Main content
Statistics and probability
Unit 7: Lesson 7
Multiplication rule for independent events- Sample spaces for compound events
- Sample spaces for compound events
- Compound probability of independent events
- Probability of a compound event
- "At least one" probability with coin flipping
- Free-throw probability
- Three-pointer vs free-throw probability
- Probability without equally likely events
- Independent events example: test taking
- Die rolling probability with independent events
- Independent probability
- Probabilities of compound events
- Probabilities involving "at least one" success
- Probability of "at least one" success
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Probability without equally likely events
Up until now, we've looked at probabilities surrounding only equally likely events. What about probabilities when we don't have equally likely events? Say, we have unfair coins? Created by Sal Khan.
Want to join the conversation?
At2:21, he says that the probability of getting heads is 60%. How does he get the 60%? Or is it just an example?(10 votes)
It's just an example of an unfair coin. He could have easily made it 51% or 70% or 12% - anything except 50%, which is a fair coin.(23 votes)
- Assuming every time you get a job interview, your chance of getting an offer is 25%, how many interviews must you get before you can be certain of getting one job offer?(14 votes)
- (1/4)^n represents the chance of getting an offer from each one of n interviews e.g. n=3 -> 3 offers from 3 interviews. But we'd be satisfied with only at least 1 offer from n interviews. This question combines concepts of "unfair coins" from this video (not 50% chance) and "at least 1 heads" (at least 1 offer) from the previous video, wherein Khan explains why it is calculated as 1 - (3/4)^n. If N represents an outcome of "no offer" and O represents "an offer" then the desired outcome with n=3 interviews is reached by OOO, NOO, ONO, NNO, OON, NON, or ONN and only fails with outcome NNN. It is therefore much easier to calculate the probability of NNN and subtract this from 1.0 = 100% to calculate the chance of not failing than it is to calculate the chances of the 7 distinct possibilities for succeeding and adding them all together, which would also work.
@glenn.searby I'm just getting into probability now, so I'm not familiar with the standard terminology, but wouldn't it make more sense to call this a 95% degree of confidence than a 5%? The 5% is the chance of being incorrect, 95% is the confidence level of correctness. If this is standard terminology, then I guess I'll just have to get used to it.(3 votes)
The Gambler's fallacy states that in a large number of throws, the probability for each event remains the same, but a gambler will tend to believe that after let's say 3 heads, the probability of getting a tail increases (which is wrong). On the other hand, if we calculate the probability to get the same thing many times in a row, this probability 1/(2^n). Therefore the probability for the other event to happen increases. An argument would be this happens because the Gambler's fallacy assumes the events are correlated when they are not. But calculating the probability for the tenth throw in a row assumes the same thing, correlated events. How does one get past this paradox?(10 votes)
That's a very interesting question. I'll try explaining as best as I could. Let's assume that we've tossed a fair coin five times already and that all were heads. Now, the paradox for the sixth try is between
a) that we tend to believe that it is more likely to get a tails now as we've got a series of heads already and because the probability of getting an all heads in a row is always less
and
b) the probability for each event remains the same (Gambler's fallacy)
Now, to get past this paradox, instead of thinking that the probability of getting all heads 6 times is very less(1/64 to be exact) and it's likely to be a tail the sixth time, we must think that we've already gotten 5 heads in a row, the probability of which is 1/32 and irrespective of the sixth try being a tails or a heads, the probability will be 1/64. (ie), the most unlikely event component of the very small probability has already occured.
