Conditional Probability
Probability of event B, given event A = P(B|A).
P (B|A) =
P (A ∩ B)
P (A)
Independent Events
Two events A and B are said to be independent if and only if
P (B|A) = P (B) or P (A|B) = P (A)
Bayes’ Rule
Total Probability / Rule of Elimination
If the events B
1
. . . B
k
are a partition of the sample space S
such that P (B
i
) 6= 0 for i = 1 . . . k, then for any event A in S:
P (A) =
k
X
i=1
= P (B
i
∩ A) =
k
X
i=1
P (B
i
)P (A|B
i
)
We find P (A) through finding the intersections between events
A and B
1
. . . B
k
and summing them together.
Example: Machines B
1
, B
2
, and B
3
make 30% 45%, and
25% of the products. Respectively, 2%, 3%, and 2% of the
products made by each machine are defective.
Given event A that the product is defective, find the
probability P (A) that it is defective.
P (A) = P (B
1
)P (A|B
1
) + P(B
2
)P (A|B
2
) + P(B
3
)P (A|B
3
)
Bayes’ Rule
Finds P (B
i
|A) (Product was randomly selected and defective,
what is the probability that it was made by Machine B
i
).
If the events B
1
. . . B
k
constitute a partition of the sample
space S such that P (B
i
) 6= 0 for i = 1 . . . k, then for any event
A in S such that P (A) 6= 0
P (B
r
|A) =
P (B
r
∩ A)
k
P
i=1
P (B
i
∩ A)
=
P (B
r
)P (A|B
r
)
k
P
i=1
P (B
i
)P (A|B
i
)
Some Discrete Probability
Distributions
Binomial Distribution
A Bernoulli trial can result in a success with probability p and
failure with probability q = 1 − p. The probability distribution
o the number of successes in n independent trials–X–is
b(x; n, p) =
n
x
p
x
q
n−x
, x = 0, 1, 2, . . . , n
Summations of Binomial Distributions
For P (a ≤ X ≤ b),
B(x; n, p) =
b
X
x=a
b(x; n, p)
For b(x; n, p), the mean is µ = np and the variance is
σ
2
= npq.
Multinomial Experiments and Distributions
If a given trial can result in any one of k possible outcomes
E
1
, . . . , E
k
with probabilities p
1
, . . . , p
k
, then the
multinomial distribution will give the probability that E
1
occurs x
1
times and E
k
occurs x
k
times in n independent
trials where
x
1
+ x
2
+ . . . + x
k
= nandp
1
+ p
2
+ . . . + p
k
= 1
The joint probability distribution is denoted by
f(x
1
, . . . , x
k
; p
1
, . . . , p
k
, n) =
n
x
1
, . . . , x
k
p
x
1
1
p
x
2
2
···p
x
k
k
Hypergeometric Distribution
Similar to binomial, but does not require independence
between trials.
Used in testing where the item tested cannot be replaced.
Find probability of x successes from the k items labeled
successes and n − x failures from the N − k items labeled
failures when a random sample size n is selected from N items.
A hypergeometric experiment possesses the following two
properties:
1. A random sample of size n is selected without
replacement from N items
2. Of the N items, k may be classified as successes and
N − k may be classified as failures.
The number X of successes of a hypergeometric experiment is
called a hypergeometric random variable.
The hypergeometric probability distribution is
h(x; N, n, k) =
k
x
N−k
n−x
N
n
,
with a range of
max{0, n − (N − k)} ≤ x ≤ min{n, k}
When both k (# of successes) and N − k (# of failures) are
larger than sample size n, the range of a hypergeometric
random variable will be x = 0, 1, . . . , n
The mean of a hypergeometric distribution h(x; N, n, k) is
µ =
nk
N
.
The variance of a hypergeometric distribution is
σ
2
=
N − n
N − 1
· n ·
k
N
1 −
k
N
Multivariate Hypergeometric Distribution
If N items can be partitioned into the k cells A
1
, A
2
, . . . , A
k
with a
1
, a
2
, . . . , a
k
elements, respectively, then the probability
distribution of the random variables X
1
, X
2
, . . . , X
k
,
representing the number of elements selected from
A
1
, A
2
, . . . , A
k
, is a random sample of size n is
f(x
1
, . . . , x
k
; a
1
, . . . , a
k
, N, n) =
a
1
x
1
a
2
x
2
···
a
k
x
k
N
n
Example:
• N = 10 individuals used for a study
• A
1
. . . A
3
= Blood types O, A, and B
• a
1
. . . A
3
= 3 with O, 4 with A, 3 with B
• n = random sample of 5 individuals
• x
1
. . . x
3
= 1 person with O, 2 with A, and 2 with B
f(1, 2, 2; 3, 4, 3, 10, 5) =
3
1
4
2
3
2
1
0
5
=
3
14
Negative Binomial and Geometric
Distributions
In a binomial distribution, we are interested in the probability
of x successes in a fixed n trials.
In a negative binomial distribution, the experiment has the
same properties, but we are interested in the probability that
the kth success occurs on the xth trial.
If repeated independent trials can result in a success with
probability p and failure with a probability q = 1 − p, then the
probability distribution of the random variable X (the number
of the trial on which the kth success occurs) is
b(x; k, p) =
x − 1
k − 1
p
k
q
x−k
, x = k, k + 1, k + 2, . . .
The probability distribution of the random variable X, the
number of the trial on which the first success occurs, is the
geometric distribution
g(x; p) = pq
x−1
, x = 1, 2, 3, . . .
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