₦airaland Forum

ladyGKilaBCrueD:
I like
can I see a pic?

No.

ladyGKilaBCrueD:
do u work out and do u have 6 packs

I work out, and I have six packs that are visible but not too prominent as I want.

pansophist:
Good. Just as you're getting an ideal look, your mood will also become better. I am currently cutting aggressively and I noticed my mood gets better with time, I can also see my cheekbone. Did you experience such?

I also took up fitness challenge since last month. Lost 5kg of pure fat. Doing the final cutting with psmf diet, it's a hard diet but very effective. True, dieting is the most important and also the hardest part. You look good mate. Keep it up.

Did you also notice that the more one exercises and gets "fitter," one doesn't succumb to stress easily as before?

Mrshape:
Lol. I will try that tomorrow I am tired too o.

I don't think I would ever try it.

@Jackpot

Here is an edited post of mine explaining your observation on the experiment of randomly selecting 100 people (I wrote this when you asked me to try the experiment):

Martinez39s:
Sure! Even though any outcome is possible, I will way most likely get at least two people with the same birthday. This is because the probability of having at least two people with the same birthday is 0.99999969275 (99.999969275% chance). The probability of not having anyone with a birthday mate is 0.00000030725 (0.000030725% chance). So you see that the event of having a selection in which no one has a birthday mate is a very very very rare one just like winning a mega lottery. I calculated these probabilities in a post of mine to you.

MrShape, naturalwaves,

From these probabilities, you can see why you keep getting people with the same birthdays in your selection—the odds are greatly stacked in your favour (you have a 99.999969275% chance of getting at least two people with the same birthday). This is the same reason why people keep getting unsuccessful in their attempt to win the lottery— the odds are incredibly stacked against them. The popular Mega Millions multi-state lottery has odds of approximately 1 in 175,711,536. This is a 0.000000569% chance.

jackpot:
It's unfair to tag what we don't understand as illogical
If I see a question that is indeterminate, I will know.

Nah! *yimu* Your question wasn't properly phrased, and it didn't make sense the way you presented it. I now understand what you were trying to ask. You should have asked "if 100 people are selected randomly what's the LIKELY number (or Likely range of number) of people who have birthday mates? This question is incredibly difficult to solve, that's if it has an answer.

For instance, do you know that some great guys think that that angle question I posted earlier was not complete? Yet, Mrshape and others dealt mercilessly with the question here.

The angle question is complete; I never thought it wasn't. Any mathematician who thinks that question isn't complete isn't that great afterall.

Do you know that questions like the following have an answer?

Let a, b, c be real numbers selected randomly from the interval (0, 1). What is the probability that the quadratic equation ax²+bx+c=0 has at least one real solution?

Just imagine where one will start. Mine is better that it can even be experimented

Ahhh! I won't even bother with this question. I can't kill myself.


MrShape, naturalwaves

jackpot:
i know, but try first

I bet you might even get up to 23 players with birthday mates if you run the experiment.

Sure! Even though any outcome is possible, I will way most likely get at least two people with the same birthday. This is because the probability of having at least two people with the same birthday is 0.99999969275 (99.999969275% chance). The probability of not having anyone with a birthday mate is 0.00000030725 (0.000030725% chance). So you see that event of having a selection in which no one has a birthday mate is a rare one just like winning a mega lottery. I calculated these probabilities in a post of mine to you.

MrShape, naturalwaves, Jackpot

@Emmaodet

Making up nonexistent problems, or exaggerating their difficulties, and blaming these "problems"/difficulties on men and patriarchy are the natural talents of feminists. Many are already professionals and veterans in this talent.

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.

The chances of winning a mega lottery in the United States of America is approximately 0 yet people still win. It would therefore be false to say that ANY person who plays the lottery will fail. The fact that a large group of people played the lottery 200 times without success isn't proof that no one can win the lottery. The odds of an event might be very large, but as long as the probability of that event isn't exactly 0 then that event can happen.

jackpot:
that is why we are assuming a random group of 100 nah.

By the way, in a random experiment of the question, the possiblity of the cases you mentioned happening is almost zero.

See, I believe that we are learning.

Download that PDF and run the experiment with the last 100 players. Tell me your result.

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?

Ans: ³⁶⁵P₁₀₀ ÷ 365¹⁰⁰ = 365! ÷ (265! × 365¹⁰⁰) = 0.00000030725

(2) What is the probability that at least two people share the same a birthday

Ans: 1 – (³⁶⁵P₁₀₀ ÷ 365¹⁰⁰) = 1 – [365! ÷ (265! × 365¹⁰⁰)] = 0.99999969275

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.

Ans: ³⁶⁵P₁₀₀ = 365!/265!

If the question has order disregarded, your answer is ³⁶⁵C₁₀₀ = 365!/(100! × 265!)

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

365¹⁰⁰ – ³⁶⁵P₁₀₀ = 365¹⁰⁰ – (365!/265!)

This question of yours doesn't make sense:

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

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,

sherylbakky:
which biochemistry and pharmacology? The same BCH n Pharmacology that we know?

Some people will just be hyping their undergraduate course in Nigeria, thereby giving themselves pious hope. I remember how a friend studying Marine Science was confidently saying that he could be employed after school in some organisation, cooperation, and company. The guy is now a photographer. I haven't contacted him in a long time. Perhaps he might have gotten employed by an organisation, cooperation, and company.