"But calculating the probability for the tenth throw in a row assumes the same thing, correlated events"
This assumption is wrong. The probability for the tenth throw in a row does NOT assume correlated events. If you see the sample space, the probability of any combination of tails and heads for ten times is 1/512. (ie) P(HHHHHHHHHH) as well as P(HHHHHHHHHT) or even P(HTHTHHTTHT) is the same 1/512. Hope this answers your question.(9 votes)
I have been looking for a good mathematical explanation as to why these probabilities are multiplied. The two answers are always; "If you add, the answer eventually is greater than 1", or the other answer is, "because that's how you do it". Now, there are all sorts of formulas and functions in probability class, and some are very complex and well explained here, but the simple multiplication of probabilities is opaque. Why? It must be harder than it looks.(6 votes)
What would the probability of a coin landing on it's side be? Let's say the coin is a quarter, with a thickness of 1.75mm and a diameter of 24.3mm. Would you calculate surface area or would this involve some physics? The stickiness of the flat surface it lands on would also affect it landing on it's edge. Thanks in advance!(2 votes)
Mathematically speaking, a coin is nothing but a cylinder. The surface area of a cylinder - excluding the top and bottom - is the circumference of the top circle (or bottom, they're equal of course) times the height of the cylinder.
So, for your coin, the area of the side of the coin is pi*24.3 mm * 1.75 mm = 133.6 mm^2. The area of its top and bottom is pi*R^2, with R = radius = half the diameter: pi*12.15^2 = 463.8 mm^2.
To calculate the actual probability of the coin landing on this side would take some fairly complicated physics though. A naive approximation would be this:
The coin has a top and bottom, each of 463.8 mm^2, and a side area of 133.6 mm^2. The chance of landing on the side area is 133.6 / (2*463.8+133.6) = 0.1259, or 12.59%. Of course the real probability is much less, since this completely disregards things like equilibrium, kinetic energy, and all that fun stuff.(6 votes)
- How would you calculate, given 13 flips of a fair coin, the probability of getting a palindromic flip?
A palindrome reads the same both ways, forwards and backwards. Examples of 13-flip palindromes are:
TTTTTTTTTTTTT
TTTHHHTHHHTTT
HHHHHHHHHHHHH
HHHTTTHTTTHHH
...
Is there an easier way to do this than calculating the probabilities of all 128 possibilities? I know you can write out and calculate for smaller numbers of flips, but what about much larger ones?
Thanks in advance.(2 votes)- This problem can be extended in the case of unfair coins also. Lets assume that the coins have :
P(H) = 60% = 3/5 and
P(T) = 40% = 2/5.
Using the same logic as Glenn's, the 1st and 13th flip, 2nd and 12th flip etc should match for palindromic flip. For the 1st and 13th flip to match, it should be either HH or TT. So we need to calculate P(HH or TT).
= P(HH or TT)
= P(HH) + P(TT)- P(HH and TT)
= [P(H).P(H)] + [P(T).P(T)] - 0
= 3/5*3/5 + 2/5*2/5
= 9/25 + 4/25
= 13/25
Hence the probability of palindromic flip in 13 coin flips
= (13/25).(13/25).(13/25).(13/25).(13/25).(13/25)
= 4826809/244140625
= 1.977%.
Please correct me if I am wrong.(4 votes)
At around7:30, I got a little bit confused. All probabilities should add up to 1, right? Is the reason that one specific example (such as the one at7:30) doesn't add up to 1 (or equal one on its own) because there is more than one combination in the sample space?(3 votes)- Here 3 coins are being flipped. Since they are unfair the calculation is slightly complicated.
P(H) = 60% = 0.6 and P(T) = 40% = 0.4
Possibilities are
HHH P(HHH) = 0.6 x 0.6 x 0.6 = 0.216
TTT P(TTT) = (0.4)^3 = 0.064
HHT, HTH, THH P(2 heads and a tail) = 3 x (0.6)^2 x (0.4) = 0.432
TTH, THT, HTT P(2 tails and a head) = 3 x (0.4)^2 x (0.6) = 0.288
Add all the probabilities = 0.216 + 0.064 + 0.432 + 0.288 = 1
We have to know which probabilities when added = 1
Here we are flipping 3 coins or the same coin 3 times so the events and the sample space is different.(3 votes)
between3:17-19, sal talked about flipping the coin many times as a trial. but, even if he did so and got more heads in the trial,what is the guarantee he would get more heads in the real event. i mean, a coin doesn;t have any memory of coming heads many times in a trial(2 votes)
Supernova: There is a statistical concept known as "The Law of Large Numbers". In layman's terms, essentially that in this case if you were to flip this coin 1,000,000 times and it came up heads 60% of the time, you could be VERY confident that this coin was biased towards heads and that the probability of flipping a heads is 60%.
