Conditional probability P(A | B) indicates the probability of event ‘A’ happening given that event B happened.

We can easily understand the above formula using the below diagram. Since B has already happened, the sample space reduces to B. So the probability of A happening becomes divided by P(B)

**Example:**

In a batch, there are 80% C programmers, and 40% are Java and C programmers. What is the probability that a C programmer is also Java programmer?

Let A --> Event that a student is Java programmer B --> Event that a student is C programmer P(A|B) = P(A ∩ B) / P(B) = (0.4) / (0.8) = 0.5 So there are 50% chances that student that knows C also knows Java

**Product Rule:**

Derived from above definition of conditional probability by multiplying both sides with P(B)

P(A ∩ B) = P(B) * P(A|B)

**Understanding Conditional probability through tree:**

Computation for Conditional Probability can be done using tree, This method is very handy as well as fast when for many problems.

**Example:** In a certain library, twenty percent of the fiction books are worn and need replacement. Ten percent of the non-fiction books are worn and need replacement. Forty percent of the library’s books are fiction and sixty percent are non-fiction. What is the probability that a book chosen at random are worn? Draw a tree diagram representing the data.

**Solution:** Let F represents fiction books and N represents non-fiction books. Let W represents worn books and G represents non-worn books.

P(worn)= P(N)*P(W | N) + P(F)*P(W | F) = 0.6*0.1 + 0.4* 0.2 = 0.14

**Exercise:**

**1)** What is the value of P(A|A)?

**2)** Let P(E) denote the probability of the event E. Given P(A) = 1, P(B) = 1/2, the values of P(A | B) and P(B | A) respectively are (GATE CS 2003)

(A) 1/4, 1/2

(B) 1/2, 1/14

(C) 1/2, 1

(D) 1, 1/2

See this for solution.

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