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### Course: Statistics and probability > Unit 8

Lesson 4: Combinatorics and probability- Probability using combinations
- Probability & combinations (2 of 2)
- Example: Different ways to pick officers
- Example: Combinatorics and probability
- Getting exactly two heads (combinatorics)
- Exactly three heads in five flips
- Generalizing with binomial coefficients (bit advanced)
- Example: Lottery probability
- Probability with permutations and combinations
- Conditional probability and combinations
- Mega millions jackpot probability
- Birthday probability problem

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# Probability & combinations (2 of 2)

Making at least 3 out of 5 free throws. Created by Sal Khan.

## Want to join the conversation?

- I'm really struggling with this. Back in the early probability videos, Sal was always setting up the problems in terms of "desired outcomes divided by all possible outcomes." Through this whole video, I kept waiting for him to divide by 32, or 2^5, since there are 32 possible outcomes for throwing 5 free throws. But now that division step is skipped entirely. Why?(47 votes)
- I was struggling with this too, here is what clarified it for me:

Suppose you throw a fair coin two times. This gives you the following outcomes and probabilities:

P(HH) = 1/2 * 1/2 = 1/4

P(HT) = 1/2 * 1/2 = 1/4

P(TH) = 1/2 * 1/2 = 1/4

P(TT) = 1/2 * 1/2 = 1/4

To calculate P(HH) you can either calculate 1/2 * 1/2 since one throw has a chance of 1/2 of getting Heads or you can count the total number of possibilites which is 4 and since they all have the same probability P(HH) is 1 divided by 4. If you want the probability of two outcomes, for example HT & TH, you just add the probabilities of these two: 1/4+1/4 = 2/4 or again count these two and divide them by the total number of possibilities, which gives you the same result, 2/4. But what if the coin is unfair and has a chance of 80% of getting Heads and 20% of getting Tails? Here are the probabilities in this case:

P(HH) = 0.8*0.8

P(HT) = 0.8*0.2

P(TH) = 0.2*0.8

P(TT) = 0.2*0.2

You can't just count the number of outcomes you are interested in and divide them by the total number here. P(HH), for example, which is 0.8*0.8 would give you 1/4 instead. But you can add the probabilities, and thats what sal did when he multiplied by 10, because there are 10 cases he was interested in with the same probability. In the example here, if you wanted to know P(HT or TH) you would calculate P(HT) + P(TH) = 0.8*0.2 + 0.8*0.2 = 0.8*0.2*2.(61 votes)

- Is there a more elegant way to calculate the probability of at least X events occurring? It seems like there has to be.

For example, if I want to calculate the odds of a student passing a 100 question multiple choice test by purely guessing, is there no better way than to sum the probability of 70 successful guesses, 71 successes, 72 successes, etc up to 100?(22 votes)- That's a great question! Yes, when the number of events is very large, we estimate the experiment with a special thing called the normal distribution, which gives us the whole answer in a single calculation. That'll be in the statistics playlist if you want to look forward to it. :)(26 votes)

- Why couldn't the answer during03:05just be 60% since it's the probability of getting three out of five? Why did he have to do all the work?(7 votes)
- just to correct your initial paragraph, we don't know the answer to the question "what is the proportion of baskets he will make out of 5 free throws?" - there is not telling that it will be 3 out of 5 (60%).

But it is possible to ask "What is the most likely number of baskets he'll make out of 5?" and we can compare the different options, and we see in the video that it's most likely to be 4 baskets, because the probability for that is the highest relative to the rest, with almost 41% chance.(0 votes)

- This is regarding two ways of looking at the reasoning:

1) Choosing 3 baskets out of 5, and in no particular order. Therefore, 5 x 4 x 3/ 3!

2) Look at BBBMM as a spelling and he's asking how many ways can you rearrange it. So that's 5! considering each letter is unique. And then dividing by 3! x 2! to account for triple counting of B, and double counting of M.

Do tell me what you (all) think. Thanks.(5 votes) - at3:58, Salman finds the number of combinations of getting 3 of 5 shots made, but I don't understand why he's doing this. He says there are 10 different combinations but arent MMMBB and MBMBM the same in terms of combinations?(6 votes)
- Good question!

The difference between combinations and permutations is harder to see in this problem. When you consider a string of characters (a word), the strings MMMBB and MBMBM would be considered equivalent combinations but distinct permutations. However, you shouldn't view the shots as a string.

