Contents of this Probability theory episode:

Random variable, Discrete and continuous random variable, Binomial distribution, Poisson distribution, Hypergeometryc distribution, Exponential distribution, Normal distribution, Uniform distribution, Probability, Average, Density function, Distribution function, Expected value, Standard deviation.

Text of slideshow

5.1.

An insurance company receives – on average – 5 home insurance claims per day.

a) What is the probability that on a day they receive fewer than the expected number of claims?

b) What is the probability that on three days in a week they receive fewer claims than expected?

X=number of claims

It can happen at any time that aliens attack Earth and then the number of home insurance claims reaches a million per day. That means X has no limit.

Y= number of days

when claims are below average

There are maximum seven days in a week, so Y has a limit.

Let's see what we know:

the probability that the number of claims in a day is below average

5.8.

The useful lifetime of a smartphone follows the exponential distribution, with an expected life of 4 years.

a) What is the probability that it works at least for 3 years?

b) What is the probability that it works longer than 3 years, but less than 5?

c) What is the probability that if it has been working for 3 years, it will die in the next 2 years?

The last question will be funny.

Let's try to figure out how it is different from the previous one.

For this, let's do some drawing.

We know it has been working for 3 years,

so it will die somewhere here.

But within 5 years.

So far, this is very much like the previous question.

To understand the difference between the two questions better, take Bob.

We try to predict the probability that Bob will die between his 70th and 71st birthdays.

The question is whether this probability is high or low. Well, it depends.

If we make the prediction at the moment of Bob's birth for the probability of him dying between his 70th and 71st birthdays, then the probability will be low.

It is low, because many things can happen to Bob until then, for example he gets hit by the bus at age 5, or gets a heart attack at age 60...

On the other hand, if Bob just turned 70, and as a present for his birthday we predict his chances of dying in the next year, then we can reassure him that this probability is pretty high.

Well, this is the difference between the two types of questions.

Both cases are about the phone dying between its 3rd and 5th year,

but in one case we ask at the moment of its birth,

in the other case we ask after 3 years of operation.

this was the first question

if we still remember

Finally, there is one more funny thing.

The exponential distribution has a strange feature.

This is a feature that Bob, unfortunately, does not have.

This feature is being forever young.

In the case of Bob, who does not have this trait, if we want to know the probability of him dying within a year, then we need to know how old he is.

His chances of dying within a year are not the same at the age of 10 or 60 or even 102. As time passes, Bob indeed has an increasing chance of dying, because he is not ageless.

However, the exponential distribution is just that.

It means it doesn't matter whether those three years have passed.

We could even lie about it.

As if the 3 years in the conditions didn't even exist.

5.7.

At a bank, in 0.3% of the cases there is no customer arriving in an hour. The number of customers follows the Poisson distribution.

a) What is the expected number of customers per hour?

b)

X = number of customers

There is no customer in 0.3% of the cases.

We need the that is in the exponent.

We can lure it down from there by taking the natural logarithm of both sides.

There is such a thing as:

So, bye-bye

This one is done.

5.81 customers are expected per hour.

Of course, there won't be exactly 5.81 customers arriving, unless we are at the morgue.

So this should be understood as 5.81 customers on average.

Let's see what is going on with the other question.

That looks less inviting.

X means the number of customers

If the customers are still alive, this could only be an integer.

5.8.

A newsstand sells 48 items per hour, and on average, 36 of those are newspapers. What is the probability that a) in 10 minutes, 2 newspapers are sold at most? b) in 5 minutes, exactly 7 items are sold? c) 4 out of the 7 items sold are newspapers?

X = number of newspapers sold in 10 minutes

X has no limit: if a group of tourists arrive from China, hungry for newspapers, then the newsstand can sell even a 1000 papers.

The average sales is 36 newspapers per hour, so it is one-sixth in 10 minutes:

X = number of items sold in 5 minutes

This is also Poisson distribution.

The expected value is 48 items per hour, thus in 5 minutes:

Z = the number of newspapers out of the 7 items sold

We don't know how many items and how many newspapers are in total.

But we know that on average, out of 48 items, 36 are newspapers: that is 75%

5.9.

