# Probability and Statistics

#### Document technical information

Format pdf
Size 123.9 kB
First found May 22, 2018

#### Document content analysis

Category Also themed
Language
English
Type
not defined
Concepts
no text concepts found

#### Transcript

```Probability and Statistics
Alvin Lin
Probability and Statistics: January 2017 - May 2017
Binomial Random Variables
There are two balls marked S and F in a basket. Select a ball 3 times with
replacement. At each trial, S is likely to be chosen with probability p, and F
is likely to be chosen with probability 1 − p. Let X be the random variable
of the experiment indicating the number of times S is chosen.
F (1 − p)
S (p)
S (p)
S (p)
F (1 − p)
F (1 − p)
S (p)
S (p)
F (1 − p)
S (p)
X=1 X=2
SSS
SSF
SFS
SFF
FSS
FSF
FFS
FFF
X
X
X
X
X
X
1
F (1 − p)
F (1 − p)
S (p)
F (1 − p)
Find the probability mass function of X, b(x; 3, p):
b(0; 3, p) = P (X = 0) = (1 − p)3
The underlying assumption is on the independence of the events:
• A1 : getting an S in the first trial
• A2 : getting an S in the second trial
• A3 : getting an S in the third trial
A1 , A2 , and A3 are mutually independent.
b(0; 3, p) = P (X = 0)
= (1 − p)(1 − p)(1 − p)
3 0
=
p (1 − p)3−0
0
b(1; 3, p) = p(1 − p)(1 − p) + (1 − p)p(1 − p) + (1 − p)(1 − p)p
3 1
=
p (1 − p)3−1
1
From 3 distinct items (trial 1, trial 2, trial 3), select 1 item. There are 3 C1
possible combinations.
b(2; 3, p) = P (X = 2)
= (p)(p)(1 − p) + p(1 − p)p + (1 − p)(p)(p)
3 2
=
p (1 − p)3−2
2
( 3 x
p (1 − p)3−x , x = 0, 1, 2, . . . , n
x
b(x; 3, p) =
0
, otherwise
The above example is an example of a binomial experiment with a binomial
random variable.
1. This experiment consists of a sequence of n smaller experiments called
trials, where n is fixed in advance of the experiment.
2
2. Each trial can result in one of the two possible outcomes (dichotomous
trials), which we generically denote by success(S) or f ailure(S). The
assignment of the S and F labels to the two sides of the dichotomy is
arbitrary.
3. The trials are independent, so that the outcome on any particular trial
does not influence the outcome of any other trial.
4. The probability of success P (S) is constant from trial to trial. We
denote this probability by p.
An experiment for which the above conditions (a fixed number of dichotomous, independent, homogeneous trials) ar satisfied is called a binomial experiment.
PMF of a binomial random variable X
The probability mass function of a binomial random variable X is:
( n x
p (1 − p)n−x , x = 1, 2, 3, . . . , n
x
b(x; n, p) =
0
, otherwise
The value of X indicates the number of S’es.
CDF of a binomial random variable X
The cumulative distribution function of a binomial random variable X is:
B(x; n, p) = P (X ≤ x)
x
X
=
b(y; n, p)
y=0
if x = 0, 1, 2, . . . , n
X ∼ Bin(n, p) denotes that X is a binomial random variable with probability
mass function b(x;n,p).
Expected value of a binomial random variable X
E(X) =
X
xb(x; n, p) = np
x∈{0,1,2,...,n}
3
Variance of a binomial random variable X
V (X) =
X
(x − E(X))2 b(x; n, p) = np(1 − p)
x∈{0,1,2,...,n}
The standard deviation of X is σ =
p
p
V (X) = np(1 − p)
Example
An aircraft seam requires 25 rivets. The seam will have to be reworked if
any of these rivets are defective. Suppose rivets are defective independently
of one another, each with the same probability. If 15% of all seams need
reworking, what is the probability that a rivet is defective?
1 − 0.15 = P (a seam does not need reworking)
= P (a seam has zero def ective rivets)
= (1 − p)25
1
1 − p = 0.85 25
1
p = 1 − 0.85 25
Find the probability that a randomly selected seam has exactly 3 defective
rivets.
25 3
p (1 − p)22
3
Example
A very large batch of components has arrived at a distributor. The batch
can be characterized as acceptable only if the proportion of defective components is at most 0.10. The distributor decides to randomly select 10 or 15
components and to accept the batch only if the defective components in the
sample is at most 1 or 2, respectively.
Consider a simpler experiment. Select 2 components and inspect them.
The probability that the two are both defective is:
(
k k−1
)(
)
N N −1
4
where k is the number of defective components in the batch. If N and k are
large:
k
k−1
∼
N −1
N
Consider sampling without replacement from a dichotomous population of
size N . If the sample size (number of trials) n is at most 5% of the population size. The experiment can be analyzed as though it were a binomial
experiment.
Trial: select a component from the batch and inspect the component for
defects.
k
p=
N
where k is the number of defective components in the batch and X is the
number of defective components in the n trials. In the real world, we do not
replace a component after inspecting it, but we can approximate it for the
purpose of this experiment.
X ∼ Bin(10, p)
P (X ≤ 2) =
=
2
X
b(x; n, p)
x=0
2 X
n x
p (1 − p)n−x
x
x=0
M ethod 1 : n = 10
p = Nk
0.01
0.02
0.1
0.2
0.25
P (X ≤ 2) 0.999 0.9985 0.9298 0.6778 0.5256
M ethod 2 : n = 10
p = Nk
0.01
0.02
0.1
0.2
0.25
P (X ≤ 1) 0.9957 0.9139 0.7361 0.3758 0.2440
M ethod 3 : n = 15
p = Nk
0.01
0.02
0.1
0.2
0.25
P (X ≤ 2) 0.9996 0.9638 0.8159 0.3980 0.2361
Which method is the best? Our goal is that the batch is accepted if p ≤ 0.10
and rejected if p > 0.10.
5
p = Nk
0.10
P (the event such that we accept the batch) high
low
The third method is the best.