Probablity of statistics

1. Probability
Trial and Event: If an experiment be repeated under essentially the same condition giving several possible
outcomes then the experiment is called a trial and the possible outcomes are known as events. Tossing a coin is
a trial and getting head or tail are outcomes.
Exhaustive outcomes: The total number of possible outcomes of any trial is known as exhaustive cases. In
throwing a dice the exhaustive number of outcomes is six.
Equally likely outcomes: Cases are said to be equally likely if we have no reason to expect any one rather than
other. Drawing a card from a pack all the 52 cards are equally likely to come. Each has a probability of
occurrence 1/6.
Mutually exclusive outcomes: Outcomes are said to be mutually exclusive if the happening of any one of them
precludes (excludes) the happening of others. or If two or more events cannot occur together, they are called
mutually exclusive event . In tossing a coin the outcomes head and tail are mutually exclusive. Drawing Kings
and Hearts from a pack are not mutually exclusive , because we can have a King of Hearts!
Independent event: If the occurrence of an event is not influenced or affected by the occurrence or not
occurrences of another event these two events are said to be independent of each other. If two coins are tossed,
head or tail in one coin and tail or head in the other coin are independent.
Dependent Event: If the occurrence of an event is influenced by the occurrence of another event then the two
events are said to be dependent to each other. Removing marbles from a bag are affected by previous events.

2. Probability
Sample space : The totality of all possible outcomes of a random experiment is called sample space.
Sample space is usually denoted by S. If a dice is thrown, the sample space is S = {1,2,3,4,5,6}.
Random variable: A variable whose values are any definite numbers or quantities that arise as a result of
chance factors such that they can not exactly be predicted in advance is called a random variable.
Definition of Probability: If a trial results in ‘n’ exhaustive, mutually exclusive and equally likely cases ‘m’
of which are favorable to a particular event A. The probability ‘p’ of the event A is defined as the ratio of
the favorable cases to the total number of case. 𝑝 𝐴 =
𝑚
𝑛
, 0 ≤ 𝑝 ≤ 1
❖ Show that (1) 𝑝 + 𝑞 = 1 (2) 0 ≤ 𝑝 ≤ 1 , 0 ≤ 𝑞 ≤ 1
Proof: Let 𝑚 be a favorable number of cases and 𝑛 be a exhaustive number of cases of an event 𝐸, then
by definition of probability 𝑝 𝐸 =
𝑚
𝑛
Since the number of cases favorable to the non-happening of the event 𝐸 are ( 𝑛 − 𝑚), the probability 𝑞
that 𝐸 will not happen is given by 𝑞 =
𝑛−𝑚
𝑛
= 1 −
𝑚
𝑛
= 1 − 𝑝 ⇒ 𝑝 + 𝑞 = 1
Obviously p as well as q are non-negative and can not exceed unit
i.e. , 0 ≤ 𝑝 ≤ 1 , 0 ≤ 𝑞 ≤ 1

3. Probability
❖ A box contains 6 red, 4 white and 5 black balls. A person draws 4 balls from the box at random . Find the
probability that among the balls drown there is at least one ball of each color .
Solution: The required event E that in a draw of 4 balls from the box at random there is at least one ball of
each color can materialise in the following mutually disjoint ways:
(1) 1 Red, 1 white, 2 black balls (2) 2 Red, 1 white, 1 black balls (3) 1 Red, 2 white, 1 black balls
Hence the required probability is given by 𝑝 𝐸 = 𝑝 1 + 𝑝 2 + 𝑝 3
=
6
1 × 4
1 × 5
2
15
4
+
6
2 × 4
1 × 5
1
15
4
+
6
1 × 4
2 × 5
1
15
4
=
240
1365
+
300
1365
+
180
1365
=
720
1365
=
144
273
=
48
91
❖ (a) Two cards are drawn at random from a well-shuffled pack of 52 cards. Show that the chance of drawing
two aces is
1
221
(b) From a pack of 52 cards, three are drawn at random. Find the chance that they are a king, a
queen and a knave.
( c ) Four cards are drawn from a pack of 52 cards. Find the probability that (1) all are diamond (2) there is
one card of each suit and (3) there are two spades and two hearts.