Think of a baseball player at the beginning of a season. Let's say he gets 10 hits in his first 20 at-bats. Would you say he is a .500 hitter? No way -- the sample size is way too small. When the number of at-bats starts getting large, you can then make the type of determination of what type of average this hitter truly is.
It's the same with the coin. The coin may be biased where it will fall on heads 60% of the time. But if you flip it 10 times, it could reasonably fall on tails 7 times. 10 times is not enough. How much is enough? Well, the more the better, but you can usually say that 1000 or more is enough to give you a true picture of probability.(4 votes)
- Why do we multiply p(a) and p(b) when we want to determine p(ab) where a and b are two independent events and p(ab) is the probability of the succession of the events a and b ?(3 votes)
Probabilities are pretty much like percentages. You need to multiply them if you'd like to combine them that way.(1 vote)
- How come for each independent event we have to multiply the probability of each events happening to get the probability of all of them happening? For example, P(H,H), why do we multiply if the events are independent? Thank you(3 votes)
2:21Video transcript
So far, we've been dealing
with one way of thinking about probability, and
that was the probability of A occurring is the
number of events that satisfy A over all of the
equally likely events. And this is all of the
equally likely events. And so in the case
of a fair coin, the probability of heads--
well, it's a fair coin. So there's two
equally likely events, and we're saying one of
them satisfies being heads. So there's a 1/2 chance
of you having a heads. The same thing for tails. If you took a die, and
you said the probability of getting an even number
when you roll the die. Well, there's six
equally likely events, and there's three even
numbers you could get. You could get 2, a 4, or a 6. So there's three even numbers. So once again, you have a
1/2 chance of that happening. And this is a really
good model where you have equally likely
events happening. Now I'm going to change
things up a little bit. So I'm going to draw a line
here because this was just one way of thinking
about probability. Now we're going to
introduce another one that's more helpful when we can't think
about equally likely events. And in particular, I'm going
to set up an unfair coin. So this right over here is
going to be my unfair coin. So that is my coin. Well, I could draw the coin. So it's a gold coin this time. It is unfair. One side of that coin is a
little heavier than the other, even though it's
meant to look fair. So it still has that picture
of some president or something on one side of it. So this is the head side. This is heads, and then,
obviously, on the back, you have tails. But as I mentioned,
this is an unfair coin. And I'm going to make
it interesting statement about this unfair coin and
one that really doesn't fit into the mold that
I set up over here, and this interesting
statement is that we have more than a 50/50
chance of getting heads or more than a 50% chance
or more than a 1/2 chance of getting heads. I'm going to say that the
probability of getting heads for this coin right
over here is 60%. Or another way to
say it, it's 0.6. Or another way to say
it, it is 6 out of 10. Or another way to
say it, it is 3/5. And this might make
intuitive sense to you and hopefully it
does a little bit, but I want you to
realize that this is fundamentally different
than what we were saying before because now we can't say that
there are two equally likely events. There are two possible events. You can either get
heads or tails. We're assuming that the
coin won't fall on its edge. That's impossible. So you're either going
to get heads or tails, but they're not
equally likely anymore. So we really can't do
this kind of counting the number of events
that satisfy something over all of the possible events. In this situation, in order
to visualize the probability, we have to kind of take what's
called a "frequentist approach" or think about it in terms
of frequency probability. And the way to conceptualize
a 60% of getting heads is to think, if we had a
super large number of trials, if we were to just flip
this coin a gazillion times, we would expect that 60% of
those would come up heads. It's unclear how I
determined that this is 60%. Maybe I ran a
computer simulation. Maybe I know exactly all
of the physics of this, and I could completely model how
it's going to fall every time. Or maybe I've actually
just run a ton of trials. I've flipped the coin a
