Conditional probability of an event is calculated applying the necessary changes to the sample space, when it is known another dependent event has already occurred.

Consider two dependent events that can occur together like the incidences of High blood pressure and diabetes. Suppose it is known that a person is diabetic, and we want to know the probability of the person suffering from High blood pressure as well. The probability in this case is called the conditional probability of the incidence of high blood pressure when the diabetes is already detected.Let us see how the conditional probability is calculated, the rules associated with it and solve few practice problems as well.

The conditional probability of an event B with reference to another event A is the probability
that event B occurs after event A has already occurred.
The notation for conditional probability P(B | A) is read as the probability of the event B given that the event A has already occurred.When it is given that the event A has already occurred, the sample space for computing the conditional probability is restricted to A.
The conditional probability of an event B, given that event A has occurred is given by
P(B | A) = $\frac{P(A \cap B)}{P(A)}$

Based on the above rule the multiplication rule for two dependent events can be given as follows:
When two events A and B are dependent the probability of both occurring is
P(A \cap B) = P(A) . P(B | A)
or
P(A \cap B) = P(B) . P(A | B).
Let A and B be two dependent events.
By theoretical definition of Probability
Probability of an event E = $\frac{n(E)}{n(S)}$
If it is given the event A has already occurred, the sample space for computing P(B | A) is restricted to the event A.

Thus we need to consider only the outcomes of the event A ∩ B as other elements in B will not contribute to this chance of occurrence. Thus, by the definition of Probability
P(B | A) = $\frac{n(A\cap B)}{n(A)}$

= $\frac{\frac{n(A\cap B)}{n(S)}}{\frac{n(A)}{n(S)}}$ Divide both the numerator and denominator by n(S).

= $\frac{P(A \cap B)}{P(A)}$.
Rewriting the above formula, the multiplication rule for two dependent events is
P(A $\cap$ B) = P(A) . P(B | A)

When the events A and B are independent, then P(B | A) = P(B) and P(A | B) = P(A).
Conditional Probabilities are often calculated using data given in table format. The table form makes it easy to identify number of elements common to both the events in question.

Solved Example

Question: Three hypothetical computer chip companies manufacture a particular kind of chip that is used in Mobile Phones. The following table details on functioning and defective chips produced during the course of manufacturing cycle by the three companies.

Company
     Number of
functioning chips

    Number of
defective chips

 Total 
      A           570
         30
  600
      B
          385
         15
  400
      C
          480
         20
  500
   Total         1435
         65
 1500

a. If a chip is selected at random, what is the probability the chip is defective.
b.If the chip selected is defective, what is probability that it is manufactured by company B?
c. If a chip selected is found to be manufactured by company A, what is the probability that it is defective?
Solution:
 
a.  Total number of chips produced = 1500
     Total number of defective chips = 65
      Probability of picking a defective chip = $\frac{65}{1500}$ ≈ 0.0433

b.   Let E be the event of picking a defective chip and F the event of picking a chip produced by company
      We need to find P(F | E)
      P(F | E) = number of defective chips produced by company B / Total number of defective chips
                   = $\frac{15}{65}$ ≈ 0.231.

c. Let E1 be the event of picking a chip manufactured by company A ad E2 the event of picking a defective chip.
    We need to find P(E2 | E1)
    P(E2 | E1) = number of defective chips produced by company A / Total number of chips produced by company A.
                    = $\frac{30}{600}$ = 0.05.
 


Solved Examples

Question 1: A card is picked from a pack of card at random. If it is known that the card picked is a spade, what is the probability that it is a face card.
Solution:
 
Let A be the event of picking a spade and B the event of picking a face card.

    We need to find P( B | A)

   P( B | A)   = $\frac{P(A\cap B)}{P(A)}$

                   = $\frac{\frac{3}{52}}{\frac{13}{52}}$ = $\frac{3}{13}$.
 

Question 2: The probability of a driver getting a parking ticket for parking in a No Parking zone is 0.10.  The probability of Roger parking his car in a No Parking Zone is 0.25. Roger parks his car in No Parking zone on a crowded day when he goes for shopping. What is the probability that he would get a parking ticket?
Solution:
 
Let A be the event of a person getting a parking ticket and B the event of Roger parking in the No Parking Zone.

     We need to find P(A | B) and it is given P(A $\cap$ B ) = 0.10 and P(B) = 0.25

     P(A | B) = $\frac{P(A\cap B)}{P(B)}$

                  = $\frac{0.10}{0.25}$ = $\frac{10}{25}$ = 0.4
 


Practice Problems

Question 1: In a restaurant 90% of the customers order Pizza and 72% of the customers order a Pizza and coke. Find the probability that a customer who orders Pizza will also order Coke. (Answer 0.8)
Question 2: A company received 1200 job applications. 480 applicants had desired education and also work exposure and 840 applicants had desired education. if the first applicant screened is known to have the desired education, what is the probability that he also had work exposure.  (Answer : $\frac{4}{7}$)
Question 3: The following table shows the Sports preferences of 500 high school students:    
Class
Base Ball Tennis
Basket Ball
Foot Ball
 Juniors     68    55      72
    42
 Seniors     75    58
      59    71

  a)  If a student is chosen, find the probability that the student's preferred sport is Basket Ball, given that he is a Junior.
  b)  If a randomly chosen student's principal interest is Tennis, find the probability that he is a Senior
       (Answer: 0.308, 0.513).