Probability Distribution (Discrete Random Variable)

Statistics and Probability • PRINCESS P. DIPASUPIL
• Special ScienceTeacher I
Probability Distribution of a
RandomVariable
Statistics and Probability
PRINCESS P. DIPASUPIL
Special ScienceTeacher I

PROBABILITY
DISTRIBUTION
the set of all possible
values of the random variable
X, together with their
corresponding associated
probabilities.

PROBABILITY
DISTRIBUTION
(𝑅𝑎𝑛𝑑𝑜𝑚
𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑋)
𝐴𝑔𝑒
𝑃(𝑥)
16 6/25
17 14/25
18 5/25

PROBABILITY
DISTRIBUTION
If X is a discrete random
variable, the probability
distribution is called a probability
mass function or pmf.
The pmf may be expressed in
tabular or graphical form.

Properties of a Probability Distribution
I. 0 ≤ 𝑃 𝑥 ≤ 1
2. Σ𝑃 𝑥 = 1

discrete probability distribution or not?
x 1 2 3 4 5
P(x) 0.10 0.20 0.25 0.40 0.05

x 1 2 3 4 5
P(x) 0.05 0.25 0.33 0.25 0.08

x 1 2 3 4
P(x) 0.2 0.1 0.4 0.3

x 1 2 3 4
P(x) 0.2 0.1 0.4 0.3
Determine 𝑃 1 + 𝑃 2 :
0.2 + 0.1 = 0.3

Determine 𝑃 3 − 𝑃 1 :
0.4 − 0.2 = 0.2
x 1 2 3 4
P(x) 0.2 0.1 0.4 0.3

Determine
P(4)+P(2).
x 1 2 3 4
P(x) 0.2 0.1 0.4 0.3

The spinner below is divided into
12 sections. Let X be the score
where the arrow will stop
(numbered as 1, 2, 3, 4, 5).
Find the probability that the
arrow will stop at 1, 2, 3, 4, and 5.
Construct the discrete
probability distribution of the
random variable X and its
corresponding histogram.

Suppose three cell phones are tested at random.We
want to find out the number of defective cell phones
that occur.Thus, to each outcome in the sample space
we shall assign a value.These are 0, 1, 2, or 3. If there is
no defective cell phone, we assign the number 0; if
there is one defective cell phone, we assign the number
1; if there are two defective cell phones, we assign the
number 2; and 3, if there are three defective cell
phones.The number of defective cell phones is a
random variable.The possible values of this random
variable are 0,1,2, and 3.

Let D = defective CP
N = non-defective CP
X = (the random variable)
number of defective cell
phones
Possible
Outcomes
Value of the
RandomVariable X
(number of defective cell
phones)
DDD 3
DDN 2
DND 2
DNN 1
NNN 0
NND 1
NDN 1
NDD 2

𝑋
number of
defective cell
phones
𝑃(𝑥)
3
1
8
2
3
8
1
3
8
0
1
8
Possible
Outcomes
Value of the
RandomVariable
X
(number of
defective cell
phones)
DDD 3
DDN 2
DND 2
DNN 1
NNN 0
NND 1
NDN 1
NDD 2

𝑋
number
of
defective
cell
phones
𝑃(𝑥)
3
1
8
2
3
8
1
3
8
0
1
8
8
8
7
8
6
8
5
8
4
8
3
8
2
8
1
8
0 1 2 3

Two balls are drawn in
succession without
replacement from an urn
containing 5 red balls and 6
blue balls. Let Z be the
random variable
representing the number of
blue balls. Find the values of
the random variable Z.

Possible
Outcomes
Value of the
Random
Variable Z
(number of
blue balls)
𝑍
number
of blue
balls
𝑃(𝑧)

𝑍
number
of blue
balls
𝑃(𝑧)

A shipment of five boxes of mussels (tahong)
contains two that are slightly spoiled. If a retailer
receives three of these boxes of mussels at
random, list the elements of the sample space
using the letters S and F for spoiled and fresh
mussels, respectively.To each sample point, assign a
value x of the random variable X representing the
number of boxes of mussels purchased by the
retailer which are slightly spoiled.

𝑋
number of boxes of
mussels which are
slightly spoiled
𝑃(𝑥)
Possible
Outcomes
Value of the
RandomVariable
X
(number of boxes
of mussels which
are slightly
spoiled)

𝑋
number of
boxes of
mussels which
are slightly
spoiled
𝑃(𝑥)

Probability Distribution (Discrete Random Variable)

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