The Conditional Probability | Complete Overview

Conditional probability

Conditional probability theory come out from the concept of taking huge risk. there are many issue now a days that stalk from the game of chance, such as throwing coins, throwing a dice, and playing cards. 

Conditional probability theory is applied in many different domains and the flexibility of Conditional probability provides tools for almost so many different needs. probability theory and samples related to the study of the probability of the happening of events.

Consider X and Y both are two events of a incidental experiment. Afterwards, the probability of the happenings of X under the circumstance that Y has already happened with P (Y) ≠ 0, is known as conditional probability and is denoted by P (X / Y).

Therefore, P (X / Y) = The probability of the happening of X, if provided that Y has already happened.

\frac{P(X\cap Y)}{P(Y)}=\frac{n(X\cap Y)}{n(Y)}

Similarly, P (Y / X) = The probability of the occurrence of Y, as X has already happened.

\frac{P(X\cap Y)}{P(X)}=\frac{n(X\cap Y)}{n(Y)}

In brief for some cases, P (X / Y) is used to specify the probability of the occurrence of X when Y occurs. Similarly, P (Y / X) is used to specify the probability of Y happening while X happens.

What is Multiplication theorem on Probability?

If X and Y both are self-supporting (independent) events of  a arbitrary experiment, then

P(X\cap Y)=P(X).P\left ( X/Y \right ) , \ \ if P(X)\neq 0

P(X\cap Y)=P(Y).P\left ( Y/X \right ) , \ \ if P(Y)\neq 0

What is Multiplication theorems for independent events? 

If X and Y both are self-supporting (independent) events connected to  a arbitrary experiment, then   P(X ∩ Y) =P(X).P(Y)

i.e., the probability of simultaneous happening of two independent events is equal to the multiplication of their probabilities. By using multiplication theorem, we have  P(X ∩ Y) =P(Y).P(Y/X)

 As X and Y are independent events, therefore P(Y/X)=P(Y)

Implies, P(X ∩ Y) =P(X).P(Y)

While events are mutually exclusive : 

If X and Y are mutually exclusive events, then ⇒ n(X ∩ Y)= 0 , P(X ∩ Y) = 0

P(X U Y)=P(X) +P(Y)

For any three events X, Y, Z which are mutually exclusive, 

P(X ∩ Y)= P(Y ∩ Z) =P(Z ∩ X) =P(X ∩ Y ∩ Z) =0

P(X \cup Y \cup Z)= P(X) +P(Y) +P (Z)

While events are independent : 

If X and Y are unconstrained (or independent) events, then

P(X \cap Y) =P(X).P(Y)

P(X \cup Y)= P(X) +P(Y) -P(X).P(Y)

Let X and Y be two events connected with a arbitrary (or random) experiment, then

(a) \ \ P(\overline{X} \cap Y) =P(Y)-P(X\cap Y)

(b) \ \ P(X \cap \overline{Y}) =P(Y)-P(X\cap Y)

If Y⊂ X, then

(a) \ \ P(X \cap \overline{Y}) =P(X)-P(Y)

(b) \ \ P(Y) \leq P(X)

Similarly if X⊂ Y, then

(a) \ \ P(\overline{X} \cap Y) =P(Y)-P(X)

(b) \ \ P(X) \leq P(Y)

Probability of occurrence of neither X nor Y is 

P(\overline{X}\cap \overline{Y})= P(\overline{X\cup Y})=1- P(X\cup Y)

Example: If from a pack of cards a single card is picked. What is the possible chance that it is either a spade or a king?

solution:

P (A) = P (a spade card) =13/52

P (B) = P (a king card) =4/52

P (either a spade or a king card) = P (A or B)

=P(A∪B)=P(A)+P(B)-P(A∩B)

=P(A)+P(B)-P(A)P(B)

=13/52+4/52-{(13/52)*(4/52)}

=4/13

Example: Someone is known to hit the target with 3 out of 4 chances, while another person is known to hit the target with 2 out of 3 chances. Find out if the probability that target are likely to be hit at all when both people are trying.

