Probability & Statistics (11 of 62) The Probability Function - Flipping Coins - General Formula 2

[@rolandastaurage]

should be (where n>=k>=0)

[@sciencelearnremember]

I agree as well.

[@RowanHumphreys]

this stumped me too.

[@mainakdutta905]

yes,it should be

[@inkymagician3996]

i instantly went in comments to double check :D

[@nolimangulabnan6101]

I noticed as well

[@slokisofttech5958]

you are great sir, your videos are awesomely knowledgeable.

[@EP-rq3pn]

Thank you very much, professor! I would love to go deeper and will watch the rest of the playlist. Have you by chance made any videos that elaborate on why using factorial works in the first place? I'm really interested in knowing

[@MichelvanBiezen]

I am not sure, but I am making a note of it and will place it under the viewer request videos.

[@gothicsire]

Best probability and statistics professor ♡♡

[@MichelvanBiezen]

Thank you. Glad you found our videos. 🙂

[@andrewpersaud4144]

using the pascal and binomial way you showed last video for small n. very handy

[@akbarmirzo]

"and that's how we do that" yes that's how we do that

[@mayuripopat2758]

Thank you so so much..you made all the concepts of probability much easier...you are an wonderful teacher..once again thank you...

[@GenuinePeacefulTimes]

I think your inequities symbols are in the wrong order shouldn't it be k is greater than zero and less than n?

[@munishyadav456]

The purpose of dividing is to reverse a multiplication which has taken place, which should not have taken place. Generally, the multiplication "should not" have taken place because two outcomes are not "distinct", i.e. they are the same. For example if I ask how many distinct ways two coins can turn out, you will say there are 2 ways for the first coin, and 2 ways for the second so the answer is 2×2=4. However if I then tell you that one of the coins is a double-headed coin, you now know that the 2nd multiplication by 2 is in error, as there is only one distinct way that coin can turn out. So you divide by 2 to give your answer 4/2=2.

[@SaptarshiSyam]

Wonderfully explained!!

[@MichelvanBiezen]

Thank you. Glad you liked it. 🙂

[@an0n0896]

thank you. this helped me so much

[@MichelvanBiezen]

KZbin offers different settings. You can set it to a higher resolution for a better picture.

[@markjones6133]

He didn't derive the formula. He just gave us the formula. Please give the derivation of the formula.

[@nicouh]

yeah, thats what i came here for too... i already knew the result, but that does not mean i understand it... :|

[@wirito]

If you guys are watching these videos I assume you have taken discrete mathematics. You should have seen the proof of this formula there.

[@Peter_1986]

The derivation for this formula is actually pretty simple, and surprisingly intuitive. Think of it this way: the number of ways that you can arrange _k_ heads is really just the Choose function, because the Choose function tells you in how many arrangements you can choose the _k_ heads - for example, if you have 2 heads and 1 tails then you can arrange the 2 heads in 3 different ways, which are of course HHT, HTH and THH. This would be the same thing as "3-Choose-2" = 3, and this is where the Choose function comes from. Then the _denominator_ tells you the number of _all possible_ outcomes, which are 2 outcomes for each of the _n_ flips, which therefore gives you 2^n. So the equation in the video simply tells you "the ratio between all the arrangements of the heads and all the possible outcomes".

[@MrWinter2]

Thank you so much! Another wonderful video!

[@fishplug10]

Awesome videos thank you!

[@jaijeffcom]

Is that symbol for n over k read "n choose k?" I would have enjoyed seeing the proof of that formula. Doesn't that relate to binomial expansion and Pascal's triangle?

[@Peter_1986]

The numerator "n-choose-k" tells you how many ways you can get heads (or tails), and the denominator 2^n tells you the total possible number of heads-tails arrangements (which is the sample space). Say for example that you want to calculate the probability of getting 2 heads in 4 tosses; if you start with simply checking the probability of getting two heads on the _two first_ tosses --- in other words, the heads-tails arrangement HHTT --- then you will get the probability (1/2)^4 = 1/16 for that particular arrangement, since each toss has a probability of 1/2 to happen both for heads and tails. _But,_ you can _also_ get 2 heads in several _other_ arrangements, and you can calculate the number of those arrangements by using the Choose function, which in this case would be "4-choose-2" = 6, which says that there are 6 ways to arrange the 2 heads after 4 tosses. So the probability of getting 2 heads would then be 6/16 = 3/8, because, _out of 16 _*_possible_*_ arrangements,_ you can arrange the heads in 6 different ways (HHTT, TTHH, HTHT, THTH, HTTH and THHT).

[@aneeessh]

excellent class sir

[@MichelvanBiezen]

Many many thanks

[@kevindave277]

Rocked my world.

[@MichelvanBiezen]

Glad you found our videos! 🙂

[@hasnainjaved6567]

if I take 1 coin n=1 , head=1 according to formula n-k where is taill (1-1)=0?? anyone ? help if you understand . Thanks

[@MichelvanBiezen]

It is not n-k but (n-k)! and 0! = 1

[@Clifffffffffford]

Thanks!

[@TheGainadl]

I want to demonstate that after flipping 100 coins the probability of having between 40 and 60 number of heads is higher than 0.7..... And I end up here watching this high school thing that I already knew .... I am the only one?

[@milosmladenovic7822]

Ha, no bro! Almost the same here..

[@zenwoo7759]

I was wondering where did i hear his voice, man has a voice that sounds like GRU from despicable me.