Contents of this Probability theory episode:

Random variable, Binomial distribution, Hypergeometric distribution, Poisson distribution, Probability, Average, Random variable with limit, Random variable without limit, Expected value, Standard deviation.

It is time to see how the three most important discrete distributions, namely the hypergeometric, the binomial and the Poisson distributions work.

Let's see a story for each of them.

On average, 24 customers arrive to the bank in an hour.

What is the probability of exactly 2 people arriving within 7 minutes?

What is the probability of maximum 2 people arriving within 7 minutes?

What is the probability of minimum 2 people arriving within 5 minutes?

On average, 24 customers arrive in an hour, but this is only the average. So, it can happen that in one hour, nobody comes, and in the next, 50 customers show up. The number of customers has no limit, it could be anything. It is not likely though that in the next 7 minutes 7 billion customers come in, but who knows.

If we recall the example about road crashes: there can be a maximum of 7 days in a week when crashes occur, however, there can be 7 billion crashes in a week. So the number of customers is like the number of crashes.

So, this is a POISSON DISTRIBUTION, which means we need the expected value.

If there are 24 customers arriving every hour, then it is 24/60=0.4 per minute. In 7 minutes it is seven times that: 2.8.

We would not expect the same number of customers in a period of 5 minutes and in a period of 7 minutes, so the expected values will be different.

If there are 24 customers arriving every hour, then it is 24/60=0.4 per minute, and in 5 minutes it is five times that number, namely 2.

___ means that ____

This is a bit too much, so let's use the complementary for the calculation.

In a given season the probability of rain is 0.2 each day. What is the probability that there are 3 rainy days in a week?

X=number of rainy days

This definitely has a limit, as there can only be 7 rainy days in a week, at most.

At a highway checkpoint they find that out of 100 cars, on average, 12 cars commit some sort of violation. If they stop 10 cars randomly, what is the probability that

a) exactly two cars commit violations?

b) a maximum of two cars commit violations?

c) a minimum of two cars commit violations?

d) two consecutive cars commit violations?

X=cars committing violations

They stop 10 cars, so it would be surprising to find that 13 committed violations.

The number of violating cars therefore is limited to a maximum of 10.

p=the probability of a car committing some sort of violation.

If 12 out of 100 cars violate the rules, that means 12% are violators, so

This is a bit too much, so let's use the complementary for the calculation.

A car commits some violation with a probability of p=0.12. It is the same for the other car, too.

At a road checkpoint they find violations at 8 cars per hour, on average. What is the probability of finding violations at

a) three cars in 15 minutes?

b) a maximum of two cars in a half hour?

X=cars committing violations

X is the number of cars committing violations, just as before.

Previously it meant how many cars are offending out of 10 stopped cars, but now it means how many cars commit an offense within 15 minutes.

Even in the worst case scenario, there is a maximum of 10 offending cars out of 10 cars. However, the number of offending cars in 15 minutes could be anything.

So now X has no limit, therefore it is POISSON DISTRIBUTION.

If there are 8 cars committing a violation in an hour, then in 15 minutes we need a quarter of that: 8/4=2

In a half hour, twice as many offending cars could be expected.

Finally, here comes a case where all three distributions will make an appearance.

We need to make some room for that.

A bolt of fabric has a minor defect at every 10 yards on average.

a) What is the probability that a 6-yard piece is perfect?

b) If 30 yards of fabric is cut into 6-yard pieces, what is the probability

that there will be exactly two pieces with defects?

c) 120 yards of fabric was cut into 6-yard pieces, resulting in 9 pieces with defects. If we pick out 5 pieces, what is the probability of 2 of them having defects?

X=number of defects

If we want the piece to be perfect, than the number of defects is presumably zero.

On average, there is 1 defect per 10 yards.

But it does not mean that they weave fabric like "oh, here is 10 yards finished again, let's put in a defect".

The defects are located in a random fashion, so there could be 2 defects in a 10-yard section, or even 13, or any number.

So this is a POISSON DISTRIBUTION, and if on average, there is 1 defect in 10 yards, then in 6 yards it is:

Y=pieces with defects

There can be any number of defects, but there can be maximum 5 pieces with defects, so Y has a limit.

Here, p is the probability of a piece having a defect.

Let's see the probability of a piece having a defect.

The previous question was about the probability of a piece being perfect.

well, the defective:

If 120 yards of fabric is cut into 6-yard pieces, there will be 20 pieces.

As it happened, 9 out of these had a defect.

We pick 5 pieces.

Z=pieces with defects

Random variable, Binomial distribution, Hypergeometric distribution, Poisson distribution, Probability, Average, Random variable with limit, Random variable without limit, Expected value, Standard deviation.

Probability theory episode