Instead, think of it this way. I have shots 1, 2, 3, 4, and 5. Out of these 5 shots, I have to choose 3 that will be good. In this case, what we're really looking for is how to choose 3 numbers out of these 5 numbers. The correct interpretation in this problem is that 135 and 513 are equivalent combinations but distinct permutations. In either case, the person is making shot 1, shot 3, and shot 5. So what we're really doing is counting the number of ways we can choose a "group" or "committee" of 3 numbers from 5 numbers. This is simply:

₅𝐶₃ = 10

You can think of this as a "string" or "word" but not in the way you suggested. The given fact is that our 5-letter word will have 3 indistinguishable M's and 2 indistinguishable B's. The question is how many distinct permutations can I make? Notice here that we have to view this as a permutation problem when we think of it as a word. In particular, we have 5! ways to arrange the letters if they were all distinguishable. But for each one of these arrangements, there are 3! ways we arrange the M's and 2! ways we arrange the B's so we have over-counted by 3! • 2!. So we divide our preliminary count by this:

5!/(3! • 2!) = 10

Notice that this is exactly the same quantity (and expression) as ₅𝐶₃.

Comment if you have questions!(7 votes)

- I have a question that I think has kind of been asked below, but I'm still unable to wrap my mind around the answer. If the shooter makes 80% of his baskets, then why does he only have a 41% shot of making any combination of 80% of 5 baskets? It seems to me that the probability of making 80% of 5 baskets (4/5) should be pretty close, if not exactly 100%. Bear in mind, I understand the math, I'm trying to understand this conceptually.(5 votes)
- I built this formula in Excel so I could manipulate some of the variables. I thought that maybe the problem would be answered by the Law of Large Numbers, i.e., that the error was in the small "sample size" and the percentage of getting 80% of baskets would approach 100% as the number of attempts went up. Turns out the opposite is true. Now I think that maybe the answer lies in the "exactly" that so many people above had had problems with. So for example, if you add up the shooters chance of getting 3, 4, and 5 baskets, you get about 71%. So now I'm wondering if this formula doesn't "build in" the probability that the shooter will get something near his 80% average. So that if we had the shooter go for 20 baskets, there's only 22% chance he will make
*exactly*16/20 baskets, but there's an 84% probability he will make 14, 15, 16, 17, or 18 baskets. I'm I on the right path here?(4 votes)

- Why is it shown on this video that is more likely for someone (whose probability of making it is 80% for each FT) to make it in 4 out of 5 FT (40.96%) than to make it in 3 out of 5 (20.48%)? Shouldn't it get less likely to make it in more FT?(3 votes)
- 80% of 5 is 4.

Therefore, we would "expect" the person to make 4 of 5 shots, its the most likely exact number. Getting exactly 4 shots is more likely than getting exactly 3 shots.

You're probably thinking more in terms of "making at least x shots", in which case the more shots that must be made, the lower the probability. It's important to keep these two notions distinct - probability at an exact value vs over a range of values.(12 votes)

- Hi Sal,

From4:50to5:45,I want to know why are we even considering 5C3 because if we are taking probability of 3 successes in 5 throws then shouldn't (0.8x0.8x0.8)(0.2x0.2) be sufficient alone as the answer?Why are we considering arrangement of getting 3 successes in 5 throws which is 5C3 as combinations shouldnt matter?Please explain because the questions is asking us the probability of 3 successes in 5 throws and it doesnt say "IN HOW MANY WAYS" can we get 3 successes in 5 throws.(4 votes)- So it is asking the chance of making 3 shots out of 5. .8*.8*.8*.2*.2 is just one variation.

If you think of a list of all possible outcomes of making 5 shots, there are 32 total possibilities. It might help to think of the question as asking, of those 32 outcomes, what is the probability of getting the ones that have 3 successes and 2 failures. then, every outcome has a percent chance of happening, so you would add the relevant percents up.

Let's try it with 3 shots that have 2 successes. with 3 shots there are 8 possibilities. I am going to list them all and their probability of happening.

BBB = .8^3

BBM = .8^2 * .2

BMB = .8^2 * 2

BMM = .8 * .2^2

MBB - .8^2 * .2

MBM = .8 * .2^2

MMB = .8 * .2^2

MMM = .2^3

If you added up all the probabilities you should get 1, which shows, or at least helps to show these are all possibilities.

