- Geometric random variables introduction
- Binomial vs. geometric random variables
- Geometric distribution mean and standard deviation
- Geometric distributions
- Probability for a geometric random variable
- Geometric probability
- Cumulative geometric probability (greater than a value)
- Cumulative geometric probability (less than a value)
- TI-84 geometpdf and geometcdf functions
- Cumulative geometric probability
- Proof of expected value of geometric random variable
Cumulative geometric probability (greater than a value)
Probability for a geometric random variable being greater than a certain value.
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- I think Sal has got this wrong. The final sentence of the question is "Find the probability that Emelia registers more than 4 vehicles before she registers an SUV."
Surely this means that at least the first 5 vehicles mustn't be SUVs, making the answer (0.88)^5, not (0.88)^4?
This table should clarify the problem:
Vehicle More than 4 Probability
SUV No 0.12
non-SUV, SUV No 0.88 * 0.12
2 non-SUVs, SUV No 0.88^2 * 0.12
3 non-SUVs, SUV No 0.88^3 * 0.12
4 non-SUVs, SUV No 0.88^4 * 0.12
>=5 non-SUVs, SUV Yes 0.88^5(58 votes)
- You have a good point. There's a tricky issue with wording. Since V represents the number of vehicles registered until the first SUV (and so including the first SUV), V - 1 represents the number of vehicles registered before an SUV (and so excluding the first SUV).
So the probability that Emelia registers more than 4 vehicles before she registers an SUV is really P(V > 5), not P(V > 4). So you're right that the answer would be (0.88)^5 instead of (0.88)^4. Sal's answer of (0.88)^4 would have been correct if he had written the final sentence as "Find the probability that Emelia registers more than 4 vehicles until she registers the first SUV."
Have a blessed, wonderful day!(17 votes)
- I feel like the wording was bogging me down. A better way to word the crux of the problem is that
P(V>4) = 1 - P(V=1) - P(V=2) - P(V=3) - P(V=4).
P(V=1) = the prob. the first car is an SUV
P(V=2) = the prob. the second car is an SUV..
P(V=3) = ...
P(V=4) = ...etc etc(10 votes)
- I too feel the same.
it can not be (0.88)^4 or (0.88)^5.(2 votes)
- As others have pointed out, Sal had made a mistake here. The mistake is really in this snippet:
P(V not ≤ 4) = P(first 4 cars are not SUVs)
These 2 things are not equal. Imagine a situation where Emilia sold 4 non-SUVs, followed by an SUV.
While this satisfies the right side of the equation, the left side is not satisfied (specifically the
V not = 4part).
The right side of the equation should be
P(first 5 cars are not SUVs)to be equivalent to the left side.
Thus, the final formula should be:
P(V > 4) = (0.88)^5.
- I have now submitted a request to Khan Academy to update the video with correct information.
P(first 4 cars not SUVs)
But it shall be
P(first 5 cars not SUVs)
- This answer is correct if the questions is:
what is the probability that Emelia registers exactly 5 Non SUVs before she register an SUV.
But here the question is different. In my point it should be 1-(p(v=1)+p(v=2)+p(v=3)+p(v=4))(3 votes)
- I was a bit confused by the equivalent statement that Sal claimed at3:11. I went through the math and found the following:
Let's call p the probability that the vehicle is an SUV. In this case,
p = 0.12 and
q = (1-p) = 0.88.
Now, the question is asking for P(V>4)
or P(V=5) + P(V=6) + P(V=7) + ....
But we know:
P(V<=4) + P(V>4) = 1
P(V>4) = 1 - P(V<=4)
= 1 - [P(V=1) + P(V=2) + P(V=3) + P(V=4)]
= 1 - [p + (1-p).p + (1-p)^2 . p + (1-p)^3 . p]
Weird, right? This is the same answer that Sal provided in the video. Did a bit of digging and noticed the cdf of geometric probability distribution is:
P(V<=y) = 1 - (1-p)^y
In other words, if we want to find P(V<=4), we can simply plug in the cdf:
P(V<=4) = 1 - (1-0.12)^4 = 1 - 0.88^4 = 0.4003
P(V>4) = 1 - 0.4003 = 0.5997 which is still the same answer.(4 votes)
- What about
P(V not <= 5) < 0.88^5
or even better
P(V not <= 5) <= 0.88^5 * 0.12
- Why would it be(0.88)^4? I feel like I'm missing something here.
Originally, my thought process was that it would be:
0.88+(0.88^2)+(0.88^3)+(0.88^4). Why wouldn't it work? It's showing the individual probabilities of each registered vehicle before an SUV (The probability of the first vehicle not an SUV, the probability of the second, third, fourth...) Maybe it's my comprehension, but this just doesn't make sense to me. I know the value of it turns out to be greater than 1, but it just seems like it would work.(3 votes)
- I still cannot understand conceptually P(X>4) = P(first 4 cars not SUV)^4
The result could be FFFFS, FFFFFS, or FFFFFFFFFFF....S
Then why P(X>4) = P(Fail)^4?(2 votes)
- Look at the sample space of the first four cars:
FFFF FFFS FFSF FFSS
FSFF FSFS FSSF FSSS
SFFF SFFS SFSF SFSS
SSFF SSFS SSSF SSSS
There's only one way to not have an SUV among the first four cars, namely FFFF.
