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## Statistics and probability

### Course: Statistics and probability > Unit 9

Lesson 5: Binomial random variables- Binomial variables
- Recognizing binomial variables
- 10% Rule of assuming "independence" between trials
- Identifying binomial variables
- Binomial distribution
- Visualizing a binomial distribution
- Binomial probability example
- Generalizing k scores in n attempts
- Free throw binomial probability distribution
- Graphing basketball binomial distribution
- Binompdf and binomcdf functions
- Binomial probability (basic)
- Binomial probability formula
- Calculating binomial probability

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# Binomial probability example

We can use the binomial distribution to find the probability of getting a certain number of successes, like successful basketball shots, out of a fixed number of trials. We use the binomial distribution to find discrete probabilities.

## Want to join the conversation?

- Why do you use combinations here? As I understand it, combinations do not consider order, but rather unique ways things are arranged. So ABC is the same as BAC (CAB, CBA, BCA, CBA) but not the same as ABD.

Are not all Successes the same and all Misses the same? This sounds to me like a variation of permutation. Is this because we only have two options, Success and Miss? Or is each attempt at a free throw a unique thing and should be counted independently?

For example (S=Success, M=Miss):

I see it as SSMMMM, SMSMMM, SMMSMM, SMMMSM, SMMMMS all being the same thing, which I would consider a permutation and not a combination.

But is it actually:

[S1, S2, M3, M4, M5, M6,] [S1, M2, S3, M4, M5, M6,] [S1 ,M2, M3, S4, M5, M6,]

[S1, M2, M3, M4, S5, M6,] [S1, M2, M3, M4, M5, S6] where each attempt is a new thing?

Am I even remotely on the right track here? I've watched the videos on permutations and combinations and understand them quite well. If combinations are found by the total number of permutations divided by #of ways to arrange (Like 6 ways to arrange ABC), what would the permutations and # of ways to arrange Successes and Misses be? I get this formula from the first Combinations video.(24 votes)- Hey ill try to explain in a way that U'll understand better .

Consider an actor who has made 6 movies (say 1,2,3,4,5,6) now you want the p(exactly2awards for the 6 movies) . Here the case where he wins the award for movie1&3 is clearly different from the case where he wins due to movies 4&5 . Similarly in this question he succeeding in 1st and 3rd and he succeeding in 3rd and 5th event is clearly different. Thus the number of ways in which this event can occur is increased and so is the probability . Like having many ways of doing the same event here has actually increased the probability of its occurrence. Hope it helps . If you still have a doubt please comment below I'll be glad to help.;)(18 votes)

- Why do we use combination formula here instead of just calculating probability of 2 shots in 6 attempts as (0.7^2)x(0.3^4)? I don't know why we care about the different sequences in which making 2 shots in 6 attempts when calculating the probability. I don't understand how number of ways that it could happen could increase the probability as they are independent events. Thank you(12 votes)
- I think this is because we are looking for the total probability of getting 2 scores out of 6 attempts. (0.7^2)x(0.3^4) calculates the probability of one specific way, rather than the total probability. To get the total probability, we need to multiply the probability of one specific way by the total possible number of ways (here 15).(11 votes)

- How would you do a problem like this if instead of "exactly" 2, you did "less than or equal to" or "more than or equal to" 2 shots made?(11 votes)
- Sean,

If we wanted to calculate the probability of two or more "scores," then we need to calculate each individual probability for the other events (X>=2,3,4,5,6) and add them together. Since the probabilities for "score" and "miss" are not equal, we need to calculate the probabilities the same way Sal did in this video. For example, when calculating the probability for three "scores" in six attempts, the exponents in the factors for 0.7 and 0.3 should add up to six. So, for three "scores" in six attempts, the 0.7 would be raised to the third power and the 0.3 would be raised to the third power. We already have the probability of two "scores" in six attempts at 0.059. I calculated the probability of three "scores" in six attempts to be 0.185. The probability of four "scores" in six attempts is 0.324. (If you think about the fact that this shooter makes seventy percent of his free throws, this probability makes sense.) The probability of five "scores" and six "scores" is 0.303 and 0.118, respectively. Adding these up, I get a proability of making two or more "scores" in six attempts to be 0.989, or approximately 99%. If you calculated the probability for one "score" in six attempts and subtract it from 1, you would get the probability for two or more "scores" in six attempts.(11 votes)

