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# Free throw binomial probability distribution

Sal uses the binomial distribution to calculate the probability of making different number of free throws.

## Want to join the conversation?

• At why is it 0.3 to the 6th power? • 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.
• Where are you getting "6 choose 2 equals 15"? Why is it not explained? Am i missing something. • 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!
• At when i solved for the equation 0.7^0 multiplied 0.3^6 my answer was 0.000729 not 0.001. • 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? • 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.
• 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

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. • 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).
• 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? • How are you calculating 6 choose 0 = 1 and 6 choose 6 = 1? This is where I am confused. • 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... • 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 :/ • 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. 