Revision Notes: Probability | Mathematics (Maths) for JEE Main & Advanced PDF Download

Important Formulas

1. Mathematical definition of probability:

Note: 
(i) 0 ≤ P(A) ≤ 1
(ii) Probability of an impossible event is zero
(iii) Probability of a sure event is one.
(iv) P(A) + P(Not A) = 1 i.e. P(A) +  = 1

2.  Odd for an event:
Then odds in favor of  and odd in against of

3. Set theoretical notation of probability and some important results:
(i) 
(ii) 
(iii) 
(iv) A ⊆ B ⇒ P(A) ≤ P(B)
(v)
(vi) P(A∩ B) ≤ P(A) P(B) ≤ P(A ∪ B) ≤ P(A) + P(B)
(vii) P(Exactly one event)
(viii) = 1 – P(A∩ B) = P(A) + P(B) – 2P (A∩B) = P(A + B) – P(A∩ B)
(ix) P(neither A nor B)
(x) When a coin is tossed n times or n coins are tossed once, the probability of each simple event is 1/2ⁿ
(xi) When a dice is rolled n times or n dice are rolled once, the probability of each simple event is 1/6ⁿ
(xii) When n cards are drawn (1 ≤ n ≤ 52) from well shuffled deck of 52 cards, the probability of each simple
(xiii) If n cards are drawn one after the other with replacement, the probability of each simple event is 1/(52)ⁿ
(xiv) P(none) = 1 – P (at least one)
(xv) Playing cards

• Total cards: 52 (26 red, 26 black)
• Four suits: Heart, diamond, spade, club (13 cards each)
• Court (face) cards: 12 (4 kings, 4 queens, 4 jacks)
• Honor cards: 16 (4 Aces, 4 kings, 4 queens, 4 Jacks)

(xvi) Probability regarding n letters and their envelopes: 
If n letters corresponding to n envelopes are placed in the envelopes at random, then 

• Probability that all letters are in the right envelopes = 1/n! 
• Probability that all letters are not in the right envelopes =  
• Probability that no letter is in the right envelope
• Probability that r letters are in the right envelope

4. Addition Theorem of Probability: 
(i) When events are mutually exclusive 
i.e. n(A ∩ B) = 0 ⇒ P(A ∩ B) = 0 
∴ P(A ∪ B) = P(A) + P(B)
(ii) When events are not mutually exclusive i.e. P(A ∩ B) ≠ 0 
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B) or P(A + B) = P(A) + P(B) – P(AB) 
(iii) When events are independent i.e. P(A ∩ B) = P(A) P(B) 
∴ P(A + B) = P(A) + P(B) – P(A) P(B)

5.  Conditional probability: 
P(A/B) = Probability of occurrence of A, given that B has already happened
P(B/A) = Probability of occurrence of B, given that A has already happened

Note: If th e outcomes of the experiment are equally likely, then 

(i) If A and B are independent events, then P(A/B) = P(A) and P(B/A) = P(B) 
(ii) Multiplication Theorem: 
P(A ∩ B) = P(A/B). P(B), P(B) ≠ 0 or P(A ∩ B) = P(B/A) P(A), P(A) ≠ 0 
Generalized: P(E₁∩ E₂∩ E₃∩ … ∩ E_n) 
= P(E₁) P(E₂/E₁) P(E₃/E₁∩ E₂) P(E₄/E₁∩ E₂∩ E₃) … If events are independent, then 
P(E₁∩ E₂∩ E₃ … ∩ E_n) = P(E₁) P(E₂) … P(E_n) 

6.  Probability of at least one of the n Independent events: If P₁, P₂, … P_n are the probabilities of n independent events A₁, A₂, … A_n then the probability that at least one of these events will happen is 1 – [(1 – P₁) (1 – P₂) … (1 – P_n)]

7.  Total probability: Let A₁, A₂, … A_n be n mutually exclusive & set of exhaustive events. If event A can occur through any one of these events, then the probability of occurrence of A

8.  Bayes’ Rule: Let A₁, A₂, A₃ be any three mutually exclusive & exhaustive events (i.e. A₁∪ A₂∪ A₃ = sample space & A₁∩ A₂∩ A₃ = φ) of a sample space S and B is any other event on sample space then,

(i) Probability distribution: 
(i) If a random variable x assumes values x₁, x₂, … x_n with probabilities P₁, P₂, … P_n respectively then 

• P₁ + P₂ + P₃ + … + P_n = 1 
• Mean E(x) = SP_ix_i 
• Variance = ∑x²P_i – (mean)² = ∑(x²) – (E(x))² 

(ii) Binomial distribution: If an experiment is repeated n times, the successive trials being independent of one another, then the probability of r success is ⁿC_r P^r q^n-r at least r success is  where p is probability of success in a single trial, q = 1 – p

• Mean E(x) = np 
• E(x2) = npq + n2p2 
• Variance E(x2) – (E(x))2 = npq 
• Standard deviation