Now, if you wanted to choose all with two baskets that includes BBM, BMB, and MBB. so that's 3 of the 8 possibilities, but each one has a .8^2 * .2 percent chance of happening.

I really hope this made sense, if not let me know.(6 votes)

- Is there an easier way to solve a computationally intensive problem like this :
`There are 17 students in a class: 9 boys and 8 girls.`

If the teacher picks a group of 8 at random, what is the probability that everyone in the group is a boy?

The way I've been solving them is by doing :`9/17 * 8/16 * 7/15 * 6/14 * 5/13 * 4/12 * 3/11 * 2/10`

and then simplifying. It takes forever and I was wondering if there was a short cut.(1 vote)- Your solving method is correct. Try canceling a factor from any numerator and the same factor from any denominator, and then repeating this process until there are no more possible cancellations of factors. This usually simplifies the calculations significantly.

After simplifying, the process of multiplying the factors that remain in the numerators and multiplying the factors the remain in the denominators can be made more efficient, by noting that factors can be multiplied in any order and grouping.

In this particular problem, a possible result after cancellations could be 9/(17*13*2*11*5).

Then 9/(17*13*2*11*5) = 9/(221*10*11) = 9/24,310.(4 votes)

- hello can u help me with a q

a man stands infront of a fireing squed there are 5 shuters each have 0.5 chance of hiting what

the chance of the person to get hit

thank you(3 votes)- The chance of getting hit = one minus the chance of not getting hit.

It is actually easier to figure it out that way.(2 votes)

