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# Generalizing k scores in n attempts

Sal generalizes 2 scores in 6 attempts to k scores in n attempts.

## Want to join the conversation?

- Could someone please (1-f)^n-k more? I just kind of get n-k but not 100%.(7 votes)
**f**= probability of a made shot

so

*(1 - f)* = probability of a missed shot**n**= # of attempts**k**= # of made shots

so**n - k**= # of missed shots

So the expression is just the probability of a missed shot to the # of missed shots.(17 votes)

- What if you wanted to know what was the probability of making AT LEAST 2 free throws? How would the formula change?(4 votes)
- You just have to sum the relevant probabilities.

In this case it would be easiest to add up the probabilities of getting 0 and 1 free throw, which gives you the probability of NOT making AT LEAST 2 free throws (i.e. making less than 2 free throws). You would then subtract that from 1 to get the probability of making AT LEAST 2 free throws.

P(AT LEAST 2 free throws) = 1 - P(less than 2 free throws)

P(less than 2 free throws) = P(0 free throws) + P(1 free throw)

= 6C0 • 0.7⁰ • 0.3⁶ + 6C1 • 0.7¹ • 0.3⁵

= 1 • 1 • 0.000729 + 6 • 0.7 * 0.00243

= 0.000729 + 0.010206

= 0.010935

P( AT LEAST 2 free throws) = 1 - 0.010935 =**0.989065**

i.e. about 99%

Note: 6C0 and 6C1 mean '6 choose 0' and '6 choose 1'.(20 votes)

- Hi All

I tried to calculate the probability for getting 10 Heads out of tossing the fair coin 20 times using the formula for Binomial Distribution and the answer I got is 17.6 % apprx. But why does it appear at first glance that I have a 50% chance of getting Heads for 10 times out of 20 coins tosses.? Am I missing something here? what mistake am I doing?(2 votes)- You're taking the (correct) assumption that a fair coin will land on heads 50% of the time, and extrapolating (falsely) that it means that is must land on heads 10 out of 20 times.

Take your incorrect extrapolation and run it against an odd number of trials to better see your logic mistake. A coin flipped 3 times will not land on heads 1.5 times, since that is impossible.

That it lands on heads 10 (half) of the time is the most likely event. Assuming your math was correct, at 18%. The next most likely event would be that it lands on heads 9 or 11 times. The next most likely event 8 or 12 times.... 7 or 13 times... the least likely event it lands on heads 0 or 20 times... but it's still POSSIBLE that it lands on heads 20 times. That's another way to see why 50% was likely incorrect... there's just too many possibilities for one of them to command 50%(5 votes)

- I really miss the fast/slow play feature of the videos

could you bring that back(3 votes)- You can still use it, but it is slightly tricky. You have to convert to "fully player" by clicking on the "Title of the video/article" and then after letting it load go to the bottom of the video & click "full player".

Let me know if this doesn't work.(2 votes)

- fk is success and (1-f) is failure right?(2 votes)
- If f stands for success then 1-f stands for failure

we raise f to the power of k because f is multipled k times. So we if we are multiplying f by k times then we are remaining with n-k times for the failure, which is 1-f

so for success we have f^k

for failure we have (1-f)^(n-k)(3 votes)

- Hi, Mauro here. In the event n=k, we have (n-k)!=0 in the denominator. How do we get rid of these guys? If I flip a coin twice and want to know P(exactly 2 heads),I get 2C2*0.50^2*0.50^0. Empirically, I know it's going to be 0.25, in which case 2C2=1. But why? I hate mindless memorization. Thanks.(2 votes)
- When n = k, you're essentially calculating the probability of all trials being successful (or all failures, depending on the context). The factorial of 0, denoted as 0!, is defined to be 1. This is a mathematical convention that ensures formulas like the binomial coefficient formula remain valid even in edge cases like n = k.

So, when you calculate (nCk) and both n and k are the same, you're looking at n! / k!(n − k)! = n! / n!0! = 1. This holds true even for your coin flip example: the probability of getting exactly 2 heads out of 2 flips is calculated as (2C2) ⋅ (0.5)^2 ⋅ (0.5)^0 = 1 ⋅ 0.25 ⋅ 1 = 0.25. Understanding this principle helps move beyond memorization to a more intuitive grasp of why the mathematics works as it does.(1 vote)

- What makes this different from the Combinatorics & Probabilities video? Or are they the same? I'm just confused because I feel like I watched similar examples in the combinatorics & probabilities video when calc free throws.(2 votes)
- The video uses binomial probability, which is a type of combinatorics. However, combinatorics is a broader field that includes more than just calculating probabilities of specific outcomes. It also involves counting, permutations, and combinations without necessarily involving probabilities. The video specifically focuses on the binomial probability aspect, which is a subset of combinatorics applied to probability theory.(1 vote)

- How will the above method differ if the question was probability of making at least 2 scores in a row?(2 votes)
- Calculating the probability of making at least 2 scores in a row involves a different approach than simply calculating the probability of a certain number of successes out of a series of trials. For sequences of successes (e.g., streaks of scores), you typically need to consider patterns of outcomes and cannot directly apply the binomial formula. The complexity increases because the positioning of the scores matters, unlike in the binomial scenario where only the count of successes and failures is relevant.(1 vote)

- Why use
`K`

and`N`

? and could you use this with anything in life?(2 votes)- could you be more specific?.

