AFM 13.6

Binomial TheoremAnd

Probability

Conditions of Binomial experiment

• Each trial has exactly 2 outcomes• There must be a fixed number of trials• The outcomes of each trial must be

independent• The probabilities in each trial are the

same

The Binomial Theorem and Probability

• LANDSCAPING Managers at the Eco-Landscaping Company know that a mahogany tree they plant has a survival rate of about 90% if cared for properly. If 10 trees are planted in the last phase of a landscaping project, what is the probability that 7 of the trees will survive?

• Let S represent the probability of a tree surviving.• Let D represent the probability of a tree dying.

Since this situation has two outcomes, we can represent it using the binomial expansion of (S + D)5. The terms of the expansion can be used to find the

probabilities of each combination of the survival and death of the trees.

• The probability of a tree surviving is 0.9. So, the probability of a tree not surviving is 1 - 0.9 or 0.1.

• The probability of having 4 trees survive out of 5 can be determined as follows.

5 5 4 1 3 2 2 3 1 4 5( ) 1 5 10 10 5 1S D S S D S D S D S D D

• Use since this term represents 4 trees surviving and 1 tree dying.

• Substitute 0.9 for S and 0.1 for D =5(0.9)^4(0.1)

=0.3281

• Thus, the probability of having 4 trees survive is about

4 15S D

4 15S D

1

3

Five mahogany trees are planted. What is the probability that at least 2 trees die?

• The third, forth, fifth, and sixth terms represent the conditions that two or more trees die. So, the probability of this happening is the sum of the probabilities of those terms.

• P(at least 2 trees die)

• = 0.0729 + 0.0081 + 0.00045 + 0.00001• = 0.0815

• The probability that at least 2 trees die is about 8%.

3 2 2 3 1 4 510 10 5S D S D S D D

3 2 2 3 4 5• 10 0.9 0.1 10 0.9 0.1 5 0.9 0.1 0.1

• Eight out of every 10 persons who contract a certain viral infection can recover. If a group of 7 people become infected, what is the probability exactly 3 people will recover from the infection?

• There are 7 people involved, and there are only 2 possible outcomes, recover R or not recovery N. These events are independent, so this is a binomial experiment.

• When is expanded, the term represents 3 people recovering and 4 people not recovering from the infection. The coefficient of is C(7, 3) or 35.

• P(exactly 3 people recovering)

• The probability that exactly 3 of the 7 people will recover from the infection is 2.9% .

3 435 0.8 0.2 R 0.8, N 1 0.8

35 0.512 0.0016

0.028672

3 4R N

7R N

12 out of every 16 persons who contract a certain viral infection can recover.

If a group of 7 people become infected, what is the probability that exactly 5 will recover from

the infection.• There are 7 people involved and there are

only 2 outcomes, and these events are independent, therefore it is a binomial

• 0.75 is the percent for recovery• 0.25 not recovered (1-.75) or (100-75)

3 4(7,5)(0.75) (0.25)c 0.31146 31.1%

8 out of every 10 persons who contract a certain viral infection can recover.

If a group of 7 people become infected, what is the probability that at least 5 will recover from the infection.

• P( 5 recover)

• P( 6 recover)

• P( all 7 recover)

• Now add together: 0.31146 0.31146 0.13348 0.7564

5 2(7,5)(0.75) (0.25)c

6 1(7,6)(0.75) (0.25)c

7 0(7,7)(0.75) (0.25)c

0.31146

0.31146

0.13348

75.6%