JEE Main Registration 2025 Closed, What is Next, Updates

Conditional Probability - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Conditional Probability is considered one of the most asked concept.

  • 33 Questions around this concept.

Solve by difficulty

It is given that the events A\; \; and\; \; B are such that P(A)=\frac{1}{4},\; P(A\mid B)=\frac{1}{2}\; \; and\; \; P(B\mid A)=\frac{2}{3}.  Then P(B) is:

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is :

Concepts Covered - 1

Conditional Probability

Conditional Probability

Conditional probability is a measure of the probability of an event given that another event has already occurred. If A and B are two events associated with the same sample space of a random experiment, the conditional probability of the event A given that B has already occurred is written as  P(A|B), P(A/B) or \mathrm{P\left ( \frac{A}{B} \right )}.

The formula to calculate P(A|B) is

\mathrm{P}(\mathrm{A} | \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} \text { where P(B) is greater than zero. }
For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}. Let A = face is 2 or 3 and B = face is even number (2, 4, 6).

Here, P(A|B) means that the probability of occurrence of face 2 or 3 when an even number has occurred which means that one of 2, 4 and 6 has occurred. 

To calculate P(A|B), we count the number of outcomes 2 or 3 in the modified sample space B = {2, 4, 6}: meaning the common part in A and B. Then we divide that by the number of outcomes in B (rather than S).

\large \mathrm{P}(\mathrm{A} | \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} =\frac{\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})} }{\frac{\mathrm{n}( \mathrm{B})}{\mathrm{n}(\mathrm{S})} }\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{\frac{(\text {the number of outcomes that are } 2 \text { or } 3 \text { and even in S})}{6}}{\frac{( \text {the number of outcomes that are even in S) } }{6}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{\frac{1}{6}}{\frac{3}{6}}=\frac{1}{3}


Properties of Conditional Probability

Let A and B are events of a sample space S of an experiment, then

Property 1  P(S|A) = P(A|A) = 1

Proof:

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\mathrm{P}(\mathrm{S} | \mathrm{A})=\frac{\mathrm{P}(\mathrm{S} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A})}=1\\\\\mathrm{Also,\;\;\;\;\;\;\;\;\;\;\;\;\;}\mathrm{P}(\mathrm{A} | \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A})}=1\\\\\mathrm{Thus,\;\;\;\;\;\;\;\;\;\;\;\;\;}\mathrm{P}(\mathrm{S} | \mathrm{A})=\mathrm{P}(\mathrm{A} | \mathrm{A})=1

 

Property 2  If A and B are any two events of a sample space S and C is an event of S such that P(C) ≠ 0, then

                    P((A ∪ B)|C) = P(A|C) + P(B|C) – P((A ∩ B)|C)

In particular, if A and B are disjoint events, then

                    P((A∪B)|C) = P(A|C) + P(B|C)

Proof:

\\\mathrm{\;\;\;\;\;\;\;}\mathrm{P}((\mathrm{A} \cup \mathrm{B}) | \mathrm{C})=\frac{\mathrm{P}[(\mathrm{A} \cup \mathrm{B}) \cap \mathrm{C}]}{\mathrm{P}(\mathrm{C})}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{\mathrm{P}[(\mathrm{A} \cap \mathrm{C}) \cup(\mathrm{B} \cap \mathrm{C})]}{\mathrm{P}(\mathrm{C})}\\\text{(by distributive law of union of sets over intersection)}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{C})+\mathrm{P}(\mathrm{B} \cap \mathrm{C})-\mathrm{P}(\left ( \mathrm{A} \cap \mathrm{B} \right ) \cap \mathrm{C})}{\mathrm{P}(\mathrm{C})}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{C})}{\mathrm{P}(\mathrm{C})}+\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{C})}{\mathrm{P}(\mathrm{C})}-\frac{\mathrm{P}[(\mathrm{A} \cap \mathrm{B}) \cap \mathrm{C}]}{\mathrm{P}(\mathrm{C})}

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\mathrm{P}(\mathrm{A} | \mathrm{C})+\mathrm{P}(\mathrm{B} | \mathrm{C})-\mathrm{P}((\mathrm{A} \cap \mathrm{B}) | \mathrm{C})\\\\\text{When A and B are disjoint events, then}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\mathrm{P}((\mathrm{A} \cap \mathrm{B}) | \mathrm{C})=0\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{P}((\mathrm{A} \cup \mathrm{B}) | \mathrm{F})=\mathrm{P}(\mathrm{A} | \mathrm{F})+\mathrm{P}(\mathrm{B} | \mathrm{F})

 

Property 3 P(A’|B) =1 - P(A|B) , if P(B) ≠ 0

Proof:

                   From Property 1, we know that P(S|B) = 1

⇒                P((A ∪ A’)|B) = 1                               (as A ∪ A’ = S )     

⇒                P(A|B) + P(A’|B) = 1                          (as A and A’ are disjoint event)

⇒                P(A’|B) = 1 - P(A|B) 

Study it with Videos

Conditional Probability

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top