Download Probability

The probability of event A, P(A),
is the ratio of the number of outcomes that include event A
to the total number of possible outcomes.
(Formula 7.1)
P( A) 
Number of outcomes that include A
Total number of outcomes
What is the probability of selecting the number 5
out of a sample space of 10 numbers (1-10)?
P(5) =
1
= .10 = 10%
10
What is the probability of selecting an odd number 1, 3, 5, 7, or 9
out of a sample space of 10 numbers (1-10)?
P(odd number)
=
P(1 or 3 or 5 or 7 or 9)
=
P(1) + P(3) + P(5) + P(7) + P(9)
=
1
1
1
1
1
   
10 10 10 10 10
=
.10 + .10 + .10 + .10 + .10
=
.50
=
50%
What is the probability of selecting a number that is a multiple of 3
out of a sample space of 10 numbers (1-10)?
P(multiple of 3) = P(3 or 6 or 9) = P(3 outcomes)
=
3
10
=
.30
=
30%
The probability of the compound event X or Y, where X and Y are
mutually exclusive events, equals:
(Formula 7.2)
P(X or Y) = P(X) + P(Y)
What is the probability of selecting a number less than 3 or greater than 7
from a sample space of numbers from 1 through 10?
P(X) = P(number < 3) = P(1 and 2) = P(2 outcomes)
=
2
10
=
0.20
P(Y) = P(number > 7) = P(8, 9, and 10) = P(3 outcomes)
=
3
10
=
Therefore
P(X or Y) = P(X) + P(Y)
=
0.20 + 0.30
=
0.50
=
50%
0.30
The probability of the compound events X or Y, where X and Y
are not mutually exclusive events, equals:
(Formula 7.3)
P(X or Y) = P(X) + P(Y) – P(X and Y)
What is the probability of selecting a number greater than 7 or selecting an
even number in a sample space of numbers from 1 through 10?
P(X) = P(number >7) = P(8, 9, and 10) = P(3 outcomes)
=
3
10
=
0.30
P(Y) = P(even number) = P(2, 4, 6, 8, and 10) = P(5 outcomes)
=
5
10
=
0.50
P(X and Y) = P(even number >7) = P(8 and 10) = P(2 outcomes)
=
2
10
=
0.20
Therefore
P(X or Y) = P(X) + P(Y) – P(X and Y)
=
0.30 + 0.50 – 0.20
=
0.60
=
60%
Given statistically independent events X and Y, the probability of
their joint occurrence is:
(Formula 7.4)
P(X and Y) = P(X)  P(Y)
There are 10 red balls, 10 yellow balls, and 10 green balls in a drum. What
is the probability of drawing a red ball, replacing it in the drum,
and then drawing a green ball?
P(X) = P(drawing a red ball) = P(10 outcomes)
=
10/30 =
0.3333
P(Y) = P(drawing a green ball) = P(10 outcomes)
Notice that the sample space remains 30, since the red ball was replaced.
=
10/30 =
0.3333
Therefore
P(X and Y) = P(X)  P(Y)
=
0.3333  0.3333
=
0.1110888
=
11.11%
Given events X and Y, which are non-independent, the probability
of both X and Y occurring jointly is the product of the probability of
obtaining one of these events times the conditional probability of
obtaining the other event, given that the first event has occurred.
(Formula 7.5)
P(X and Y) = P(X)  P(Y | X) or P(Y)  P(X | Y)
An urn contains 20 red balls, 20 green balls, and 20 yellow balls,
thoroughly mixed. What is the probability of drawing a red ball,
not replacing it, and then drawing a green ball?
P(X) = P(drawing a red ball) = P(20 outcomes)
=
20
60
=
0.333333
P(Y) = P(drawing a green ball) = P(20 outcomes)
However, note the sample space is no longer 60, since the red ball was not replaced.
=
20
59
=
0.338983
Therefore
P(X and Y) = P(X)  P(Y | X)
=
0.333333  0.338983
=
0.1129942
=
11.30%
In how many ways can we select a different subset of things or
individuals from a population of such things or individuals? The
combinations (C) of (N) things or individuals taken (n) at a time.
N Cn 
(Formula 7.6)
N!
n!( N  n)!
Ricky went to the ice cream shop to get a triple-dip ice cream cone. There
are 10 different flavors of ice cream for him to choose from. How many
possible combinations of triple-dip ice cream cones could he make?
10
C3 
10!
10! 10  9  8  7  6  5  4  3  2  1 3628800



 120
3!(10  3)! 3!7! 3  2  1  7  6  5  4  3  2  1
30240
-or-
10  9  8  7!
10  9  8  7! 720
=

 120
3  2  1 7!
3  2 1 7!
6