Added: 3 years ago
From: khanacademy
Views: 82,681
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  • there is no such thing as can't in khan academy

  • Handwriting... o_o

  • mind was literally blown at the universe drawing at the very end, great work!

  • statistics is so fucking tedious

  • Thank you SO MUCH for taking the time to post this!!

  • Not sure if its just me, but could there be a mistake with the fair coin P[H = 4/6] equation? Looks like Sal missed multiplying the binomial probability equation by its 'no success' term, which is (1/2)^ (n-k) , where n = 6, and k = 4??

  • @alexpui82 No, the probability of success and failure is the same. So, [(1/2)^k]*[(1/2)^(n-k)] = (1/2)^n

  • You don't have to be a genius to get all that, it depends on how much time you spend on it.

  • The handwriting is a little unclear to read..

  • @Alnasrlina this is math, not history. u don't have to understand the handwriting if u get the steps.

  • legend

  • i was lost at 4:20

  • I don't need all that , can I just borrow your brain for one day "Exam's day" ?

  • very gud vid dude!!!!got it!!! ;)

  • Man that was a difficult to follow

  • Unfair coins? Odd. Great vid.

  • Dude! you are a genius.

  • Itz cool..... I understand it.......

  • If you are confused .. Look at the binomial distribution first

  • You taught me what my teacher was unable to.

    Thank you for that!

  • imediatley when you introduced the "fair" and "unfair" approach, you lost me. I think many people will agree with me.

  • Can you correct my train of thought here ... given a fair coin, if you flipped it 6 times, i thought the probability of getting 4 out of 6 heads exactly was equal to the probability of getting heads 4x and getting tails 2x ... since there are only two options, the probability of getting 4 heads exactly is (1/2)^4 ...  i'll continue the rest of the message in a subsequent message

  • @finess4eva

    likewise, the probability of getting two tails is (1/2)^2, hence i thought if given a fair coin and you flipped it 6x, the probability of getting 4/6 heads exactly was (1/2)^4*(1/2)^2, which is 1/64

    though i get why 15/64 is valid if you use the combination formula I was wondering why you felt the combination formula was appropriate for figuring out what is the probability of getting 4 of 6 heads exactly if you flip a coin 6x, given a fair coin, i look forward to hearing from you

  • @finess4eva Isn't it because, 1/64 is true for one defined sequence however there isn't just one sequence where 4/6 heads occur. So we need to figure out how many different ways 4 heads can appear in a sequence of 6 flips. Thats why we use the combination formula.

  • @finess4eva this is because you can get Heads and Tails in different order. For eg. HHHHTT, HHHTHT, HHHTTH and etc.

    If based on your interpretation, P(1H) = 1/64, P(2H) = 1/64, P(3H) = 1/64, P(4H) = 1/64, P(5H) = 1/64, P(6H) = 1/64. Total = 6/64. Then where is the rest of 58/64 of the throws?

  • @sad74 This was 3 years ago -.-

  • maybe a remake of video?

  • dam there's like shitloads of formulas for each topic that's wat really confuses me

  • Is this the 9th video on probability?

  • IM STILL CONFUSED

  • lol... I actually thought that the unfair coins were 8 "2-sided Hs" and 2 "2-sided Ts". And, came out with the probability of 1.

    :-P

  • Wow. Tnx for this dude. I never really pay attention during classes so I always end up finding lessons in youtube. I liked the way you explained it. Very understandable. Good work.

  • Comment removed

  • It is actually easier to use a tree diagram lol

  • Thanx alot prof

  • i also still in confused.

  • i love you >>> :)

    you save a life

  • nice i prefer not using probability formulas cause i think their confusing and dont allow you to gain intuition you done a great job sal :)

  • probablility is confusing!

  • Trust me you're not alone... I failing it also...

  • i'm so failing my final exam =_=...

  • The result is ok. however this is the formal way:

    P(F|4/6H)=P(4/6H|F)*P(F)/P(4/6­H)

    P(4/6H|F), P(F) you know

    P(4/6H) use the total probability theorem.

    and that's it.

  • Wrong! The p(B) is not p(B|C)+p(B|D) but

    p(B)=P(B|C).P(C)+P(B|D).P(D) if C and D are mutually exhaustive. Total probability theorem....

  • Thank's a lot, very nice lectures:)

    I hope you keep up the good work and

    present us more stuff in this topic.

  • Hey,

    I have a midterm exam for tomorrow and you helped me so much dude, especially with that PDF video, please keep up

    Greetings from Turkey (see how far you reach)!

  • Sal, calc has scientific mode. Could have saved you a few keypresses lolll

  • Awesome videos btw, i recommended them in a few places.

  • brilliant

  • Thank you Sal

  • you are awesome youtube is lucky.

  • Great Help !! Thanks

  • Hai Sal, Can you please make a video on Probability Density Function and Gaussian Distribution? Thanks mate.

  • yes, a little insight into Gauss is needed!

    Your work is amazing, Sal, you make all the material accessible and interesting.

  • thx ur vids rly help me in honors precal

  • i love probability

  • If you have time can you go into more statistics related stuff like bell curves? I would find more data analysis stuff quite useful. Also, the probability stuff was a great help, thanks!

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