4. Binomial distribution
Binomial distribution: Let an experiment be repeated for 𝑛 independent trials each with one of two possible
outcomes, success or failure. The number of success 𝑥 in a trial is a discrete random variable which can
assume values 0,1,2, . ; . , 𝑛. Let 𝑝 be the probability of success and 𝑞 be the probability of failure in a single
trial so that 𝑝 + 𝑞 = 1.If the probability of success, 𝑝 remains same for trial to trial, than the distribution of 𝑥
is known as Binomial distribution and its probability function is given by
𝑝 𝑥 = 𝑛
𝑥
𝑝𝑥𝑞𝑛−𝑥 , 𝑥 = 0, 1, 2, … … . 𝑛
Mean and Variance of Binomial Distribution: The Probability function of Binomial Distribution is given by
𝑝 𝑥 = 𝑛
𝑥
𝑝𝑥
𝑞𝑛−𝑥
, 𝑥 = 0, 1, 2, … … . 𝑛
Mean = E(X)= σ𝑥=0
𝑛
𝑥. 𝑝(𝑥) = σ𝑥=0
𝑛
𝑥. 𝑛
𝑥
𝑝𝑥
𝑞𝑛−𝑥
= 0 + 1. 𝑛
1
𝑝1𝑞𝑛−1 + 2. 𝑛
2
𝑝2𝑞𝑛−2 + 3. 𝑛
3
𝑝3𝑞𝑛−3+………… +𝑛. 𝑛
𝑛
𝑝𝑛𝑞𝑛−𝑛
= 𝑛𝑝𝑞𝑛−1 + 2.
𝑛.(𝑛−1)
2
𝑝2𝑞𝑛−2 + 3.
𝑛.(𝑛−1)(𝑛−2)
1.2.3
𝑝3𝑞𝑛−3 +……..+𝑛. 𝑝𝑛
= 𝑛𝑝(𝑞𝑛−1 + (𝑛 − 1)𝑝𝑞𝑛−2 +
(𝑛−1)(𝑛−2)
2
𝑝2𝑞𝑛−3 +……..+𝑝𝑛−1)
= 𝑛𝑝(𝑞𝑛−1 + 𝑛−1
1
𝑝𝑞𝑛−2 + 𝑛−1
2
𝑝2𝑞𝑛−3 +……..+𝑝𝑛−1) = 𝑛𝑝(𝑞 + 𝑝)𝑛−1= 𝑛𝑝1𝑛−1 = 𝑛𝑝

5. Binomial distribution
So Mean = 𝑛𝑝
Variance = 𝐸[𝑥 − 𝐸(𝑥)]2
= 𝐸(𝑥2
) − { 𝐸(𝑥)}2
= 𝐸{𝑥(𝑥 − 1)} + 𝐸(𝑥) − { 𝐸(𝑥)}2
Now 𝐸 𝑥 𝑥 − 1 = σ𝑥=0
𝑛
𝑥(𝑥 − 1). 𝑝(𝑥) = σ𝑥=0
𝑛
𝑥(𝑥 − 1). 𝑛
𝑥
𝑝𝑥𝑞𝑛−𝑥
= 0 + 0 + 2.1. 𝑛
2
𝑝2𝑞𝑛−2 + 3.2. 𝑛
3
𝑝3𝑞𝑛−3 + 4.3. 𝑛
4
𝑝4𝑞𝑛−4+………… +𝑛(𝑛 − 1). 𝑛
𝑛
𝑝𝑛𝑞𝑛−𝑛
= 2.
𝑛.(𝑛−1)
2
𝑝2
𝑞𝑛−2
+ 3.2
𝑛.(𝑛−1)(𝑛−2)
1.2.3
𝑝3
𝑞𝑛−3
+ 4.3
𝑛.(𝑛−1)(𝑛−2)(𝑛−3)
1.2.3.4
𝑝4
𝑞𝑛−4
……..+𝑛(𝑛 − 1). 𝑝𝑛
= 𝑛. (𝑛 − 1)𝑝2(𝑞𝑛−2 + (𝑛 − 2) 𝑝𝑞𝑛−3 +
(𝑛−2)(𝑛−3)
2
𝑝2𝑞𝑛−4……..+𝑝𝑛−2)
= 𝑛. (𝑛 − 1)𝑝2(𝑞𝑛−2 + 𝑛−2
1
𝑝𝑞𝑛−2 + 𝑛−2
2
𝑝2𝑞𝑛−3 +……..+𝑝𝑛−1)
= 𝑛. (𝑛 − 1)𝑝2(𝑞 + 𝑝)𝑛−2= 𝑛. (𝑛 − 1)𝑝21𝑛−2 = 𝑛. (𝑛 − 1)𝑝2
So Variance = 𝐸 𝑥 𝑥 − 1 + 𝐸 𝑥 − 𝐸 𝑥 2 = 𝑛. 𝑛 − 1 𝑝2 + 𝑛𝑝 − 𝑛𝑝 2 = 𝑛𝑝 − 𝑛𝑝2
= 𝑛𝑝 1 − 𝑝 = 𝑛𝑝𝑞
∴ Standard Deviation = + Variance = + 𝑛𝑝𝑞