Thus 𝑃(𝑋 > 4) = (𝑃(F))⁴
– – –
As you kind of mention, 𝑃(𝑋 > 4) = 𝑃(𝑋 = 5) + 𝑃(𝑋 = 6) + 𝑃(𝑋 = 7) + ...
and so on, forever.
We know that 𝑃(𝑋 = 𝑛) = (𝑃(F))ⁿ⁻¹⋅(1 − 𝑃(F))
Thus, 𝑃(𝑋 > 4) = (𝑃(F))⁴⋅(1 − 𝑃(F)) + (𝑃(F))⁵⋅(1 − 𝑃(F)) + (𝑃(F))⁶⋅(1 − 𝑃(F)) + ...
This is a geometric series, where the first term is 𝑎 = (𝑃(F))⁴⋅(1 − 𝑃(F))
and the common ratio is 𝑟 = 𝑃(F)
Because 0 < 𝑟 < 1, we know that this series converges to
𝑎∕(1 − 𝑟) = (𝑃(F))⁴⋅(1 − 𝑃(F))∕(1 − 𝑃(F)) = (𝑃(F))⁴(2 votes)
- I don't get it why P( V > 5) is the same thing as P(V not <= 5)?(2 votes)
- Its the same thing because the <= statement refers to less than or equal to so V not <= 5 means V is not less than or equal to 5 and must be greater than 5.(2 votes)
1. (0.88)^4 Sals answer
2. (0.88)^5 considering the lowest number after 4 is 5 cars and this is where she stops
3. (0.88)^5 x (0.12) same as answer 3, but this is when she saw sixth car which is an suv and stopped
4.cumulative: 1-((p(v=0) +….p(v=4)). Pretty much same as Sals answers but 1-0.59 to get probability of more than 4 cars
I like answers 3, 4(2 votes)
- [Instructor] Emelia registers vehicles for the Department of Transportation. Sports utility vehicles, also known as SUVs, make up 12% of the vehicles she registers. Let V be the number of vehicles Emelia registers in a day until she first registers an SUV. Assume the type of each vehicle is independent. Find the probability that Emelia registers more than four, more than four, vehicles before she registers an SUV. So, let's just first think about what this random variable V is. So, it's the number of vehicles Emelia registers in a day, until she registers an SUV. So, for example if the first person who walks in the line or through the door, has an SUV and they're trying to register it, then V would be equal to one. If the first person isn't an SUV, but the second person is, then V would be equal to two, so forth and so on. So, this right over here is a classic geometric random variable, right over here. So, geometric random variable. We have a very clear success metric for each trial. Do we have a SUV or not? Each trial is independent, they tell us that. They are independent. The probability of success in each trial is constant. We have a 12% success for each new person who comes through the line. Now the reason this is not a binomial random variable, is that we do not have a finite number of trials. Here, we're gonna keep performing trials. We're gonna keep serving people in the line, until we get an SUV. And so, what we have over here, when they say find the probability that Emelia registers more than four vehicles before she registers an SUV. This is the probability that V is greater than four. So, I encourage you like always, pause this video and see if you can work through it. And we're gonna assume, she's not just gonna leave her, I guess her desk, or wherever the things are being registered. She's not going to leave the counter until someone shows up registering an SUV. So, we will just keep looking at people, I guess we could say over multiple days, forever. She'll work for an infinite number of years, just for the sake of this problem, until an SUV actually shows up. So, try to figure this out. Alright, I'm assuming you've had a go and some of you might said, well, isn't this going to be equal to the probability that V is equal to five, plus the probability that V is equal to six, plus the probability that V is equal to seven, and it just goes on and on and on forever. And this is actually true. And you say, well, how do I calculate this? I'm just summing up an infinite number of things. Now the key realization here, is that one way to think about the probability that V is greater than four, is this is the same thing as the probability that V is not less than or equal to four, these two things are equivalent. So, what's the probability that V is not less than or equal to four? This might be a slightly easier thing for you to calculate. Once again, pause the video and see if you can figure it out. Well what's the probability that V is not less than or equal to four? Well that's the same thing as the probability of first four customers, or first four, I guess people. First four, I'll say customers, or I'll say first four cars. The customer's cars, not SUVs. So, this one is feeling pretty straightforward. What's the probability that for each customer she goes to, that they're not an SUV? Well that's one minus 12%, or 88%, or 0.88. And if we want to know the probability of the first four cars are not SUVs. Well that's 0.88 to the fourth power. And so that's all we have to calculate. And so let's get our calculator out. And say I'm going to get, oops. I'm going to get 0.88 and I'm going to raise it to the fourth power and I get, and I'm just going to round it to, the nearest, let's see, do they tell me to round it? Okay, I'll just round it to the nearest, I guess 100th, well, I'll just write it as 0.5997. Is approximately equal to, 0.5997. If you wanted to write this as a percentage it would be approximately 59.97%. So a little bit better than half, than a 50% shot, a little less than a two-thirds shot, that she is going to have to see more than four customers until she sees an SUV.