- When I first heard him ask the question, I thought that the only calculations we had to do were to multiply the probabilities together, like: 0.7^2 *0.3^4. In the way that the question is worded, why do we need to find the combination and multiply that as well? If all we care about is the probability of 2 scores in 6 tries, and if when the scores happen doesn't matter, then why multiply by the combination?(9 votes)
- The combination gives the number of different ways that the result can happen. For these sorts of probabilities another way you could write it is:

P(SSMMMM OR SMSMMM OR SMMSMM OR SMMMSM OR SMMMMS OR MSSMMM OR MSMSMM OR MSMMSM OR MSMMMS OR MMSSMM OR MMSMSM OR MMSMMS OR MMMSSM OR MMMSMS OR MMMMSS)

and then the answer would be the sum of each of these probabilities.

If you know you are going to do the sum of a group of identical values then an alternative is to use multiplication:

P(Exactly 2 Scores in 6 Attempts) = 15 * 0.7^2 * 0.3 ^4

OR

6C2 * 0.7^2 * 0.3^4

since 6C2 = 15(9 votes)

- The question states 'What is the probability of getting exactly 2 scores in 6 attempts", it does not specify in which attempt. Why do we care about the order of score/miss attempts? Why do we have to multiply (0.7)^2*(0.3)^4 on 15 combinations? thanks for examplanations.(9 votes)
- It isn't so much that we care about the order, but the probability of getting 2 successful scores in 6 attempts depends on how many ways you can get those 2 successful scores. You could score twice and then fail 4 times, or fail four times and then succeed twice, or any mixture of successes and failures. The fact that it can happen so many ways increases the probability of seeing 2 scores out of 6.

Try this with a simple example, tossing 2 coins. You can get 4 different outcomes:

HH (heads 1st, heads 2nd), HT, TH, or TT. The probability of seeing exactly 1 Head is 2/4 because you count both ways it can happen and then multiply by the probability of each outcome. The outcome itself is (0.5)(0.5) = 0.25 since a head has prob = 0.5 and tail has prob = 0.5. Then multiply by the 2 outcomes that have one Head to get 2(0.25) = 0.5. Half of the time you toss two coins, you will see exactly one Head. Take out 2 coins and try it. Toss them 50 times, and count the number of times you see 0 Heads, 1 Head, and 2 Heads. 1 Head should happen twice as often as 0 or 2 Heads. It may not be exact because this is experimental data, but it should be close.(4 votes)

- At2:34, Sal starts multiplying each of the separate probabilities. I have a more fundamental question: Why do we "multiply" the separate probabilities in any such problem? That multiplication does not make so much intuitive sense to me(5 votes)
- Well, let's start with a simpler example. I flip a coin twice. What's the probability that I will get two heads? Obviously, the probability that you'll get a head the first time is 1/2, and the second probability is 1/2 also. What could we do to these numbers to get the answer? Subtraction doesn't work, clearly, because 1/2 - 1/2 = 0. Addition doesn't work because 1/2 + 1/2 = 1. We know that there is more than 0 and less than 1 chance that you'll get two heads. Division won't work either, because 1/2 / 1/2 = 1 also.

So what's left? Multiplication is the simple answer. 1/2 * 1/2 = 1/4.

Now clearly that wasn't exactly a proof. We didn't exhaust all the options. But here's another way of thinking about it: list all the options for one throw:

H

T

Now for 2 throws:

HH

HT

TH

TT

Now there's TWICE as many options, and only one of them results in HH. The more flips you add, the less likely "all heads" will be. Out of the basic operations, only multiplication makes sense - it makes fractions smaller after time.

Answering that question was harder than I thought it would be! If you still don't understand, try following these links:

http://mathforum.org/library/drmath/view/74065.html

https://www.quora.com/Why-do-we-multiply-the-probability-of-independent-events(8 votes)

- Why not 15/32 because there are 32 Permutaions out which 15 qualify. ?