solution:

 probability of  target hit by first person = P (A) =3/4

probability of  target hit by second person = P (B) =2/3

The two events are not mutually exclusive, since both persons hit the same target   =P (A or B)

=P(A∪B)=P(A)+P(B)-P(A∩B)

=P(A)+P(B)-P(A)P(B)

=3/4+2/3-{(3/4)*(2/3)}

=11/12

Example: If  A  and B are two events such that P(A)=0.4 , P(A+B)=0.7 and P(AB)=0.2 then P(B) ?

solution: Since we have P(A+B)=P(A) +P(B) -P(AB)

=> 0.7=0.4+ P(B)-0.2

=> P(B) =0.5

Example: A card is selected at arbitrary from a pack of cards. What is the possibility of the card being a red color card or a queen.

Solution: Required probability  is

P(Red + Queen)-P(Red ⋂ Queen)

=P(Red) +P(Queen)-P(Red ⋂ Queen)

=26/52+4/52-2/52=28/52=7/13

Example: If the probability of X failing in the test is 0.3 and that the probability of Y is 0.2, then find the probability that X or Y failed in the test?

Solution: Here P(X)=0.3 , P(Y)=0.2

Now P(X ∪ Y)= P(X) +P(Y) -P(X ⋂ Y)

Since these are independent events, so

P(X ⋂ Y) =P(X) . P(Y)

Thus required probability is 0.3+0.2 -0.06=0.44

Example: The chances to fail in Physics are 20% and the possibility to fail in Mathematics are 10%. What are the possibilities to fail in at least one subject ?

Solution: Let P(A) =20/100=1/5, P(B) =10/100=1/10

Since events are independent and we have to find 

P(A ∪ B)=P(A) +P(B) -P(A). P(B)

=(1/5)+(1/10)-(1/5). (1/10)= (3/10)-(1/50)=14/50

So the chance of fail in one subject is (14/50)X 100=28%

Example: The probability of solving a question by three students are 1/2,1/4, and 1/6 respectively. What  will be the possible chance of answering the question?

Solution:

(i) This question can also be solved by one student

(ii) This question can be answered by two students concurrently.

(iii) This question can be answered by three students all together.

P(A)=1/2, P(B)=1/4, P(C)=1/6

P(A ∪ B ∪ C)= P(A) + P(B) +P(C)- [P(A).P(B)+P(B).P(C)+P(C).P(A)] + [P(A).P(B).P(C)]

=(1/2)+(1/4)+(1/6)-[(1/2).(1/4)+(1/4).(1/6)+(1/6).(1/2)] +[(1/2).(1/4).(1/6)] =33/48

Example: A random variable X has the probability distribution

X12345678
P(X)0.150.230.120.100.200.080.070.05
Conditional probability: Example

For the events E ={X is prime number} and F={X<4},  find the probability of P(E ∪ F) .

Solution:

E ={ X is a prime number}

P(E) = P(2) +P(3)+ P(5) +P(7) =0.62

F ={X < 4}, P(F) =P(1)+P(2)+P(3)=0.50

and P(E ⋂ F) = P(2)+ P(3) =0.35

P(E ∪ F) =P(E)+P(F) – P(E ⋂ F)

      = 0.62+0.50 – 0.35 = 0.77

Example: Three coins are tossed. If one of them appears tail, then what would be the possible chance that all the three coins appear tail ?

Solution: Consider E is the event where all the three coins appears tail and F is the event where a coin appears tail. 