In the video N means total attempts possible(it can be scoring, or flipping a coin or rolling a dice)

K means number of attempts we want( for example 3 out of 6 or 1 out of 5 )(1 vote)

- At7:21, when a customer places an order with candy's On-line supermarket, a computerized accounting information system automatically checks to see if the customer has exceeded his/her credit limit. past records indicate that the probability of customers exceeding their credit limit is 0.5. suppose that, on a given day 20 customers place orders.

1. What is the probability that at least two customers will exceed their limits?(2 votes)- To find the probability that at least two customers will exceed their credit limits, we first consider the probability of the opposite scenario—either none or one customer exceeding their limit—and subtract this from 1 to find our desired probability.

Given: Probability of exceeding the limit p = 0.5, Number of customers n = 20.

Probability of no customer exceeding the limit:

20C0 ⋅ (0.5)^0 ⋅ (0.5)^20

Probability of exactly one customer exceeding the limit:

20C1 ⋅ (0.5)^1 ⋅ (0.5)^19

P(At least two) = 1 − [P(None) + P(One)]

= 1 − [(20C0) ⋅ (0.5)^20 + (20C1) ⋅ (0.5)^20]

= 1 − [1 ⋅ (0.5)^20 + 20 ⋅ (0.5)^20]

= 1 − [21 ⋅ (0.5)^20]

Calculate this value for the exact numerical probability.(1 vote)

## Video transcript

- [Voiceover] So the last video, we studied the circumstance where I had a 70% free throw probability. I have a 70% chance of
making any free throw, which is actually higher than my actual free throw probability, which might be a surprise to you. But we said in that circumstance, if it's a 70% chance of making it, well, that means that
you have a one minus 70%, or 30%, chance of missing. And we said if you took six attempts, the probability of you getting exactly, making two of the baskets, exactly two scores ... I called them "scores"
instead of "making it," just because I wanted "making" and "miss" to have different
letters in that video ... We said, well, there's six
choose two different ways of making two, exactly two
out of the six free throws, and then the probability
of any one of those ways is going to be making it twice, which is 0.7 squared, and missing it four times, so 0.3 to the fourth power. This was just one particular situation, but we could generalize based on the logic that
we had in that video. In fact, let's do that. So if I were to generalize it, if I were to say the probability, the probability ... It's the exact same logic of exactly ... Exactly ... Now let's say k ... Let me do this in a color,
an interesting color. So let me do it in this
orangeish-brown color. K shots, or exactly k scores ... I'll call making a free throw a score. We'll just assume you got a point for it. So exactly two k scores ... in n attempts ... Let's just say in n attempts. In n, and let me go back
to that green color. N attempts ... N attempts is going to be equal to ... Well, how many ways can
you pick k things out of n, or n choose k? N choose k ... N choose k. Actually, let's just
generalize it even more. Let's just say that you have, your free throw probability is p. So let's say p is ... So for this situation right over here, since we generalized it fully, let's say that p is the
probability of making a free throw. Actually, since I already have a p here, let me just say f is
equal to the probability of making a free throw. Or you could say, your
probability of scoring, if you call a score making a free throw. So if f is your probability
of making a free throw ... So if you want n scores, then this is going to be ... This is going to be ... Well, it's going to be f to the n power, and then you're going to have ... and then you're going to
miss the remainder ... Sorry, f to the k power, because you're making exactly k scores. So f to the k power, and then the remainder, so the n minus k attempts, you're going to miss it. It's going to be that
probability of missing, and the probability of missing
is going to be one minus f, so it's going to be times one minus f to the n minus k power. To the n minus k power. And just, if you like, or I encourage you, pause the video and just make sure you
understand the parallels between this example where I had ... f was 70%, our f was 70%. One minus f ... Or f was .7 and one minus f would be .3, and we were seeing, how do we get two scores in six attempts? And here we're saying,
k scores in n attempts. This is just a general
way to think about it. The whole reason why I'm
setting this up this way is, it's interesting to now think about the probability distribution for a random variable that's defined by the number of scores
in your n attempts, or the number of scores
in your six attempts. And actually, since I've
been pushing the limit, or I've been doing longer
videos than I intend to, I will do that in the next video.