5 people, 1 won't share birthday and since we are looking at a total of 100 students, you will get 20 students in all as what you desire.

Unfortunately, in this case the probability is marginal and the number you require is about 3 in a 10 million samples or something like that.

Cc: jackpot, Mr. Shape

jackpot:


Okay, here's the info. I had to look for the sheet of paper that I used


2 shared 13th January
2 shared 5th February
2 shared 9th August
2 shared 31st August
3 shared 2nd October
2 shared 16th October
2 shared 16th December
Total of 15.


The rest 85 didn't have birthday mates.


I hope all is clear now?

Thanks for the clarification.
This is my points exactly.
Based on this example, 15 people have shared birthdays but the number of shared birthdays is just 7. That's the difference I was talking about.

The question requires.... Number of unshared birthdays from the group and not number of students with an unshared birthday.

Thanks Mr. Shape for talking about the birthday paradox. I had to quickly read on it on wiki and then I was lucky to get this video. Please, let us take time to watch it. The dude in the video even used 100 people to illustrate from time stamp 5:33 Please, try and watch.
CC: Martinez39s, jackpot.


https://www.youtube.com/watch?v=ofTb57aZHZs

Ps: The dude also got 0.00003 as his answer.

jackpot:

Among the 15 players, 6 pairs of players shared 6 different birthdays while 3 players shared the same birthday.

So, 6×2+3=15.

The rest 85 players do not have birthday mates.

See what you wrote now Jackpot
You initially said 15 have shared birthdays.
Now, 6 pairs =12
Hiw did they share different birthdays again. I think you didn't understand my question.

I actually meant that, for the number that have their birthdays shared, what is the distribution like? E. G if 15 people share birthdays, 4 can be April 30 and the remaining 11 can be May 1. That's the sort of info I requested.

jackpot:
no, 6+1=7. I get your point about shared birthdays but that isn't the angle of the question.

But I mean how many students in a group of 100 do we expect to not share their birthdays with others in the group? In other words, how many do not have birthday mates?

Like in the data, 85 didn't have birthday mates.

I am coming. Let me check what you posted again but you got the basic drift anyway. shared birthdays may be 7 but 15 players may have a shared birthday.

Mrshape:

The question how many birth dates are shared can't be answers with does information, it is a lot of possibilities

I mean from the figure she supplied about the 15 people. She has answered that and I saw that it is 13 birthdays that were shared amongst the 15 people. I asked because I wanted to make her realize that the numbers cannot be equal.

jackpot:


You have 100 students only in a class. What is the expected number of students in the class that do not have birthday mates?

Unfortunately, this is not the question. I still maintain that you're confusing those 2 things. Question requires an unshared birthday and not individuals with an unshared birthday. That's one part you need to iron out to start with.

jackpot:

Among the 15 players, 6 pairs of players shared 6 different birthdays while 3 players shared the same birthday.

So, 6×2+3=15.

The rest 85 players do not have birthday mates.

That means that 15 players share birthdays but only 13 birthdays are shared.

jackpot:
the birthdays belong to somebody na

What's the difference?

This is the difference.

20 students can have shared birthdays but
Shared birthdays can be just 1 if the 20 were born on exactly same day.
20 students can even have 3 shared birthdays if for example 4 of them have same day, 13 have another day and 3 have another day. That still gives 20 students with a shared birthday but only 3 shared birthdays.

jackpot:



Okay, for the first 100 players in the PDF whose link is shown, I found out that
15 share birthdays while 85 didn't share birthdays.

However, this gives just an idea of the problem but is not the answer at all

You should also be careful not to confuse "unshared birthdays" with "number of students with unshared birthdays" The question requires the former.

jackpot:
Sure.

From that data, 85 have distinct birthdays but 15 must have at least a birthday mate. Note that as you noticed, this is different from saying that 15 share the same birthday.
A lot. I just considered the first 100 so that I will not deviate from the question.

You've not answered my question. From the 15 people with shared birthdays, how many birthdays are shared
15 people can share just 1 birthday.

jackpot:



I think that the answer you got is the probability of 100 persons having distinct birthdays.


I don't think E=NP should work here. If you multiply by 100, it's still a negligible number and it will tend to invalidate your earlier probability (because I think that your probability answer is saying that it is almost impossible not to see a shared birthday among 100 persons). Check again.

No, my answer says it is almost impossible to see an unshared birthday which means that the probability of having a shared birthday is higher.

jackpot:



Okay, for the first 100 players in the PDF whose link is shown, I found out that
15 share birthdays while 85 didn't share birthdays.

However, this gives just an idea of the problem but is not the answer at all

The question is not as basic as you see. There is a large difference between 15 students sharing birthday and number of shared birthdays. 15 students can share just 1 birthday. 15 students can share 3 birthdays etc etc . How many birth days are shared in the document?

jackpot:
I think that you are solving a different question

The question says expected number of unshared birthdays among 100 students?

