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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.

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• 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. • 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.;)
• 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 • 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).
• 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? • 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.
• 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? • 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
• 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. • 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:
• At , 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 • 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
• Why not 15/32 because there are 32 Permutaions out which 15 qualify. ?
2 ^ 6 gives 32. • 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).   