2 ^ 6 gives 32.(3 votes)- Given the previous video, I follow your logic - though 2^6 = 64. So, there are 64 possible outcomes. 6C2 gives us 15 desired outcomes. Therefore, P(make exactly two shots) should be 15/64.

The problem is that this calculation of probability implicitly assumes that make or miss is equally likely (like a fair coin). However, Sal made it clear in stating the problem that make has a 0.7 chance, while miss has a 0.3 chance.

Let's revisit a fair coin example and calculate it the way Sal did in this video. For the sake of clarity, let's try to find a similar probability ... P(exactly 2 heads in 6 tosses). So, first, I calculate the probability of a particular two-head outcome = (1/2)^6, or 1/64. Then I do a combination calculation (6C2) to determine the number of outcomes matching my condition - 15. Therefore I get a probability of 15/64.

In the free throw example, the particular outcomes are not equally likely. The probability of a particular two-shot-four-miss outcome is (0.7^2 * 0.3^4). Then we multiply that probability by the number of matching outcomes (15).(11 votes)

- I've noticed that n C k is just ( n P k ) / ( k! ). Right????(5 votes)
- You're right. You can think of nCk as:

(# of permutaions of k items in n items) / (# of arrangements of k items)

= nPk / k!(2 votes)

- hi, what if it was "at least 2" instead of "exactly 2"? how would you solve it?(2 votes)
- There are a couple ways.

You could figure out the chances of making 2, 3, 4, 5, or 6 shots and then add them all together, but the better way is to calculate the probability of making 0 and 1 and then subtracting their sum from 1, because if you didn't make 0 or 1 then you must have made at least 2.(5 votes)

- The final answer (0.05935) that is written down at the end of the video is WRONG. If you look at the calculator (at about9:21), it shows 0.059535, not 0.05935. This doesn't really matter if you're just rounding to the nearest percent anyway, but I just thought I'd point it out. I can't call it a 'typo', maybe a 'write-on-a-screen-o?'(4 votes)