The probability that the newsstand cannot sell even a single item in 15 minutes is

a) How many do they sell per hour, on average?

b) What is the probability of selling 10 in a half hour?

c) At most, for how long cannot they sell any items with the probability of 0.6?

X = number of items sold

The sales of these newsstands always follow the Poisson distribution.

cannot sell even a single item with a probability of

It is expected to be able to sell 6 items in 15 minutes.

Then, we can assume selling four times of that in an hour: 24 items.

Now, in a half hour, it is expected to sell 12 items:

We have no idea for how long they cannot sell anything. Let's call this time t.

Well, this is splendid, the only question is that what do we have now.

What we got is the expected value of the Poisson distribution in t time.

In other words, we can expect that this many items are bought in t time.

But how long is t?

In 15 minutes, 6 items are bought, in t time, 0.511 items.

At most, 1.2775 minutes pass such that there is still at least 60% chance of nobody buying anything.

5.10.

In a certain month, out of 30 days, 12 days have rain. What is the probability that there are 3 rainy days in a week?

X = number of rainy days

30 days total, out of which 12 are rainy:

We check 7 days, and out of that, 3 days must be rainy:

Only there is a little trouble.

Out of the 30 days, 12 are rainy, on average. This means that this year it could rain on 25 days, or only on 5 days.

So, we have no idea how many rainy days there are, we only know the average.

But X has a limit, since there could only be 7 rainy days in a week, and that means this follows the binomial distribution.

5.11.

In a book, on average there are 80 typos in 100 pages. What is the probability that there are 7 typos on 10 consecutive pages?

X = number of typos

Total number of pages is 100, on which there are 80 typos:

We check 10 pages, and there must be 7 typos:

Only there is a little trouble.

100 pages contain 80 typos on average, that means there could be 150 or even only 25, for instance.

So, we have no idea how many typos there are, we only know the average.

Out of the 10 pages we check, maximum 10 can have typos, but there can be any number of typos.

X is the number of typos, so it has no limit.

typos are expected on 1 page

5.12.

On a certain exam, usually 60% of students fail. If 10 students are taking the exam on a day, what is the probability that

a) at most 2 students pass?

a) at least 2 students pass?

X = how many pass

Since 10 are taking the exam, definitely no more than 10 will pass.

Let's see what we know:

But, here we have a little problem.

X means how many pass, so then this p should be the probability of somebody passing the exam.

That means X and p should always refer to the same thing.

There is no problem if p is the probability of a student failing, but in that case X should mean the number of students who fail.

Since now X means how many pass, p is the probability of someone passing the exam.

5.13.

Random variable X has a uniform distribution, with an expected value of 10, and standard deviation of .

How high are these probabilities: , and and ?

The expected value of the uniform distribution is:

And its standard deviation:

This is a linear system.

If we add the two equations up,

Now let's see the distribution function.

5.14.

A fire station on average receives an alarm every two hours. What is the probability that

a) at most 2 alarms are received in 8 hours?

b) after receiving an alarm at 800, the next alarm will be between 930 and 1000?

X = number of alarms

The number of alarms follows a discrete distribution, and let's see...

in 8 hours, there could be any numbers, so it is Poisson.

Alarms are received usually every two hours, so in 8 hours it is expected to receive 4 alarms.

Y = elapsed time between alarms

The elapsed time is continuous and exponential.

Usually 2 hours pass between alarms, so the expected value is 2 hours.

EXP

5.15.

The number of calls to a customer service center follows the Poisson distribution, the time between the calls is a random variable with exponential distribution, and the probability that a call comes in within 5 minutes is:

a) What is the average number of calls per hour?

b) What is the probability that in a half hour at least three calls come in?

c) What is the probability that at least 10 minutes pass between two calls?

X = number of calls

a call comes in within 5 minutes

If in 5 minutes , then in an hour

And in a half hour

Y = elapsed time between calls

This tends to change all the time as we go along.

Let's see what it is right now.

If 12 calls come in within a half hour,

then usually 5 minutes pass between calls:

 

Problem with Poisson and Binomial distributions

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Random variable, Discrete and continuous random variable, Binomial distribution, Poisson distribution, Hypergeometryc distribution, Exponential distribution, Normal distribution, Uniform distribution, Probability, Average, Density function, Distribution function, Expected value, Standard deviation.

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