@Jackpot
In my last reply to you, I didn't mention that I haven't been following this thread thoroughly. Is the rephrased question you sent to me complete? Perhaps you left out the fact that certain number of people shared birthdays in a certain way. Perhaps, your rephrased question is indeed complete, and therefore irrational.

Here is my last reply:
https://www.nairaland.com/1147658/nairaland-mathematics-clinic/270#89041840

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?

Your question doesn't make a shred of sense, or it is incomplete. If I have 100 students in a class, I need to know their birthdays before I can know the expected number of students that have no birthday mates because there are 365¹⁰⁰ possible ways a group of 100 students can be assigned birthdays.

Let's pick four possibilities with a class of 100:
1) If no one has a birthday mate, the expected number of students with no birthday mates is 100.
2) If only two students share the birthday and others don't have birthday mates, the expected number of students with no birthday mate is 98.
3) if three students share the same birthdays, two students share another birthday, and the rest have no birthday mates, the expected number of students with no birthday mates is 95.
4) If all students share the same birthday, the expected number of students with no birthday mates is 0.

So you see, you need to know the birthdays of each student to know the expected number of students with no birthday mates.

Edit: I could list many more possibilities. In fact, I can list a possibility to arrive at any expected number (from 0 to 100) of students with no birthday mates.

jackpot:
I think that you are solving a different question

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

I said it, but others were telling me I need probability. Naturalwaves found the probability of selecting a group of 100 in which at least two share a birthday assuming we have 365 days in a year. His solution doesn't make sense because number of birthdays can't be a decimal, neither can it be a probability.

Nevertheless, I am yet to understand your question.
•• If you are looking for the number of possible ways in which no two people in a group of 100 share the same birthday (assuming we have 365 days in a year) then your answer is

³⁶⁵P₂₆₅ = 365!/265!

•• If you seek the number of possible ways in which at least two people in a group of 100 share the same birthday (assuming we have 365 days in a year) then your answer is

365¹⁰⁰ – (³⁶⁵P₂₆₅) = 365¹⁰⁰ – 365!/265!

I repeat, I still don't understand your question except I presume what you could possibly mean.

xendra:
congratulations OlawaleBammie


if I tell you where I work you will beg me for money

but its too easy to find so I can't mention here, that's the luck you have

don't say things you don't know about people just to get attention and it will be well with you.

peace.

Did you read the op? He wasn't insulting you or trying to pick a fight.

GreatResearcher:
What would u say about the guy who. sent davido baby mama 150k just to get her reply his chats?

dejt4u:
you can your expected values from probability.. It is even simpler in this case since our N is 100

Well, I am not deep into probability (and combinatorics.) I assumed the question solely had to deal with permutation.

naturalwaves:
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
:
:
:
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: Jackpõt, Dèjt4u, Martiñez39s, Mrshãpe, Rićhiez, mathêfaro, ifadã123, Goalñaldo

This is also one of the best advanced calculators available: https://keisan.casio.com/calculator

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

Jeweltz:
I don't think so. I still got guys from this forum on my wait-list who has been consistently sending me recharge- ranging from #1500#2000 and More. More like to say since January, I Neve use my money buy card o . They're all waiting for my validation.always praising me on Whatsapp lol. The wicked me only respond their chats when I need help

I feel bad for those boys. What simping.

Jeweltz:
What did I just read?

A guy is sending card to women to unblock him on social media.

Please send me money to chat with mee. I'll open my DM

Too late. GreatResearcher1 has gotten smarter and better.

A feminist-like strawmanning engineered to make women the victims while vilifying males as opressors, and subtly elevating the female gender above the male gender. Such bleating and whining over a nonexistent problem (or at least, a problem not perpetrated by the overwhelming majority men) is typical of feminists, especially when they cry victim.

If you have equal rights and opportunities as men, but you still prefer to be a victim, and attribute blame to men, society, etc. over a nonexistent problem, you are on your own.

48 hrs.

whitelotus:
Some people were arguing that one should just get married at 25 and everything will fall into place.

The stupidity those people are displaying is a dangerous one.

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

Do you actually mean the number of possible ways in which the students can have their birthdays without any two sharing the same birthday? If that is what you mean, your answer is

³⁶⁵P₁₀₀ = 365!/265!

It doesn't pay to be bluepilled. It pays to be redpilled. Take the redpill today. Op, in addition to these links, I will release my redpill series and topics starting from 1st of September.

https://www.nairaland.com/5695941/many-memes-redpill-guys-get

https://www.nairaland.com/5150034/boys-get-here-mumu-must


The posts below me should be heeded.

ichidodo:
[color=#1980BC] CIA and FBI combined cannot match up to a woman when it comes to manipulation, espionage and subterfuge... especially when potency to lead to marriage is involved... That's why we must involve our mothers or elderly aunties to investigate our potential mates as it only takes a woman to catch a woman...Next time don't be a simp...[/color]

Biglittlelois:
Lol all these guys going for churchy ladies thinking they are all saints, Pele, you've taken the first step which is separation, you can as well divorce her and have peace of mind.

palsenator:
With respect bro, there are verified incidents where the laws have been suspended. It's unexplainable but they happen. That we. you have not experienced it does not mean they don't happen.
I don't know the angle to come from because I want to believe you ain't an atheist. Please respond to this and I will send you some verified incidents to prove my point.

I am a atheist. Please show me with proof these "verified" incidents in which the laws of nature were suspended.