## Video transcript

- [Voiceover] Let's say that you know your probability of making a free throw. You know that the probability... The probability, and let's
say of scoring a free throw because make and miss,
they both start with M and that can get confusing, so let's say the probability of scoring... scoring a free throw, is equal to, is going to be, say 70%. If we want to write it as a percent or we could write it as 0.7 if we write it as a decimal. Let's say the probability of
missing a free throw, then and this is just going to come straight out of what we just wrote down, the probability of missing, of missing a free throw,
is just going to be 100% minus this. You're either going to make or miss, you're going to score or miss, I don't want to use make in this because they both start with M so this is going to be a 30% probability, or if we write it as a decimal, 0.3 One minus this is 0.7. These are the only two possibilities, so they have to add up to 100%, or they have to add up to one. Now, let's say that you were
going to take six attempts. What we are curious
about, is the probability of exactly two scores in six attempts. So let's think about what that is and I encourage you to get inspired at any point in this video you should pause it and you should try to work through what
we're asking right now, so this is what we want to figure out, the probability of exactly two scores in six attempts. So, let's think about the way, let's think about the particular ways of getting two scores in six attempts and think about the probability for any one of those particular ways, and then we can think about how many ways can we get two scores in six attempts? So, for example, you could get you could make the first two free throws, so it could be score, score and
then you miss the next four. So score, score, and then it's
miss, miss, miss, and miss. So what's the probability of
this exact thing happening? This exact thing? Well, you have a 0.7 chance of making, of scoring on the first one, then you have a 0.7 chance
of scoring on the second one, and then you have a 0.3 chance
of missing the next four. So, the probability of
this exact circumstance is going to be what I am writing down. Hopefully you don't get
the multiplication symbols confused with the decimals, I'm trying to write them a little bit higher. Times 0.3, and what is
this going to be equal to? Well, this is going to be equal to... This is going to be 0.7 squared times 0.3 to the one, two, three, fourth-- to the fourth power. Now, is this the only way to
get two scores in six attempts? No, there's many ways of getting
two scores in six attempts. For example, maybe you miss the
first one, the first attempt and then you make the
second attempt, you score, then you miss the third attempt, and let's just say you
make the fourth attempt, and then you miss the next two. You miss, and you miss. This is another way to get
two scores in six attempts, and what's the probability
of this happening? Well, you'll see, it's
going to be exactly this, it's just we're multiplying
in a different order. This is going to be 0.3 times 0.7 you have a 30% chance of
missing the first one, a 70% chance of making the second one, and then times 0.3, a 30%
chance of missing the third, times a 70 percent chance
of making the fourth, times a 30% chance for each of the next two misses if you wanted the exact circumstance, this is once again going to be 0.7 and if you just rearrange the
order that you're multiplying, this is going to be 0.7 squared times 0.3 to the fourth power, so for any one of these particular ways to get exactly two scores in six attempts, the probability is going to be this. So, the probability of
getting exactly two scores in six attempts, well it's going to be any one of these probabilities times the number of ways you can get two scores in six attempts. Well, how... If you have out of six attempts, you're choosing two of
them to have scores, how many ways are there? Well, as you can imagine, this is a combinatorics problem, so you could write this as you could write this,
let me see how I could... You're going to take six attempts. You could write this as six choose, what we're trying to, you're
picking from six things, six attempts, and you're
picking two of them, or two of them are
going to need to be made if you want to meet these circumstances. This is going to tell us the number of different ways you can make two scores in six attempts. Of course, we can write this as kind of a binomial coefficient notation. We can write this is as six, choose two and we can just apply the formula for combinations, and if this
looks completely unfamiliar I encourage you to look up
combinations on Khan Academy and then we go into some detail on the reasoning behind the formula that makes a lot of sense. This is going to be equal to six factorial over two factorial times six minus two factorial. Six minus two factorial, I'm going to do the
factorial in green again, and what's this going to be equal to? This is going to be
equal to six times five times four times three times two, and I'll just throw in the one there although it doesn't change the value, over two times one. And six minus two is four, so that's going to be four factorial, so this right over here is four factorial so times four times three
times two times one. Well, that and that is going to cancel, and the six divided by two
is three, so this is 15. There's 15 different
ways that you could get two things out of six, I
guess, is one way to say it or there's 15 different ways that you could get two things out of six. Another way of thinking about it is there's 15 different ways to make two out of six free throws. Now, the probability for each of those is this right over here. The probability of exactly
two scores in six attempts, this is where we deserve a
little bit of a drumroll. This is going to be six choose two times 0.7 squared. This is two, you're going to make two and then it's 0.3 to the fourth power. These will necessarily add up to six. So this right over here was a three, then this right over here would be a three and then this would be six minus three or three right over here. Now, what is this value? Well, it's going to be equal to we have our 15, three times five so we have this business right over here. It's going to be 15 times let's see, in yellow, 0.7 times 0.7 is going to be times 0.49 and let's see, three to the fourth power would be 81. But, I am multiplying four decimals, each of them have one space to the right of the decimal point so this is, I'm going to have four spaces to the right of the decimal so 0.0081 so there you go, whatever this number is, and actually I might as
well get a calculator out and calculate it, so this is going to be this is going to be... Let me... So, it's 15 times 0.49 times 0.0081 and we get 0.059535 so this is going to be equal to, let me write it down. Actually, I wish I had a
little bit more real estate right over here, but I'll
write it in a very bold color. This is going to be, well, actually, I'm kind
of out of bold colors. I'll write it in a
slightly less bold color. This is going to be equal to 0.05935 if we wanted the exact
number, or we could say this is approximately, if we
round to the nearest percentage this is approximately
a six percent chance, six percent probability of
getting exactly two scores in the six attempts. I didn't say two or more, I just said exactly two
scores in the six attempts. Actually, it's a fairly low probability because I have a pretty
high free throw percentage. If someone has this high
of a free throw percentage it's actually reasonably unlikely that they're only going to make two scores in the six attempts.