The answer I got is the probability that there won't be an unshared birthday.
Multiply the answer by 100 to get your answer. That's a negligible figure as it is far far less than 1.

Mechanics96:
.


I saw your solution, I solved it same way too.

Thanks for adding this �

Great!

ifada123:

Let's go.

Probability of first person number having birth day with another is 364/365
Second 363/365
Third 362/365
......... 265/365
So
Prob of not have same of 100 persons will be =

(First ) and (second ) and (third)and....×(100th no)
=(First not) x (second not) x (third not)x.....(100th not)
(364/365)×(363/365)×(362/365).......x(265/360)
=(1/365)^100 x 364 P100
=

Almost everyone will share same birth day

The last number for the 100th was supposed to be 266 and not 265. Your solution is similar to mine just that you were rushing when solving.

You were supposed to compute as (365!)/(265!*365¹⁰⁰).

.

dejt4u:
always proud of you. You're a good teacher. Keep it up bro!

Thanks bro. I do appreciate you very much as well.

Mechanics96:


Exactly what I have solved here, *just added that the probability of having at least a pair of students whose birthday is the same will be:
1-0.0000003073 which is 0.99999... And that's approximately 1.

So if we pick out a pair of students at random assuming that they are sure to share birthdays (probably 1) then we have 98 students left, out of which there may still be some pairs of same birthdays. There's got to be how to get a specific number of students who are likely not to share birthdays... Maybe someone can help figure this out.

If the probability of having 2 students with the same birthday like you said is almost 1 which I agree with. THEN, the probability of not having 2 same birthdays will be extremely close to 0.

Moreover,the fact that there are 100 physical students does not mean that to get the figure from the group, you must have a digit or something. Your probability can even tend far far towards 0. Check my solution on the previous page.

Martinez39s:
Jackpot never mentioned anything about probability. How did probability get involved in the solution?

Lmao. She used term expected number in a group of 100 which points towards probability. If you need to get the number out of 100,just multiply the answer by 100. E. G if you have (1/4) =0.25
as a probability which means 1 out of every 4. If you now want to get the 1,you will need to multiply 4 by 0.25.So,multiply 100 by the answer to get the figure. You won't still get up to even 1. What does that say? It means that the probability that no 2 persons will share same birthday out of the 100 is negligible.

Goalnaldo:
so the answer is not complete yet?

Look up now. The solution is there. Let me tag you on it so you can get it.

dejt4u:
but in reality, this is not valid

How is it not valid bro in reality?? If you have a group of 100 people, the probability that none will share same birthday is almost non existent. The probability that at least 2 people will share the same birthday will be quite high. Therefore, it is very valid in reality.

jackpot:
Maybe we should try to simulate it using real data. This is the [url]https://img.fifa.com/image/upload/hzfqyndmnqazczvc5xdb.pdf [/url]for 2018 fifa world cup data with the players birthdays.

We can decide to pick any consecutive 100 players. Either from top, middle or bottom.

Let's go!

Did you see the solution I posted?

dejt4u:
I think I like this

Question is quite tricky and requires a high thought process. Will get a calculator to compute final answer soon.

jackpot:
Among a group of 100 random students, what is the expected number of students who don't share birthdays with others in the group?

Note: by birthday, it means same month and day, and not necessarily year.


Assumptions: none of the students were born on February 29 and that number of days in a year is 365.

Mrshape, dejt4u, martinez39s, Richiez, mathefaro, naturalwaves, other mathematicians please help

SOLUTION
We need to take a step by step approach in order to get this.
Since 29th of February is not included, this means that we are dealing with just 365 days.
Having said that, what is the expected probability of the 1st student? That is going to be 365/365 days
What is the expected probability of the 2nd student? 364/365
Third? 363/365
Fourth....362/365
Fifth..................361/365
This is how it will go on till we get to the 100th student which will be 266/365 but we cannot do this as it will be too cumbersome.
Now, let us sort the denominator. Every component in the denominator is 365 i.e 365/365 * 364/365 * 363/365................*266/365.
This means that the denominator will be 365¹⁰⁰.

How about the numerator? We have 365! but we are not going to be needing the whole of it. We only need the area from 365 to 266. How do we get this?
All we need do is to divide 365! by 265!.

Thus, the desired solution to the problem is;
(365!) / (265! * 365¹⁰⁰).

Modified:
So, I have gotten a calculator,
(365!) / (265! * 365¹⁰⁰). = 3.0725 * 10^-7 approximately.
That is around 0.00000030725 approx.
As you can see, the probability is very slim as it is almost unlikely that out of 100 students, none will share a birthday which makes a lot of sense.

cc: Jackpot, Dejt4u, Martinez39s, Mrshape, Richiez, mathefaro,ifada123, Goalnaldo

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