million times, and I said, wow, 60% of those, 600,000
of those, came up heads. And then, we could make a
similar statement about tails. So if the probability
of heads is 60%, the probability
of tails-- well, there's only two
possibilities, heads or tails. So if I say the probability
of heads or tails, it's going to be equal
to 1 because you're going to get one of
those two things. You have 100% chance of
getting a heads or a tails, and these are mutually
exclusive events. You can't have both of them. The probability
of tails is going to be 100% minus the
probability of getting heads, and this, of course, is 60%. So it's 100% minus 60%, or
40%, or as a decimal, 0.4, or as a fraction, 4/10, or as
a simplified fraction, 2/5. So, once again,
this probability is saying-- we can't say
equally likely events. We could say that,
if we're going to do a gazillion of
these, we would expect, as we get more and more and more
trials, more and more flips, 40% of those would be heads. Now, with that out of
the way, let's actually do some problems with this. So let's think about the
probability of getting heads on our first flip and
heads on our second flip. So, once again, these
are independent events. The point has no memory. Regardless of what I
got on the first flip, I have an equal chance
of getting heads on the second flip. It doesn't matter if I got
heads or tails on the first. So this is the probability of
heads on the first flip times the probability of heads
on the second flip, and we already know. The probability of heads on
any flip is going to be 60%. I'll write it as a decimal. It makes the math a little
bit easier, 0.6, 0.6, and we can just multiply. I'll do it right over here. So this is 0.6 times 0.6. Now, it's always good
to do a reality check. One way to think about it
is I'm taking 6/10 of 6/10, so it should be a little
bit more than half of 6/10 or probably a little
bit more than 3/10. And we've explain
this in detail where we talk about
multiplying decimals, but we essentially just
multiply the numbers, not thinking about
the decimals at first. 6 times 6 is 36. And then you count
the number of digits we have to the right
of the decimal. We have one, two to the
right of the decimal. So we're going to
have two to the right of the decimal in our answer. So it is 0.36, and
that makes sense. We're taking 60% of 0.6. We're taking 0.6 of 0.6, a
little bit more than half of 0.6. And, once again, it's a
little bit more than 0.3. So this also makes sense. So it's 0.36. Or another way to
think about it is there's a 36%
probability that we get two heads in a row,
given this unfair coin. Remember, if it was a fair
coin, it would be 1/2 times 1/2, which is 1/4, which
is 25%, and it makes sense that this is more than that. Now, let's think about a
slightly more complicated example. Let's say the probability
of getting a tails on the first flip, getting
a heads on the second flip, and then getting a
tails-- I'm going to do this in a new
color-- and then getting a tails on the third flip. So this is going to be equal
to the probability of getting a tails on the first flip
because these are all independent events. If you know that you had
a tail on the first flip, that doesn't affect
the probability of getting a heads
on the second flip. So times the probability
of getting a heads on the second flip,
and then that's times the probability of getting
a tails on the third flip. And the probability of getting
a tails on any flip we know is 0.4. The probability of getting
a heads on any flip is 0.6, and then the probability of
getting tails on any flip is 0.4. And so, once again, we
can just multiply these. So 0.4 times 0.6. There's actually a couple of
ways we can think about it. Well, we could literally say,
look, we're multiplying 4 times 6 times 4, and then
we have three numbers behind the decimal point. So let's do it that way. 4 times 6 is 24. 24 times 4 is 96. So we write a 96,
but remember, we have three numbers
behind the decimal point. So it's one to the right
of the decimal there, one to the right of
the decimal there, one to the right
of decimal there. So three to the right. So we need three to the right
of the decimal in our answer. So one, two-- we need one more
to the right of the decimal. So our answer is 0.096. Or another way to think about it
is-- write an equal sign here-- this is equal to a 9.6% chance. So there's a little bit
less than 10% chance, or a little bit less
than 1 in 10 chance, of, when we flip this
coin three times, us getting exactly a
tails on the first flip, a heads on the second flip,
and a tails on the third flip.