Question 1

Where are you getting "6 choose 2 equals 15"? Why is it not explained? Am i missing something.

Accepted Answer

In general, n choose k = n! / [k!(n-k)!], where ! denotes factorial, and the factorial of a whole number is the product of the whole numbers from 1 up to that number.

So 6 choose 2 = 6! / [2!(6 - 2)!] = 6! / (2! * 4!) = (1*2*3*4*5*6)/(1*2*1*2*3*4) = (5*6)/(1*2) = 15.

Have a blessed, wonderful day!

Question 2

At 1:56 why is it 0.3 to the 6th power?

Accepted Answer

A simpler way of thinking about it would be this:

What are the chances of missing a free throw? Well, the probability of making a free throw is 70%, so the probability of missing a free throw must be 30%.

So what is the probability of missing two free throws in a row? Well, 30% of the times that you go to attempt a free throw, you will miss. And, of those times, 30% of the times when you go to shoot again, you will miss a second time. Thus, the probability of missing twice in a row is 30% * 30%, or 0.3 * 0.3, which equals 0.09 (or 9%).

The same logic can be applied to 6 misses. The probability of you missing 6 times in a row would therefore be 0.3 * 0.3 * 0.3 * 0.3 * 0.3 * 0.3 = 0.3^6

I hope this helped.

Question 3

At 2:30 when i solved for the equation 0.7^0 multiplied 0.3^6 my answer was 0.000729 not 0.001.

Accepted Answer

Your calculation is correct but he is rounding to the nearest 1/1000. Thus, 0.000729 ~ 0.001.

Edit: Corrected my comment from 1/100 to 1/1000; thank you @Hodorious

Question 4

In the first video (no. of heads) we only take combinatorics and not the individual probability  (p(H) = 0.5 p(T) =0.5). In this video, we take both P(Score)= 0.7 P(MIss)=0.3 and the combinatorics i.e. the different ways in which you could score and miss. What am I missing?

Accepted Answer

Good question.  When the two events have equal probability, we can just use combinatorics, because every outcome is equally likely.  Say we flip a coin six times.  Then p(HHHTTT) = p(HHHHHH) = p(TTHHTT) = ... etc = (0.5)^6 = 1/(2^6) = 1/64
But when the two possibilities have different probabilities, we need to use those probabilities.  So if p(S)= 0.7 and p(M)=0.3, then p(SSSSSS) = (0.7)^6, p(SMSMSM) =  (0.7)^3 *  (0.3)^3, p(SSMMSS) = (0.7)^4 *  (0.3)^2.  So there are still 64 different outcomes of 6 shots, like flipping 6 coins, but they have seven different groups *with different probabilities* that can be collected -- groups with 0,1,2,3,4,5, or 6 scores out of 6 shots.  So there are 6C4 different ways he could have 4 scores and 2 misses, all with the same probability.  If you DID use the probabilities for H and T, you would still get the correct answer, of course.

Question 5

Suppose that a basketball player has a 50% free throw percentage or that I flip an unbiased coin.  I am supposing that both have the same premise.

Suppose that I flip the coin 6 times and want the probability of 3 heads in 6 flips.

Using the formula:

Probability = # of total possible winning outcomes / # of total outcomes

or

Probability = "n Choose k" X (f^k) X (1-f)^(n-k)  as seen in this video.

In my example:
n=6
k=3
f=0.5 (heads)

Solving P(X=0) ... P(X=6), I expect a discrete probability distribution.

Since the probability of getting heads is exactly 50%, I would expect P(X=3) = 50%, P(X=2) and P(X=4) to be smaller,
P(X=1) and P(X=5) to be even smaller, and P(X=0) and P(X=6) to be even smaller.

But that's not how it works out.

P(X=0) = "6 Choose 0" X (0.5)^0 X (0.5)^6 = (6!/0!6!) X (0.5)^6 = 1  X 0.015625 = 0.015625
P(X=1) = "6 Choose 1" X (0.5)^1 X (0.5)^5 = (6!/1!5!) X (0.5)^6 = 6  X 0.015625 = 0.09375
P(X=2) = "6 Choose 2" X (0.5)^2 X (0.5)^4 = (6!/2!4!) X (0.5)^6 = 15 X 0.015625 = 0.234375
P(X=3) = "6 Choose 3" X (0.5)^3 X (0.5)^3 = (6!/3!3!) X (0.5)^6 = 20 X 0.015625 = 0.3125  <-------
P(X=4) = "6 Choose 4" X (0.5)^4 X (0.5)^2 = (6!/4!2!) X (0.5)^6 = 15 X 0.015625 = 0.234375
P(X=5) = "6 Choose 5" X (0.5)^5 X (0.5)^1 = (6!/5!1!) X (0.5)^6 = 6  X 0.015625 = 0.09375
P(X=6) = "6 Choose 6" X (0.5)^6 X (0.5)^0 = (6!/6!0!) X (0.5)^6 = 1  X 0.015625 = 0.015625

Where's my mistake?

Intuitively, it makes sense that 3 heads in a row or 3 tails in a row should have a percentage much lower than 50%: (0.5)(0.5)(0.5)=0.125 or 12.5%.

But I want 3 heads (half of 6) in any order when flipping the coin 6 times.  This should mirror the coin's natural tendency to come up heads 50% and tails 50% of the time.

Accepted Answer

All of the probabilities you calculated are correct. Your only mistake is:

> "Since the probability of getting heads is exactly 50%, I would expect P(X=3) = 50%"

Can you explain why you think that this should be the case?

Taking a look at the probabilities, we see that P(X=3) is the _largest_, which aligns with our expectation based on the known probability of 50%. But just because the probability of Heads on a given flip is 50%, does not mean that for 6 flips P(X=3) should be 50% as well.

This is actually a pretty good example of why it is important to study probability: because we humans often have _bad intuition_ about how probability works.

Maybe making tree diagrams will help to understand why your intuition was mistaken. Draw a full tree diagram for 1, 2, 3, and 4 flips. For each tree, determine the probability of the most likely value (in the case of 6 flips, this would be 3 heads/3 tails).

Question 6

But then why do we call this a binomial distribution if the distribution of P(X) is not symmetric. The probability distribution graph wont be symmetric in this case. So why?

Accepted Answer

The Binomial distribution is named because each trial has two possible outcomes (success or failure). It does not need to be symmetric.

Question 7

But how can I calculate 120c3? It´s easy when it´s only 6c0,1,2,3 and so on, but what do I do when the number is 120c3? My problem is this: 120c3*0,05^3*0.95*117...

Accepted Answer

120C3 = (120!)/(117! * 3!), right?  Think about what that would look like if you wrote it out:
(120*119*118*117*116*115*...3*2*1)/((117*116*115...*3*2*1)(3*2*1)), right?  Can you see that lots and lots of terms will cancel?  Not much will be left after you cancel...

Question 8

A calculator claims it can randomly and independently generate a digit from 0-9. For any four digits generated, the probability of 2 zeros is 0.03. Is the calculators claim correct? Show your working.

Im no sure how to exactly do this one :/

Accepted Answer

You're in the right section: binomial probability.
You need to use binomial expansion to work out the probability of two out of four zeroes. Where, if the claim is correct, the probability of a zero is 0.1
You can use a formula very similar to the one Sal uses in this video:
P(exactly k zeroes in n digits) = 𝑛C𝑘 · fᵏ · (1-f)ⁿ⁻ᵏ, where f is the 0.1 above.

Spoiler alert: the calculator's (manufacturer's) claims are not correct.

Statistics and probability

Course: Statistics and probability > Unit 9

Free throw binomial probability distribution

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