Question 1

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?

Accepted Answer

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.

Question 2

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?

Accepted Answer

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. :)

Question 3

at 3:58, 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?

Accepted Answer

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!

Question 4

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.

Accepted Answer

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?

Question 5

Hi Sal,

From 4:50 to 5:45,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.

Accepted Answer

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.

Question 6

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?

Accepted Answer

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.

Question 7

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

Accepted Answer

The chance of getting hit = one minus the chance of not getting hit.

It is actually easier to figure it out that way.

Question 8

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.

Accepted Answer

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.

Course: Statistics and probability > Unit 8

Probability & combinations (2 of 2)

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