Conditional Probability

Let A and B be two events associated with the same random experiment. The probability of occurrence of A under the condition that B has already occurred (and P (B) ≠ 0) is called Conditional Probability and is denoted by P (A/B). P (A/B) signifies the probability of occurrence of event A given that event B has already occurred.

Thus, we define Conditional Probability as

P (A/B) = P (A∩B) / P (B), where P (B) ≠ 0

If we have to calculate P(B/A), i.e. we have to calculate the probability of occurrence of event B given that event A has already occurred, then the formula is changed to

P (B/A) = P (A∩B) / P (A).

P (A∩B) is the number of outcomes common between event A and event B. This can also be written as P (A or B), and the relation can be represented as

P (AUB) = P (A) + P (B) – P (A∩B)

If P (A∩B) = 0, then events A and B are mutually exclusive, i.e. they do not occur together.

Example 3

Let A and B be events such that P (A) = 0.5, P (B) = 0.6 and P (AUB) = 0.8. Calculate

  1. P(A/B)
  2. P(B/A)

P (A) = 0.5, P (B) = 0.6 and P (AUB) = 0.8 and P (AUB) = P (A) + P(B) - P(A∩B),

P (A∩B) = 0.3

  1. P(A/B) = P(A∩B) / P(B) = 0.3 / 0.6 = 0.5
  2. P(B/A) = P(A∩B) / P(A) = 0.3 / 0.5 = 0.6

Example 4

A die is rolled. If the outcome is an odd number, what is the probability of it being a prime number?

Solution: Sample space is given by S = {1, 2, 3, 4, 5, 6}

A = Event of getting an odd number = {1, 3, 5}

B = Event of getting an odd prime number = {3, 5}

Thus, P (A∩B) = 2/6 (the probability of a number being odd AND prime), and P (A) = 3/6


P (B/A) = P (A∩B) / P (A) = (2/6) / (3/6) = 2/3

A logically consistent way of looking at the problem is to apply the basic rules of probability discussed earlier; P (A) = Number of possible favorable outcomes / Total number of outcomes

Here the number of possible odd numbers = 3, and of these there are 2 that are prime.

Therefore as above, probability of getting a prime from the odd numbers on a die is 2/3

Example 5

Using a table to calculate conditional probabilities

In a box there are 100 resistors having resistance and tolerance as shown in the table below. Let a resistor be selected from the box and assume each resistor has the same likelihood of being chosen. Define three events as follow: A as “draw a 47W resistor,” B as “draw a resistor with 5% tolerance” and C as “draw a 100W resistor.”

Try these questions

  1. A and B are events such that P (A) = 0.8, P (B) = 0.1 and P (A∩B) = 0.6. Calculate P (AUB)
    a.     1.5
    b.     0.3
    c.     0.6
    d.     0.9
    Answer: B
  2. A pair of dice is rolled. Find the probability that the sum of the two numbers that appear in each of the two die is 7.
    a.     1/7
    b.     1/6
    c.     1/5
    d.     ¼
    Answer: B
  3. Let A and B be events such that P(A) = 0.3, P(B) = 0.4 and P(AUB) = 0.5. What is the value of P (B/A).
    a.     0.6
    b.     0.67
    c.     0.7
    d.     0.8
    Answer: B