Chapter 17 Notes
A.
Bernoulli Trials:
1.
There are only two possible outcomes:
a.
The probability of success (p)
b.
The probability of failure (q)
2.
All trials are independent: What happens in one trial does not change what will
happen in the next trial.
B.
Geometric Probability Model: Completely specified by one parameter, the
probability of success (p). Usually helps determine the probability of a single success.
1.
Expected Value: μ = 1/p
2.
Example: People with O-negative blood are called “universal donors” because Onegative blood can be given to anyone else, regardless of the recipient’s blood type.
Only about 6% of people have O-negative blood. If donors line up at random for a blood
drive:
a.
How many do you expect to examine before you find someone who has Onegative blood?
μ = 1/0.06 = 16.6667 = ~16.7 people
Blood drives such as this one should expect to examine on average about 16.7 people
to find a universal donor.
b.
What is the probability that the first O-negative donor found is the 5th person in
line?
P(fifth person in line) = (0.94)4(0.06) = 0.047, ~4.7%
On the Graphing Calculator: 2nd, DISTR (VARS), geometpdf(p, n)
geometpdf(0.06, 5) = 0.047
The probability that the first O-negative donor is the fifth person in line is about 4.7%.
c.
What is the probability that there will be at least one O-negative donor among the
first five people in line?
1 – P(no O-negative donors) = 1 – (0.94)5 = 0.266
The probability that there is at least one O-negative donor among the first four is about
26.6%
d.
What is the probability that the first O-negative donor found is one of the first four
people in line?
P(one of four) = (0.06) + (0.94)(0.06) + (0.94)2(0.06) + (0.94)3(0.06) = 0.2193, ~21.9%
P(one of four) = 1 – P(none of the first four) = 1 – (0.94)4 = 0.2193
On the Graphing Calculator: 2nd, DISTR, geometcdf(p, n)
geometcdf(0.06, 4) = 0.219
The probability that the first O-negative donor is one of the first four people in line is
about 21.9%.
C. Binomial Probability Models: Specified by two parameters, the number of trials (n)
and the probability of success (p). Usually helps determine the probability of multiple
successes. Order is important.
1.
Mean: μ = np
2.
Standard Deviation: σ = √(npq)
3.
Total number of outcomes for a trial: (nx) =
n!__
x!(n – x)!
a.
On the Calculator: nCx: n, MATH, PRB, 3:nCr, enter, r, enter
4.
The probability of a binomial trial: nCx(pxqn – x)
5.
Example: People with O-negative blood are called “universal donors” because Onegative blood can be given to anyone else, regardless of the recipient’s blood type.
Only about 6% of people have O-negative blood. If donors line up at random for a blood
drive:
a.
On one particular afternoon there were 125 donors in line. How many of these
donors would be expected to have O-negative blood?
E(O-negative) = 125(0.06) = 7.5 donors
There would expected to be about 7.5 donor with O-negative blood.
b.
What is the standard deviation of the blood donors?
σ = √(125(0.06)(0.94)) = 2.66 donors
c.
What is the probability that there are 3 universal donors in the first 10 people in
line?
(10) =
10!___ = 10! = 10(9)(8) = 120
(3)
3!(10 – 3)! 3!7! 3(2)(1)
P(3 out of 10 donors) = 120(0.06)3(0.94)7 = 0.0168, ~1.7%
On the Calculator: 2nd, DISTR, binompdf(n, p, m)
binompdf(10, 0.06, 3) = 0.0168
d.
What is the probability at least 3 out of the first 5 people in line are universal
donors?
P(at least 3 out of 5) =
3
2
4
5
5C3(0.06) (0.94) + 5C4(0.06) (0.94) + 5C5(0.06) = 0.00197
On the calculator: 2nd, DISTR, binomcdf(n, p, m)
binomcdf(5, 0.06, 3) = 0.99994?
e. The binomcdf is a directional program which only works for no more than a number.
To use it for at least situations, you must (1 – binomcdf) to the opposite direction.
Instead of at least 3 out of 5 people, we must look at it as (1 – no more than 2 out of 5
people)
1 – binomcdf(5, 0.06, 2) = 0.00197