Geometric Distribution EXPLAINED with Examples

hey guys it's mark from ace tutors and

in this video

i'm going to explain a new probability

distribution called the geometric

distribution

first i'll go over this concept at a

high level then i'll explain some of the

distribution's key parameters

and finally we'll finish things off with

an example

so let's dive right in now the geometric

distribution is

very similar to the last one we

discussed the binomial distribution

so if you haven't yet watched that video

i recommend clicking here to give that a

look first before moving forward

in fact the geometric distribution is

really just a more restricted version of

the binomial distribution

used for a specific scenario the

geometric distribution requires the same

conditions to be met as the binomial one

in order to be used however the binomial

distribution is used when you are

ultimately trying to find the

probability

of getting a certain number x successes

out of your total number of trials

where the geometric distribution is used

when you're ultimately trying to find

the probability of getting your very

first success

on a specific trial number x

so for example you might be a salesman

going to sell insurance door-to-door

and you want to find the probability of

getting a successful sale on your first

house

or third house or 15th house

now that we covered the concept at a

high level let's now go over a couple

key parameters for this distribution

first the mean or expected value

this parameter is pretty straightforward

and is just mu

equals one over p where p is the

probability of getting a success

on any given trial basically what the

expected value gives you

is the amount of trials you can expect

to go through before getting your first

success

next is standard deviation

this value sigma equals the square root

of 1 minus p

divided by p or you might see it as the

square root of q

divided by p where q just equals one

minus p

or the probability of a failure on any

given trial since there are only two

outcomes

a success or a failure

finally we have probability and in order

to understand this formula

let's take a step back and remember what

our scenario looks like

for geometric distribution we want to

find the probability of getting our

first success

on a certain trial number in order to

have our first success on some trial

number x

that means that we must have had

failures on every trial before trial

number x

so if we had this many trials before our

first success

we would have had a failure followed by

another failure

followed by another failure and one more

failure

before finally getting our success

okay that's great but how can we derive

a probability from

this well since the same conditions as

the binomial distribution must have been

met

we know that each of our trials are

independent from one another

and when you have independent events you

can simply multiply the probability of

each event together

to get the overall probability of the

events occurring

since the probability of failure is q

and the probability of success is p

this gives us q times q times q times q

times p if we were to simplify this a

bit

we would have one less failure than we

have trials

and one success at the end which reduces

to

q raised to that x minus 1 times p

where x again represents the trial we

want our first success on

finally let's solidify this concept with

an example

coming home from work you always seem to

hit every single light

throughout time you calculate the odds

of making it through any light to be

0.2 on any given night

how many lights can you expect to hit

before finally making it through

one and with what standard deviation

finally what's the probability of the

third light you come across

being the first one that is green

okay this might seem like a lot so let's

unpack it a bit

in this problem our success is making it

through light without having to stop

and we determine the probability of that

occurring to be 0.2

so that's our p-value which makes our

probability

of failure or having to stop at the

light equal to 0.8

all right first we want to find out how

many red lights we can expect to hit

before finally coming across a green one

this is the job for our expected value

formula which states that mu

is equal to 1 over p and using our

p-value

we get 1 over 0.2 or 5 lights

which means that on any given night

returning from work

we can expect to come across our first

green light by the fifth one

next we need to find the standard

deviation of this distribution

once again this formula is the square

root of q

divided by p plugging in our q and p

values

we get the square root of 0.8 divided by

0.2

which ends up equaling 4.47 lights

this means that on average night after

night the number of lights you need to

go through before getting your first

green one

deviates from the mean of five by four

point four seven lights

finally we want to find the probability

of getting our first green light

on light number three for this example

our x value is three for our probability

formula

plugging in this value along with p and

q

we get 0.8 squared times 0.2

because we have to hit two red lights

before finally hitting the first green

one

as a result the probability of this

occurring on any given night

is 0.128 which kind of makes sense

if we expect to get our first green on

light number five

with a standard deviation of 4.47 lights

there is a small but not unreasonable

possibility

of hitting your first green on light

number three

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