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??
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
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?
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.
I am not sure about one part. When you work out the probability of 4/6 heads given you picked a fair coin, why do you not multiply 15/64 by 1/3 since you need to factor in the probability of picking a fair coin in the first place... Other wise isnt this calculation showing the probability of picking 4/6 heads using a fair coin and not what you need to calcuate which is given that you picked a fair coin??
well I agree with you p(B)=P(B|C).P(C)+P(B|D).P(D) but IT IS what he has done: 10:09 1/3*15/64+2/3*15*.8^4*.2^2
1/3=P(C) 15/64=P(B|C) and 2/3=P(D)
P(B|D)=15*.8^4*.2^2
Why the heck do you say he's wrong and then say theright way is to do what he in fact did! Same thing on previous comment verify, and it's clear that's what he did!
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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louisbrassyyy 6 hours ago in playlist Probability
there is no such thing as can't in khan academy
tenseman08 1 month ago 4
Handwriting... o_o
MrAaronsrun 1 month ago
mind was literally blown at the universe drawing at the very end, great work!
cooltoon666 1 month ago
statistics is so fucking tedious
ssips720 3 months ago
Thank you SO MUCH for taking the time to post this!!
djleejdetroit 4 months ago
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 5 months ago
@alexpui82 No, the probability of success and failure is the same. So, [(1/2)^k]*[(1/2)^(n-k)] = (1/2)^n
FrankMattS 4 months ago
You don't have to be a genius to get all that, it depends on how much time you spend on it.
derealson 6 months ago
The handwriting is a little unclear to read..
Alnasrlina 6 months ago
@Alnasrlina this is math, not history. u don't have to understand the handwriting if u get the steps.
Deepskies1 1 month ago
legend
HoffyD 8 months ago
i was lost at 4:20
AnotherBrownKid 9 months ago 3
I don't need all that , can I just borrow your brain for one day "Exam's day" ?
LailaSOAD 9 months ago
very gud vid dude!!!!got it!!! ;)
varun19able 9 months ago
Man that was a difficult to follow
aew782 10 months ago
Unfair coins? Odd. Great vid.
DawnBreaker1100 11 months ago
Dude! you are a genius.
kakamana22 1 year ago 8
Itz cool..... I understand it.......
arunderoyal 1 year ago
If you are confused .. Look at the binomial distribution first
ulvund 1 year ago
You taught me what my teacher was unable to.
Thank you for that!
Search111add 1 year ago
imediatley when you introduced the "fair" and "unfair" approach, you lost me. I think many people will agree with me.
dvssk8er3 1 year ago
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 1 year ago
@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 1 year ago
@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.
Skies678 1 year ago
@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 1 year ago
@sad74 This was 3 years ago -.-
DawnBreaker1100 11 months ago
maybe a remake of video?
haloapa 1 year ago
dam there's like shitloads of formulas for each topic that's wat really confuses me
xxGkSmasher 1 year ago
Is this the 9th video on probability?
vrolok4768 1 year ago
IM STILL CONFUSED
rezaeijavan 1 year ago
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how many videos are on youtube? click to my chanell to find yout
3runrob 1 year ago
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
LordAlda 1 year ago
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.
hexen3x3 2 years ago
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I am not sure about one part. When you work out the probability of 4/6 heads given you picked a fair coin, why do you not multiply 15/64 by 1/3 since you need to factor in the probability of picking a fair coin in the first place... Other wise isnt this calculation showing the probability of picking 4/6 heads using a fair coin and not what you need to calcuate which is given that you picked a fair coin??
milesyounghamilton 2 years ago
Comment removed
milesyounghamilton 2 years ago
It is actually easier to use a tree diagram lol
rinwhr 2 years ago
Thanx alot prof
MrRomanticHeart 2 years ago
i also still in confused.
freakyfas 2 years ago 16
i love you >>> :)
you save a life
foxer071 2 years ago 2
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 :)
woo216 2 years ago
probablility is confusing!
milashkola 2 years ago 2
Trust me you're not alone... I failing it also...
KhazarX 2 years ago 2
i'm so failing my final exam =_=...
SugarySpirit 2 years ago 2
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.
jpcgandre 2 years ago
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....
jpcgandre 2 years ago
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well I agree with you p(B)=P(B|C).P(C)+P(B|D).P(D) but IT IS what he has done: 10:09 1/3*15/64+2/3*15*.8^4*.2^2
1/3=P(C) 15/64=P(B|C) and 2/3=P(D)
P(B|D)=15*.8^4*.2^2
Why the heck do you say he's wrong and then say theright way is to do what he in fact did! Same thing on previous comment verify, and it's clear that's what he did!
2CSST2 1 year ago
Thank's a lot, very nice lectures:)
I hope you keep up the good work and
present us more stuff in this topic.
pachlioptas 2 years ago 2
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)!
dvntrs 2 years ago 3
Sal, calc has scientific mode. Could have saved you a few keypresses lolll
TheEvilDetector 2 years ago 2
Awesome videos btw, i recommended them in a few places.
TheEvilDetector 2 years ago 2
brilliant
maxpurdy 3 years ago
Thank you Sal
samsamisam22 3 years ago
you are awesome youtube is lucky.
joeyjoey1122 3 years ago 3
Great Help !! Thanks
shoutinghell 3 years ago
Hai Sal, Can you please make a video on Probability Density Function and Gaussian Distribution? Thanks mate.
nash911 3 years ago
yes, a little insight into Gauss is needed!
Your work is amazing, Sal, you make all the material accessible and interesting.
dAvrilthebear 2 years ago
thx ur vids rly help me in honors precal
johnthecoolguy 3 years ago
i love probability
volintine 3 years ago
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!
DanMan7997 3 years ago 2