Two Heads, hehehe, that's what I need to understand numbers. But your really great educator, a natural born educator. I've watched your other videos (non-math related) and understand even the most complex ideas. I love the way you repeat yourself and really want your viewers to understand the concepts. You're great.
@CarineFrisch You would want to multiply the two probabilities, or simply 0.5·0.6=0.3. But if you want to calculate the probability of that happening, then selecting a fair coin is 5/8 times probability of selecting unfair coin is 3/7 (assuming coin isn't replaced), or vice versa probability of selecting unfair coin is 3/8 times probability of selecting fair coin is 5/7 (they both work out the same). Then multiply by each respective probability of landing Heads, and you get 3/8·5/7·0.5·0.6=0.08
Hi Khan, Please tell why are you not considering the case where one head is fair and other one is unfair? It's quite possible that we can get one head from fair and other from unfair coin. So you should also consider the probability of these cases.
What happens if you take a fair coin on the first pick and an unfair the second time. Should you not calculate that probability and then add to the other two P(Fair and heads) and P(Unfair and heads)
what is the difference between, getting two heads in a row knowing that the coin was a fair one, and getting 2 heads in a row whilst actually getting a fair coin? For dependent events, what is the difference between P(HH, given a fair coin) and P(Fair and HH)?
The first one means that the the event preceding the coin toss ie the picking of the fair coin has already happened, so the chances of that happening is 100% But in the second case the first event still hasn't happened, so the chances of picking the fair queen will not be 100%, and so we would need to calculate the chances and multiply it together because they are indpendent events
can u help me how to solve this A store sells two types of tables: plain and deluxe. When an order for a table arrives, there is an 80% chance that the plain table will be desired. (a) Out of 5 orders, what is the probability that no deluxe tables will be desired? (b) Assume that each day 5 orders arrive and that today (Monday) an order came for a deluxe table. What is the probability that the first day in which one or more deluxe tables are again ordered will be in three more days (Thursday)? What is the expected number of days until a deluxe table is desired? (c) Actually, the number of orders each day is a Poisson random variable with a mean of 5. What is the probability that exactly 5 orders will arrive on a given day?
My question: lets say we have a bag where there are 10 balls. 1 of em is red and the rest are white. i pull something out of the bag 5 times. whats the propability of getting red in any of these times?
There is 1/10 + 1/9 + 1/8 + 1/7 + 1/6 possibilities if we suppose you don't put back the ball again. If you put ball in again then you have 1/10 + 1/10 + 1/10 + 1/10 + 1/10 possibilities.
You do an awesome job of breaking everything down visually so that I can grasp the concept easily. Thanks SO much!
4:05 My mind on this moment: "oh no"💀
Two Heads, hehehe, that's what I need to understand numbers. But your really great educator, a natural born educator. I've watched your other videos (non-math related) and understand even the most complex ideas. I love the way you repeat yourself and really want your viewers to understand the concepts. You're great.
i like to class with khan academy :)
Great, and I just had a probability test today.
How did it go😂?
@CarineFrisch You would want to multiply the two probabilities, or simply 0.5·0.6=0.3. But if you want to calculate the probability of that happening, then selecting a fair coin is 5/8 times probability of selecting unfair coin is 3/7 (assuming coin isn't replaced), or vice versa probability of selecting unfair coin is 3/8 times probability of selecting fair coin is 5/7 (they both work out the same). Then multiply by each respective probability of landing Heads, and you get 3/8·5/7·0.5·0.6=0.08
Hi Khan,
Please tell why are you not considering the case where one head is fair and other one is unfair?
It's quite possible that we can get one head from fair and other from unfair coin. So you should also consider the probability of these cases.
you pick ONE coin from the bag; that bag can be either fair or unfair; so the 2 flips are either from an unfair coin or a fair coin;
@@AnotherGamer21 Thanks mate.
Great explanation...thank you
Whats the probability of a person pressing the like button for this vid??
Ans : almost 100%
Thanks Sal ! ur amazing
As for today, It's 529/541, So Practically It's 97.78 %😂.
Thank you so much for an amazing education!!!
What happens if you take a fair coin on the first pick and an unfair the second time. Should you not calculate that probability and then add to the other two P(Fair and heads) and P(Unfair and heads)
C Frisch incomplete answer
since its dependent shouldn't the coins being picked decreased ex. 5/8 then 4/7 then 3/6?
kawmekawmehaw you are only choosing 1 coin
sal khan was telling us about what is a probability
thanks
The example made it easy to understand, thanks Khan.
love it.
Great
Incredible
Alright, guys, he's probably not going to see your videos. If you want to contact him go to khanacademy . org or something
why did you not consider the case of unfair head and fair head coin?
because you're flipping the same coin
I don't understand the part where Prob. of HH given Fair is 50%?????
Because for fair coin the prob of H is 1/2 = 50%
what is the difference between, getting two heads in a row knowing that the coin was a fair one, and getting 2 heads in a row whilst actually getting a fair coin? For dependent events, what is the difference between P(HH, given a fair coin) and P(Fair and HH)?
The first one means that the the event preceding the coin toss ie the picking of the fair coin has already happened, so the chances of that happening is 100%
But in the second case the first event still hasn't happened, so the chances of picking the fair queen will not be 100%, and so we would need to calculate the chances and multiply it together because they are indpendent events
can u help me how to solve this A store sells two types of tables: plain and deluxe. When an order for a table
arrives, there is an 80% chance that the plain table will be desired.
(a) Out of 5 orders, what is the probability that no deluxe tables will be desired?
(b) Assume that each day 5 orders arrive and that today (Monday) an order came
for a deluxe table. What is the probability that the first day in which one or more
deluxe tables are again ordered will be in three more days (Thursday)? What is the
expected number of days until a deluxe table is desired?
(c) Actually, the number of orders each day is a Poisson random variable with a
mean of 5. What is the probability that exactly 5 orders will arrive on a given day?
the rest are fair coins 50% i didnt understand this part
Same here
Just wondering... How do you make a coin unfair so that it's not a 50/50 chance
+David Lee put somthing more heavy on one side...
Will you ever do videos on psychology???
not clear
= 124/83055 (0.149% not 100% bro)
@djancak Thank-you
My question: lets say we have a bag where there are 10 balls. 1 of em is red and the rest are white. i pull something out of the bag 5 times. whats the propability of getting red in any of these times?
Do you take the ball out OR do you put it back in after taking it out.
These are two different questions.
There is 1/10 + 1/9 + 1/8 + 1/7 + 1/6 possibilities if we suppose you don't put back the ball again.
If you put ball in again then you have
1/10 + 1/10 + 1/10 + 1/10 + 1/10 possibilities.
lol @ South Park calculator
Omg i skip this video because i follow part 7 and clueless