This presentation contains the topic as follows, probability distribution, random variable, continuous variable, discrete variable, probability mass function, expected value and variance and examples

1. PROBABILITY
DISTRIBUTION
Mr. S.L.Khairnar

2. A variable whose value is determined by
the outcomes of a random experiment is
called a random variable
RandomVariable

3. A discrete random variable is the one
which takes only integer values
Discrete RandomVariable

4. A continuous random variable is the one
which takes all values
Continuous RandomVariable

5. 1) Head orTail or “1” for Head and “0” for
Tail is discrete random variable
2) Rainfall in cms in a city for every year
is continuous random variable
Example

6. If X is a random variable taking values
𝑥1, 𝑥2, … , 𝑥 𝑛 then P(X) for
𝑋 = 𝑥1, 𝑥2, … , 𝑥 𝑛
is called Probability Mass Function
corresponding to the given experiment
Probability Mass Function

7. All the outcomes of an experiment
expressed in the form of random
variable along with their probabilities of
their occurrences form a Probability
Distribution
Probability Distribution
x 𝒙 𝟏 𝒙 𝟐 𝒙 𝟑 …. 𝒙 𝒏
P(x) 𝑷(𝒙 𝟏) 𝑷(𝒙 𝟐) 𝑷(𝒙 𝟑) ----- 𝑷(𝒙 𝒏)

8. Tossing a coin, 1 and 0 represents
random variables resp Head andTail
Example
x Head Tail Total
1 0
P(x) 1/2 1/2 1

9. Tossing two coins,
TT=0, HT orTH=1 HH=2
Example
x TT HT orTH HH Total
0 1 2
P(x) 1/4 1/2 1/4 1

10. The probability distribution for the
number on the uppermost face of a dice
is
Example
x 1 2 3 4 5 6 Total
P(x) 1/6 1/6 1/6 1/6 1/6 1/6 1

11. If X is a random variable, the function
F(x)=P(X≤x) is called cumulative
frequency distribution function for X
Cumulative Frequency Distribution
x 𝒙 𝟏 𝒙 𝟐 𝒙 𝟑 …. 𝒙 𝒏
P(x) 𝑷(𝒙 𝟏) 𝑷(𝒙 𝟐) 𝑷(𝒙 𝟑) ----- 𝑷(𝒙 𝒏)
𝐹 𝑥3 = 𝑃 𝑋 ≤ 𝑥3 = 𝑃 𝑥1 + 𝑃 𝑥2 + 𝑃(𝑥3)
𝑃 𝑋 > 𝑥3 = 1 − 𝐹 𝑥3 = 1 − 𝑃 𝑋 ≤ 𝑥3 = 1 − (𝑃 𝑥1 + 𝑃 𝑥2 + 𝑃 𝑥3 )
= 𝑃 𝑥4 + 𝑃 𝑥5 + ⋯ + 𝑃(𝑥 𝑛)

12. The probability distribution for the
number on the uppermost face of a dice
is
Example
x 1 2 3 4 5 6 Total
P(x) 1/6 1/6 1/6 1/6 1/6 1/6 1
𝐹 3 = 𝑃 𝑋 ≤ 3 = 𝑃 1 + 𝑃 2 + 𝑃 3 =
1
6
+
1
6
+
1
6
=
3
6
=
1
2
𝑃 𝑋 > 3 = 1 − 𝐹 3 = 1 − 𝑃 𝑋 ≤ 3 = 1 − 𝑃 1 + 𝑃 2 + 𝑃 3 = 1 −
1
2
=
1
2
= 𝑃 4 + 𝑃 5 + 𝑃 6 =
1
6
+
1
6
+
1
6
=
3
6
=
1
2

13. ExpectedValue andVariance
x 𝒙 𝟏 𝒙 𝟐 𝒙 𝟑 …. 𝒙 𝒏
P(x) 𝑷(𝒙 𝟏) 𝑷(𝒙 𝟐) 𝑷(𝒙 𝟑) ----- 𝑷(𝒙 𝒏)
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑉𝑎𝑙𝑢𝑒 = 𝐸 𝑋 =
𝑥∈𝑆
𝑥𝑃(𝑥) 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝐸 𝑋2
− [𝐸 𝑥 ]2
𝒙 𝑃(𝑥) 𝑥𝑃(𝑥) 𝑥 𝟐 𝒙 𝟐 𝑷(𝒙)
𝑥1 𝑃(𝑥1) 𝑥1 𝑃(𝑥1) 𝑥1
𝟐
𝒙 𝟏
𝟐
𝑷(𝒙 𝟏)
𝑥2 𝑃(𝑥2) 𝑥2 𝑃(𝑥2) 𝑥2
𝟐
𝒙 𝟐
𝟐
𝑷(𝒙 𝟐)
Total 1 E(X) 𝐸 𝑋2

14. ExpectedValue andVariance
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑉𝑎𝑙𝑢𝑒 = 𝐸 𝑋 =
𝑥∈𝑆
𝑥𝑃 𝑥 = 1 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝐸 𝑋2 − [𝐸 𝑥 ]2 =
3
2
− 12 =
3
2
− 1 =
1
2
𝒙 𝑃(𝑥) 𝑥𝑃(𝑥) 𝑥 𝟐
𝒙 𝟐
𝑷(𝒙)
0 1
4
0 ×
1
4
= 0
0
0 ×
1
4
= 0
1 1
2
1 ×
1
2
=
1
2
1
1 ×
1
2
=
1
2
2 1
4
2 ×
1
4
=
1
2
4
4 ×
1
4
= 1
Total 1
𝐸 𝑋 = 0 +
1
2
+
1
2
= 1
----
𝐸 𝑋2
= 0 +
1
2
+ 1 =
3
2
x 0 1 2 Total
P(x) 1/4 1/2 1/4 1