## Video transcript

Welcome back. I actually recorded this video
earlier today, but then I realized my microphone
wasn't plugged in. And I won't name names in
terms of who unplugged it. But anyway, back
to probability. My wife is giggling
mischievously. Anyway, so let's do a
slightly harder problem than we did before. We were dealing with fair
coins, let's deal with a slightly unfair coin. Let's say I have a coin and
it's-- actually instead of unfair coin let's
do basketball. Let's say I'm shooting free
throws and I have a free throw percentage of 80%. So when I shot a free throw,
8 out of 10 times, or 80% of the time I will make it. But that also says that 20%
of time I will miss it. So given that, if I were to
take-- I don't know-- 5 free throws, what is the probability
that I make at least 3 of the 5 free throws? Well, let's think of it this
way, what is the probability of any particular combination
of making 3 out of the 5? So what do I mean by that? Let me pick a particular
combination. Let's say it's a basket,
basket, basket, and then I miss, miss. So that would be I
made 3 out of the 5. It could be-- I don't know--
basket, miss, basket, miss, basket. And there's a bunch of them and
we'll actually try to figure out how many of them there are. But what is the probability of
this particular combination? Well, I have an 80% chance of
making this first basket. Times 80%. That's a times right there. Times 80%. And then what's my
probability of missing? Well, that's 20%, right? Times 0.2. times 0.2. So this sequels 0.8 to the
third power times 0.2 squared. What's the probability of
getting this exact combination? Well, it's 0.8
times-- then I miss. There's a 20% chance of that. So times 0.2 times 0.8
times 0.2 times 0.8. We could rearrange this because
when you multiply numbers it doesn't matter what order
you multiply them in. So this is the same thing
as 0.8 times 0.8 times 0.8 times 0.2 times 0.2. So this is also the same
thing as 0.8 to the third times 0.2 squared. The probability of getting any
particular combination of 3 baskets and 2 misses is going
to be 0.8 to third times 0.2 squared. Now what's the total
probability of getting 3 out of 5? Well, it's going to be the sum
of all of these combinations. You know, I could list them
all, but we hopefully now are proficient enough in
combinatorics and combinations to figure out how many
different ways, if we have 5 baskets and we're picking-- or
we have 5 shots and we're picking 3 of them to be the
ones that are basket shots. So what do I mean? So let's say my 5 shots-- you
know, I've shot 1, 2, 3, 4, 5. Out of these five, I'm
going to choose 3. So once again, I'm putting my
hat on as the god of probability and I will choose 3
of these shots to be the ones that happen to be the
ones that get made. So essentially, out of
5 I am choosing 3. 5 choose 3. And what does that equal to? That's 5 factorial over 3
factorial times 5 minus 3 factorial, so that's
2 factorial. And that equals 5 times 4 times
3 times 2 times 1 over 3 times 2 times 1 times 2 times 1. We can ignore all the 1's. Let's see. We get 3 times 2 times 1. 3 times 2 times 1. We can cancel that. This 1 we can ignore, and
then this 2 and then this turns into 2. So there are 10
possible combinations. These are two of them. You know, basket, basket,
basket, miss, miss. Basket, miss, basket,
miss, basket. And you know, it's a good
exercise for you to list the other 8 of them. But using just the binomial
coefficient, and hopefully you have an intuition of why that
works and I'd be happy to make more videos if you feel that
that you need more explanation. But I made a couple. There are 10 combinations. So essentially, the probability
of getting exactly 3 out of 5 baskets, if I am an 80% free
throw shot, is going to be-- switch colors. The probability of 3 out of 5
baskets is going to be equal to the probability of each of the
combinations, which is 0.8 to the third times 0.2 squared. I make 3, miss 2. And then, times the total
number of combinations. Each of these has a
probability of this much. And then there's 10 different
arrangements that I could make. There's 10 different ways of
getting 3 baskets and 2 misses. So times 10, and what
does that equal to? Let me get my high-end
calculator here. So let's see what that is. That is 0.8 times 0.8 times 0.8
times 0.2 times 0.2 times 10. Equals 20.48. So it's essentially, a 20.48%
chance that I get exactly 3 out of 5 of the baskets. Now let's make it slightly
more interesting. Let's say I don't want it as a
probability of 3 out of the 5. And this is actually something
that probably, people are more likely to ask. What is the probability of
getting at least 3 baskets? Well, if you think about it,
this is the probability. This is equal to the
probability of getting 3 out of 5 baskets, plus the probability
of getting exactly 4 out of 5 baskets, plus the probability
of getting exactly 5 out of 5 baskets. We already figured
this one out. That's 20.48%. So what's the probability of
getting 4 out of 5 baskets? Well, once again, if we want
exactly 4 out of 5 baskets, so an example could be-- I
don't know-- miss, basket, basket, basket, basket. What's the probability of
any one of the combinations where I make 4 baskets? Well, it's going to be 0.8 to
the fourth times-- and then I have a 20% chance
that one miss. You know, it could have
been basket, miss, basket, basket, basket. That's also exactly 4, but
when you multiply them, the probability of getting any
one of these particular combinations is exactly this--
0.8 to the fourth times 0.2. If I have 5 baskets, how many
ways can I pick 4 of them to be the ones that I make if I'm
once again the god of probability? So this is going to be 0.8 to
the fourth times 0.2 times 0.2 times-- out of 5 baskets, I'm
choosing 4 that I'm going to make. So this is the number of
combinations where I get 4 out of the 5. So what is 5 choose 4? That's 5 factorial over 4
factorial times 1 factorial. Well, that equals just 5. You can work that out. So let's just figure this out. This is going to be 0.8
times 0.8 times 0.8-- that's 3-- times 0.8. That equals-- did
I do that right? Let's see. 0.1. Wait. 0.8 times 0.8--
yeah, that's right. Times 0.2 times 5. So 40.96%. So this is 40.96%. So roughly, 41% chance that I
get exactly 4 out of 5 baskets. Which is interesting because
that's kind of my free throw percentage. So it's almost a little less--
you know, 2/3 shot of kind of hitting my free throw
percentage on the mark on that time. And what's the probability
of getting 5 out of 5 Well, there's only one way
of getting 5 out of 5. You have to get all 5 of them. So this is 0.8 to
the fifth power. Let me get the calculator back. So it's 0.8 times 0.-- oh,
wait-- times 0.8 times 0.8 times 0.8 times 0.8
equals 0.3276. So 32.77% shot. And then we can add them
all up because we want the probability of at least 3. So it's going to be that, the
probability of getting 5 out of 5, plus the probability of
getting 4 out of 5, which is 0.4096. Plus, the probability
of getting 3 out of 5. So that's 0.2048
equals 0.94208. So 94.21-- roughly, rounding--
% chance, which makes sense. If I have an 80% free throw
percentage on any one shot, I have a very high probability of
getting at least 3 out of 5 when I go to the
free throw line. Anyway, I'm all out of time. I'll see you in the next video.