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# Cumulative geometric probability (greater than a value)

AP.STATS:
UNC‑3 (EU)
,
UNC‑3.E (LO)
,
UNC‑3.E.2 (EK)
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    Probabilitysequence             vehicles                     before SUV?SUV                   No            0.12non-SUV, SUV          No            0.88 * 0.122 non-SUVs, SUV       No            0.88^2 * 0.123 non-SUVs, SUV       No            0.88^3 * 0.124 non-SUVs, SUV       No            0.88^4 * 0.12>=5 non-SUVs, SUV     Yes           0.88^5``
• 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!
• 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).

where
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
• I too feel the same.
it can not be (0.88)^4 or (0.88)^5.
(1 vote)
• 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 = 4` part).

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`.

D-mo
• I have now submitted a request to Khan Academy to update the video with correct information.

Sal writes
P(first 4 cars not SUVs)
=(0.88)^4≃0.5997
≃59.97%

But it shall be
P(first 5 cars not SUVs)
=(0.88)^5≃0.5277
≃52.77%
• 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))
P(V not <= 5) < 0.88^5
or even better
P(V not <= 5) <= 0.88^5 * 0.12
?
• I was a bit confused by the equivalent statement that Sal claimed at . 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
Therefore,
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]
= 0.5997
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
so
P(V>4) = 1 - 0.4003 = 0.5997 which is still the same answer.
(1 vote)
• 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.
(1 vote)
• 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?
(1 vote)
• 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
(1 vote)
• I don't get it why P( V > 5) is the same thing as P(V not <= 5)?
(1 vote)
• 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.
(1 vote)