Overview: If you want to find the probability of obtaining "successes" from independent trials, each with probability . We can use the formula:

Number of "fails"Number of "successes"

Number of combinations a group of

Probability of "success" q - Probability of "fail"

Recall that:

It is known that 30% of a cereal boxes come with a Disneyland ticket. Suppose you buy 5 of these cereal boxes. Assume the cereal boxes are independent of another.

Find the probability of finding exactly 2 Disneyland tickets among the 5 boxes.a)

P(X = 0) =

Find the probability of finding only 1 Disneyland ticket among the 5 boxes. b)

P(X = 1) =

Find the probability of finding no Disneyland tickets among the 5 boxes. c)

P(X = 0) = binomialpdf(5, .3, 0) = .1681

The function binomialpdf, takes the inputs: n , p , x; calculates the exact probability for you. To reach binomialpdf : DISTR Ex: binomialpdf(5,.3,2) = P(X = 2) = .3087

Example: How many different ways can two friends be selected from a group of five friends?

Now we can try out some examples where you would use this formula:

Binomial DistributionTuesday, May 30, 2017 3:17 PM

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Find the probability of finding two or less Disneyland tickets among the 5 boxes.d)

Notice how this question is slightly different. It's asking for the probability-of finding TWO OR LESS. That means that you could: find two ticketsOR find one ticket OR find no tickets

Remember that exclusive probabilities are additive. Using symbols and answers from a - c :

P(X ≤ 2) = P(X = 2) + P(X = 1) + P(X = 0) = 168 = 0.837

Table of probability for the example

x p( X = x )

0 .168

1 .360

2 .309

3 .132

4 .028

5 .003

To answer question (d) we first had to calculate the probabilities for x = 0 , 1 , 2 individually and thenadd them all together. There's a function in the calculator that does this!

The function binomialcdf takes the inputs n , p , x ; it calculates the probability P(X ≤ x ) you can find thefunction under : DISTR

Ex: binomialcdf(5 , .3 , 2 ) = P (X ≤ 2) = .837

e) Find the probability of finding more than 2 Disneyland tickets among the 5 boxes.

We are interested in more than 2 boxes P(X>2) so that means we found 3 tickets OR more tickets.-

Using the table and symbols: P(X 2) = P(X = 3) + P(X = 4) + P(X = 5) = P( X ≥ 3 ) =.132 + .028 + .003 = .163

Rewriting P(X > 2) as P ( X ≥ 3 ) It allows us to use The complement of P(X ≤ 2) meaning: P( X > 2 )= P( X ≥ 3) = 1 - P ( X ≤ 2) = 1 - .837 = 0.163

This is important because our function binomialcdf can only tell us the probability for P(X ≤ 2) like we used in example (d) - computing the table for every problem is time consuming so we use the complimentary rule : 1 - P(A) = P(A)' [A' = "not" A]

P( X ≥ 3) P( X ≤ 2)

0

1

23

4

5

=+ 1

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In this case we want to find less than 4 tickets P ( X < 4 ) . It means the probability of finding 3 tickets or 2 tickets or 1 ticket or 0 tickets. In symbols and using the table:

-f) Find the probability of finding less than 4 Disneyland tickets among the 5 boxes.

P ( X < 4 ) = P( X = 0 ) + P( X = 1 ) + ( X = 2 ) + P( X = 3) = P( X ≤ 3)

= .168 + .360 + .309 + .132 = 0.969

Rewriting P(X < 4 ) as P ( X ≤ 3) is also important. Remember, we don't want to rely on the calculating tables every time. Remember, our calculator can tell us the probability for P ( X < x) so in this case binomialcdf(5, .3 , 3) would give us the same answer: 0.969

In this case we want MORE THAN 3 tickets. That means I could find 4 tickets OR 5 tickets. Because we are dealing with discrete probabilities, in symbols and using the table:

-g) Find the probability of finding at more than 3 Disneyland tickets among the 5 boxes.

P ( X > 3 ) = P (X = 4) + P (X = 5) = P(X ≥ 4) = .028 + .003 = 0.031

Our calculator can only tell us how to find the P( X ≤ x ) so using this and

= 1 - P (X ≤ 3) [binomialcdf(5, .3 , 3)] = 1 - 0.969 = 0.031

complimentary rule : P( X ≥ 4 ) = 1 - P ( X ≥ 4 )'

The answer we found from the table. Once again, we used the complement rule to calculate the probability.

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