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Binomial Distribution. The binomial distribution is used when there are exactly two mutually exclusive outcomes of a trial. These trials are often referred to as Bernoulli trials E.g. Flipping a Coin (Head or Tail) Rolling a Dice – HUH ??????

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Binomial distribution

Binomial Distribution

The binomial distribution is used when there are exactly two mutually exclusive outcomes of a trial. These trials are often referred to as Bernoulli trials

E.g.

Flipping a Coin (Head or Tail)

Rolling a Dice – HUH ??????

Well we could say the outcomes are getting a 6 and not getting a 6


Binomial distribution

These outcomes are appropriately labeled "success" and "failure". E.g. getting a 6 = success The binomial distribution is used to obtain the probability of observing x successes in N trials, with the probability of success on a single trial denoted by p.E.g. for each trial chance of a 6 =The binomial distribution assumes that p is fixed for all trials.


Binomial experiment
Binomial Experiment "failure".

  • Fixed number of trials = n

  • Only two outcomes , success (p) or failure (q)

  • Each trial is independent and identical

  • Probability, p never changes in any trial and q (failure) = 1 – p and also never changes


Binomial distribution
WHY? "failure".

Tree diagrams can only go so far before they become cumbersome.

E.g. Rolling a Die in successive trials

exciting huh!


Binomial distribution

6 "failure".

3 rolls of a dice

6

6’

6

6

6’

6’

6

6

6’

6’

Paths in Red show those outcomes which have two sixes

6’

How many paths are there?

6’

What is the probability at the end of each path?


Deriving the rule
Deriving the rule "failure".

There are 3 branches

Each branch has a probability of

WHY are they the same?

So the total chance of rolling two sixes in 3 trials is


Let s look at the problem again using our new variables
Let’s look at the problem again using our new variables "failure".

Let p = rolling a 6 = 1/6 , q = not rolling a 6 =5/6

Let n = trials = 3

And let r = number of successes you desire

THE PREVIOUS CALCULATION

Would look like

No. arrangements of (ppq)


Arrangements of p and q
Arrangements of p and q "failure".

ppq = 3 arrangements from the tree diagram

What if you carried out 8 trials and wanted the probability of getting 2 sixes.

THAT IS A BIG TREE!

How many arrangements of ppqqqqq are there?

That’s right it is a combination!


So now to wrap it up
SO NOW TO WRAP IT UP! "failure".

  • The rule is now complete

    And the good news

    The Calculator has a function to do it all for you…Binompdf which is in the DISTR Menu

    Just enter in (n, p, r)