F= {HHT, HTH, THH, HTT, THT, TTH, TTT}

and E = {TTT}

Required probability = P(E/F)=P(E ⋂F )/P(E)=1/7

Total probability and Baye’s rule

The law of total probability :

For the sample space S and n mutually exclusive and exhaustive events E1 E2 ….En related with a random experiment. If X is a specific event which happens with the events E1 or E2 or …or En, then 

P(X)=P(E_{1}).P(\frac{X}{E_{1}})+P(E_{2})P(\frac{X}{E_{2}})+……..+ P(E_{n}).P(\frac{X}{E_{n}})

Baye’s rule : 

Consider S be a sample space and E1, E2, …..En be n incongruous (or mutually exclusive) events such that

\sum\limits_{i = 1}^nE_{i}=S \ \ and \ \ P(Ei) \ \ >0 \ \ for \ \ i=1,2,……, n

We can think of Ei’s as the factors leading to the outcome of the an experiment. The probabilities P(Ei), i = 1, 2, ….., n are called known as prior (or earlier) probabilities. If the assessment outcomes in an result of event X, where P(X) > 0. Then we have to perceive the possibility that the perceived the event X was due to cause Ei, that is, we look for the conditional probability P(Ei/X) . These probabilities are known as posterior probabilities, given by Baye’s rule as

P(E_{i}/X)=\frac{P(E_{i}).P(X/E_{i})}{\sum\limits_{k = 1}^nP(E_{k})P(X/E_{k})}

Example: There are 3 Boxes which are known to contain 2 blue and 3 green marbles ; 4 blue and 1 green marbles and 3 blue and 7 green marbles respectively. A marble is drawn at random from one of the boxes and found to be a green ball. Then what is the probability that it was drawn from the Box containing the most green marbles.

Solution: Consider the following events :

A ->marble drawn is green;

E1 -> Box 1 is chosen;

E2 Box 2 is chosen

E3 Box 3 is chosen.

P(E1)=P(E2)=P(E3)=1/3 , p(A/E1)=3/5

Then

P(A/E2)=1/5, P(A/E3)=7/10

Required probability =P(E3/A)

=\frac{P(E_{3})P(A/E_{3})}{P(E_{1})P(A/E_{1})+P(E_{2})P(A/E_{2})+P(E_{3})P(A/E_{3})}  =\frac{7}{15}

Example: In an entrance test there are multiple choice questions. There are four probable correct answers to each question of which one is right. The possible chance that a pupil perceives the right answer to a particular question is 90%. If he gets the right answer to a particular question, then what is the probable chance that he was predicting.

Solution: We define the following events :

A1 : He knows the answer.

A2 : He might not know the answer.

E: He is aware of the right answer.

P(A1) =9/10, P(A2) =1-9/10=1/10, P(E/A1)=1,

P(A_{1}) =\frac{9}{10} \ \, P(A_{2}) =1-\frac{9}{10}=\frac{1}{10} \ \ , \ \ P(\frac{E}{A_{1}}) \ \,

Then \ \ P(\frac{E}{A_{2}})=\frac{1}{4}

So the expected probability

Conditional Probability
Conditional Probability

Example: Bucket A contains 4 Yellow and 3 Black Marbles and Bucket B contains 4 Black and 3 Yellow Marbles. One Bucket is taken at random and a Marble is drawn and noted it is Yellow. What is the probability that it comes Bucket B.

Solution: It is based on Baye’s theorem. 

Probability of picked Bucket A , P(A)=1/2

Probability of picked Bucket B , P(B)=1/2

Probability of Yellow Marble picked from Bucket A  =P(A). P(G/A)=(1/2)x (4/7)=2/7 

Probability of Yellow Marble picked from Bucket B = P(B).P(G/B)=(1/2)x(3/7)=3/14

Total probability of Yellow Marbles= (2/7)+(3/14)=1/2

Probability of fact that Yellow Marbles is drawn from Bucket B  

P(G/B)={P(B).P(G/B)}/{P(A).P(G/A)+P(B).P(G/B)}={(1/2)x(3/7)}/{[(1/2)x(4/7)]+[(1/2)+(3/7)]} =3/7

Conclusion:

 In this article we mainly discuss on the Conditional Probability and Bayes theorem with the examples of these the direct and dependent consequence of the trial we discuss so far now in the consecutive articles we relate probability to random variable and some familiar terms related to probability theory we will discuss, if you want further reading then go through:

Schaum’s Outlines of Probability and Statistics and Wikipedia page.

For further study, please refer our mathematics page.

About DR. MOHAMMED MAZHAR UL HAQUE

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