Precalculus (2018 edition)
- Compound probability of independent events
- Probability without equally likely events
- Independent events example: test taking
- Die rolling probability with independent events
- "At least one" probability with coin flipping
- Free-throw probability
- Three-pointer vs free-throw probability
- Independent probability
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.
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- 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:
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.(122 votes)
- if you took the possibilities for each spot it would be
because the first 6 can be either heads or tails and the back six must copy the first six to stay a palindrome. you then know the middle one can be either. so it would be
2*2*2*2*2*2*2*1*1*1*1*1*1/2*2*2*2*2*2*2*2*2*2*2*2*2 = 2^7/2^13 = 1/64(62 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)
- You can never be certain. Since we can never achieve total certainty, we instead arbitrarily assign a 'degree of confidence' where we say the chance that we're wrong is small enough that we don't care about it.
An example for a commonly chosen degree of confidence (5%):
Change of an offer = 1/4
Chance of no offer = 3/4
Chance of no offer after n interviews = (3/4) to the nth power.
As n -> infinity, the chance of no offer tends toward 0, but never reaches it, so we can see that we never reach total certainty of having had an offer, because we can't go to infinity interviews.
If we'd been to 11 interviews, we'd have a (3/4)^11 = 0.04.22 = 4.22% chance of not having an offer. We've met our chosen standard, so we would say that we were certain of getting at least job offer after 11 interviews with a 5% degree of confidence.
Hope this helps.(80 votes)
- 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)
- 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
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)
- 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
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)
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.