# Probability that a geyser erupts

Lets say you have a geyser that has a 2/3 probability of erupting in a 50 minute interval? What is the probability that it will erupt in a 20 minute interval?

The way I tried to solve it that a 20 minute interval is 2/5 of a 50 minute interval, so the probability is 2/3 * 2/5 = 4/15, but apparently this is worng. Where did I go wrong and what is the right method to solving it?

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Are you quite sure it's the probability, not eruption rate? –  Alex Dec 21 '13 at 1:36
The problem with this question is that we only know that the geyser erupts exactly once every $75$ minutes. We either need to make assumptions about the rate of eruptions (probably Poisson) or have some additional information about the eruptions distribution. –  Alex Dec 21 '13 at 1:44
@Alex: We don't know it erupts exactly every $75$ minutes. It could be a number of distributions. The answer changes depending on which you assume. Your post assumes a different (very reasonable for a math problem, less so for the real world, but probably the one the problem setter intended) one. –  Ross Millikan Dec 21 '13 at 4:10
@RossMillikan: I mean it erupts exactly once anywhere between $0$ and $75^{\text{th}}$ minute. That's what I understand by 'exactly one every $75$ minutes. –  Alex Dec 21 '13 at 4:26
@Alex: If you mean that after erupting it randomly chooses a delay in the range $0$ to $75$ that doesn't work. The expected delay will be of order $37.5$ min, so the chance of erupting in a $50$ minute span will be quite high. If you guarantee an eruption every $75$ minutes, it has to wait exactly $75$ minutes every time. The exponential distribution can also give a probability of $\frac 23$ of an eruption in a $50$ minute window, but doesn't guarantee an eruption within $75$ minutes. As I said in my edit, I think this is the one the problem setter intended. I wish it had been specified. –  Ross Millikan Dec 21 '13 at 4:33

Let us assume the event (random variable) of geyser erupting within 50 minutes follows an exponential distribution

In other words P(X<=50minutes)= $1-e^{-\lambda (50)}$ = 2/3

Find lamda from this, The lamda is somewhere around 0.021972.

Now find P(X<=20 minutes) = $1-e^{-\lambda * (20)}$ = 0.35561.

Divide the 50 minute interval into five 10-minute intervals. Let Q be the probability that the geyser does not erupt in each of the 10 minute intervals. Thus, $Q^5 = \frac{1}{3}$ and $Q \approx 0.8027$. For 20 minutes it is just $1−Q^2$ or 35.56%.

Thanks

Satish

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Nice answer! Actually, there seems to be an easier way to do this. Divide the 50 minute interval into five 10-minute intervals. Let Q be the probability that the geyser does not erupt in each of the 10 minute intervals. Thus, $Q^5$ = $1/3$ and Q = 0.8. For 20 minutes it is just $1 - Q^2$ or 36%. –  user2612743 Dec 21 '13 at 2:58
That's a very clever alternative way of solving it and indeed it comes out to precisely the same value of 0.3556059... @Satish, perhaps you should add this to your answer. –  Garrett Dec 21 '13 at 3:36
@Garry, I was amazed at the solution myself. I will add it, I am having my dinner. I shall do it when I have time –  satish ramanathan Dec 21 '13 at 3:37

Your calculation would be fine if the geyser erupted exactly every $75$ minutes and the probability comes from the interval chosen. The probability over an interval of length $t$ would be $\begin {cases} \frac t{75}& t \lt 75\\1 & t \ge 75 \end {cases}$

You are probably expected to assume that the probability of eruption in a short interval of time $dt$ is $p\ dt$ and that the eruptions are independent. Then the chance of no eruption over an interval of length $t$ is $e^{-pt}$. Use the data you are given to find $p$, then evaluate $e^{-20p}$

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I think your reading is a reasonable one as well, especially with the image of "Old Faithful" from Yellowstone which the popular image says is nicely periodic (it is not). –  Ross Millikan Dec 21 '13 at 2:06
In fact, Old Faithful now follows a bimodal distribution, with the peaks centered on $65$ and $91$ minutes. –  Ross Millikan Dec 21 '13 at 4:14

The answer is highly dependent on what your model is for geyser eruption. Suppose you know it erupts exactly once per hour, and the unreliable person you asked says it erupted 15 minutes ago. You are 2/3 confident that this is correct, but 1/3 confident it was actually 5 minutes ago. This gives a 2/3 probability of eruption in the next 50 minutes, and a 0 probability of eruption in the next 20 minutes.

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OK, I guess I'm late but will try nevertheless. All we know is that the probability of the event $P(S<75)=1$. This isn't enough to solve the problem, but given the situation (volcanic eruption) this follows some sort of exponential decay. So the mean $\frac{1}{\lambda}$ number of events in this period is $1$ if we take the time frame $=1$. The standardizing constant $\alpha$ is found from the equation $$\int_{0}^{1} \alpha e^{-t}dt=1\\ \alpha=\frac{1}{1-e^{-1}}$$ due to memorylessness property. Under these conditions the probability is $$P(E)=\alpha \int_{0}^{\frac{20}{75}}e^{-t}dt$$

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