6. Poisson distribution
Poisson distribution: If 𝑋 be a Poisson variable then the probability function of 𝑋 is given by
𝑝 𝑥 =
𝑒−𝑚𝑚𝑥
𝑥!
, 𝑥 = 0, 1, 2, … … . 𝑛
Mean and Variance of Poisson distribution : The Probability function of Poisson distribution is given by
𝑝 𝑥 =
𝑒−𝑚𝑚𝑥
𝑥!
, 𝑥 = 0, 1, 2, … … . 𝑛
Mean = E(X)= σ𝑥=0
∞
𝑥. 𝑝(𝑥) = σ𝑥=0
∞
𝑥.
𝑒−𝑚𝑚𝑥
𝑥!
= 0 + 1.
𝑒−𝑚𝑚1
1!
+ 2.
𝑒−𝑚𝑚2
2!
+ 3.
𝑒−𝑚𝑚3
3!
+………… ∞
= 𝑚𝑒−𝑚
(1 +
𝑚
1!
+
𝑚2
2!
+………… ∞) = 𝑚𝑒−𝑚
. 𝑒𝑚
= 𝑚
∴ Mean= 𝑚
Variance = 𝐸[𝑥 − 𝐸(𝑥)]2 = 𝐸(𝑥2) − { 𝐸(𝑥)}2 = 𝐸{𝑥(𝑥 − 1)} + 𝐸(𝑥) − { 𝐸(𝑥)}2
Now 𝐸 𝑥 𝑥 − 1 = σ𝑥=0
∞
𝑥(𝑥 − 1). 𝑝(𝑥) = σ𝑥=0
∞
𝑥(𝑥 − 1).
𝑒−𝑚𝑚𝑥
𝑥!
= 0 + 0 + 2.1.
𝑒−𝑚𝑚2
2!
+ 3.2.
𝑒−𝑚𝑚3
3!
+ 4.3.
𝑒−𝑚𝑚4
4!
+………… ∞

7. Math and it’s solution
= 𝑚2𝑒−𝑚(1 +
𝑚
1!
+
𝑚2
2!
+………… ∞) = 𝑚2𝑒−𝑚. 𝑒𝑚 = 𝑚2
So Variance = 𝐸 𝑥 𝑥 − 1 + 𝐸 𝑥 − 𝐸 𝑥 2 = 𝑚2 + 𝑚 − 𝑚2 = 𝑚
∴ Variance = 𝑚
Example5: If 20% of the bulbs produced by a machine are defective, determine the probability that, out of
5 bulbs chosen at (a) exactly 2 bulb will be defective (b) None of the bulbs will be defective (c) at most 1
bulbs will be defective.
Solution: The Probability function 𝑃(𝑋) of 𝑥 (defective bulbs) in Binomial Distribution is given by
𝑝 𝑥 = 𝑛
𝑥
𝑝𝑥𝑞𝑛−𝑥 , 𝑥 = 0, 1, 2, 3, 4, 5. Now n=5, p=20%=0.2 and q=1-p=0.8
∴ 𝑝 𝑥 = 5
𝑥
0.2 𝑥 0.8 5−𝑥
Now (a) The probability of exactly 2 bulb will be defective = 𝑝 2 = 5
2
0.2 2 0.8 5−2 = 0.20
(b) The probability of none of the bulbs will be defective = 𝑝 0 = 5
0
0.2 0 0.8 5−0 = 0.33
(c) The probability of at most 1 bulbs will be defective 𝑝 0 + 𝑝 1 = 0.33 + 5
1
0.2 1 0.8 5−1 = 0.74

8. Math and it’s solution
Find the mean and variance from the following table:
Solution: Calculation table:
Mean, 𝜇 = σ x P(x) = 10.85
And Variance = σ x𝟐P(x) − 𝜇2 = 131.95 − 10.85 2 = 14.23
Experience of the employees in a textile company (in Year) : x 5 10 12 15 20
Probability : P(x) 0.20 0.30 0.30 0.15 0.05
Experience : x Probability : P(x) x P(x) x𝟐
P(x)
5
10
12
15
20
0.20
0.30
0.30
0.15
0.05
1
3
3.6
2.25
1
5
30
43.2
33.75
20
෍ x P(x) = 10.85 ෍ x𝟐
P(x) = 131.95