Okay I've asked some website about this and the answer wasn't enough for me

Okay I've asked some website about this and the answer wasn't enough for me.

I've got a friend who is willing to try his luck on some exams, one of which is 40 questions with 4 answers and only 1 right (.25 of getting it right and .75 of getting it wrong)
What are the odds of him getting it right ?

The only answer I had was someone who calculated the event of getting specifically 20/40

0.25^x where x is the number of answers he gets right by choosing completely randomly.
Can we /thread?

No.

My question is about the sum of every event after 20/40, 20 included.
so 21,22,23,... until 40.

You didn't answer it. Just got me the possibility of answering 20 right, the events are supposed to happen at the same time,

Once you get me that with the explanation of which rule you followed, /thread it m8

.25^20 is only the possibility of doing 20/40

>What are the odds of him getting it right ?
you mean at least 20/40?

21/40.

That's not right, I believe the correct answer would be the sum of

(0.25^20) + (0.25^21) + ... (0.25^40)

Am I the only one who has no idea what op is asking?

Let me rephrase the shit :

Suppose you have an exam tomorrow, you don't plan on studying but on trying your luck.
It's a MCQ (multiple choice questions) there's 4 of them and the teacher said there is only 1 correct answer.
What would be the probability of not failing the exam, meaning doing 20 correct guesses at least.
or 21, 22, 23, etc.

Fucking lazy secondhand speaker.

Enjoy failing

>41
>1.21265960236e-12

I guess I could work on my English while you try to correct that, and give me a real probability

What's wrong with his answer? it gives you the correct result.

The range(a,b) function gives you all integers between a and b not including b. That's the reason for using 41 instead of 40.

kek you fucking wrote down the entire number from an image
l2python, this is your probability of success, answering completely randomly.

so 40 questions and 4 answers is 160 questions.
160!=471472363599206132240694321176194377951192623045460204976904578317542573467421580346978030238114995699562728104819596262106947389303901748942909887857509625114880781313585012959529941660203611234871833992565791817698209861793313332044734813700096000000000000000000000000000000000000000
or
4.7147236359921*10^284
Possible combinations.
if 25% of these combinations are correct tat would mean that you have a 1 in 1.178680908998*10^284 chance of answering randomly and getting a perfect score.
Satisfied?

The probability of getting at least [math]20[/math] correct answers is
[eqn]\sum_{i=20}^{40} \left(\frac14\right) ^i \left(\frac34\right) ^{40-i} \binom{40}i[/eqn]

And if thats wrong you can do 4!*40 which gets 960 combos and 1 quarter of that is 240 so 1 in 240 maybe idk

Funny how you're being such a dick about it when you have it wrong. What does your program give for the probability of getting at least 0 correct answers, or at least 6 correct answers? Do those numbers seem reasonable to you?

Jesus, there is so much wrong with this.
OP already knows there is only one answer per question, so it's not 160!, you're implying some retard would check every single box in this scenario, and you're still doing it wrong to calculate that.
There are exactly 4^40 combinations because nobody would bother putting more than one answer per question.

you ruined it and made me smarter at the same time gz

[eqn]\frac{346 01337270 7911438797}{604462909 807314587 353088} \approx 0.000572431[/eqn]

So what are the odds of getting at least 20/40 multiple choice questions right and each question has a .25 chance of being randomly correct

First, enumerate the possibilities.
1: You get at least 20 questions correct
2: You get fewer than 20 questions correct.

Therefore the sample space is 2. Exactly 1 of these possibilities leads to the desired result, therefore the probability of success is 1/2.

Let's find out bitch.

0 correct answers or more is fucking retarded since that has a probability of 1, but I'll show it to you anyway.

You have a shitty 25% chance to begin with. I don't understand what the fuck you're expecting the odds to be.

...

Lol, you still don't get it. Calculate 0 with your program, not by knowing the answer.

Use range(0,41).

You are assuming here that the items are not weighted. Who knows, maybe the teachers has weighted some of the test items to be more valuable, for example sometimes right answer is 1 point sometimes 2.

Also - do you really think he'll get every single question wrong 2/3 of the time if he has a 1/4 chance of getting any given question right? Use some common sense and gain some humility, man.