9.  Truth of the statement: 
(i) If two persons A and B speak the truth with probabilities P₁& P₂ respectively and if they agree on a statement, then the probability that they are speaking the truth will be given by
(ii) If A and B both assert that an event has occurred, the probability of occurrence of which is α, then the probability that the event has occurred  given that the probability of A & B speaking truth is p1,p2 respectively.
(iii) If in the second part, the probability that their lies coincide is β, then from the above case, the required probability will be

Problem-Solving Tactics

Following are some extra methods which may be useful to solve probability questions: 
Venn Diagrams: It is a diagram in which the sample space is represented by a rectangle and the element of the sample space by points within it. Subsets (or events) of the sample space are represented by the region within the rectangle, usually using circles.
For example, consider the following events when a die is thrown, 
A = {odd numbers} = {1, 3, 5} 
B = {even numbers} = {2, 4, 6} 
C = {prime numbers} = {2, 3, 5} 
Let us see how Venn diagrams are to be applied by using them to prove some results as follows:
Theorem 1: For any two events A and B, A ⊆ B ⇒ P(A) ≤ P(B). 

Proof: From the adjoining diagram, we have 
A ∪ (B – A) = B and A ∩ (B – A) = f 
∴ P(B) = P[A ∪ (B – A)] [∵ A ) (B  A) = f] 
⇒ P(B) = P(A) + P(B  A) [∵ P(B  A) e 0]
⇒ P(A) ≤ P(B) Proved

Proof: Let A and B be two compatible events. Then A ∩ B ≠ φ. From the adjoining Venn diagram. it is clear that:
(A – B) ∩ (A ∩ B) = fand (A – B) ∪ (A ∩ B) = A
⇒ P(A - B) P(A ∩ B) = P(A)
⇒ P(A - B) = P(A) - P(A ∩ B) Proved

Remarks: This result may be expressed as

Proof: We have ,P(A ∪ B ∪ C) = P[(A ∪ Β) ∪ C] = P(A ∪ B) + P(C) –P [(A ∪ B) ∩ C]
= P(A ∪ B) + P(C) –P[(A ∩ C) ∪ (B ∩ C)] (Distributive Law) 
= [P(A) + P(B) – P(A ∩ B)] + P(C)– P[(A ∩ B) ∪ (C ∩ C)] [Addition law] 
= P(A) + P(B) –P(A ∩ B) + P(C) –P[(A ∩ C)∪ (B ∩ C)]
= P(A) + P(B) + P(C) –P(A ∩ B) – [P(A ∩ C) + P(B ∩ C) –P[(A ∩ C) ∩ (B ∩ C)]
= P(A) + P(B) + P(C) –P(A ∩ B) –P(A ∩ C) + P(B ∩ C) + P[A ∩B∩ C] Proved

Probability Tree Diagrams
Calculating probabilities can be hard. Sometimes you add them, sometimes you multiply them and often, it is hard to figure out what to do. That’s when tree diagrams come to the rescue!
Here is a tree diagram for two tosses of a coin:

How do you calculate the overall probabilities?

So, there you go. When in doubt, draw a tree diagram, multiply along the branches and add the columns. Make sure all probabilities add to 1 and you are good to go!

Solved Examples

Que1: If cards are drawn at random from a pack of 52 playing cards without replacement then the probability that a particular card is drawn at the nth draw is:
(a) 1/(53-n) 
(b) 1/52
(c) n/52 
(d) n/(53 – n)

Ans: (b)
P(special card at n^th drawn)

Ans: (c)
n = 4 (r person)
P = 1/6 (correct ans. by one)
x = number of correct ones

Ans: (b)
x = get spade
Total spades in 52 cards = 13
P = P(he fails exactly the first two times)

Ans: (d)
5 is to be chosen from 9 people there is a couple in group of 9
P = P(couple chosen)
q = P(couple don’t chose)

Ans: (d) 
P(n) = probability of shown in fall

Ans: (a)
P(A + B) = P(A) + P(B) – P(AB)
and P(A), P(B) > P(AB)
∴ P(A + B) > P(A), (AB)
and P(A), P(B), P(AB), P(A + B) ≥ 0
∴ option (a) is correct
P(AB) ≤ P(A) ≤ P(A + B) ≤ P(A) + P(B)

Que7: An integer x is chosen from the first 50 positive integers. The probability that,  is:
(a) 1/25
(b) 2/25
(c) 1/10
(d) None of these

Ans: (c)
x ∈ {1, 2,……,50}


x₁ = 2. 0871 and x₂ = 47. 912
In region integer y ∈ {3, 4,……,47}
So only for x = 1, 2, 48, 49, 50;
y is greater than 0 or

Ans: (a)
P = 1/2 events
P = probability of win test match
L = Lose match; W = win match

There is 5 match series.
Total possibility = 25
P(India’s second win occurs at the 3rd test)
P(LWW) + P(WLW)

Ans: (b)
Given that , P(A) = 0.4,  = 0.6
P(the event A happens at least one
= 1 – P(none of the event happens)
= 1 – (0.6) (0.6) (0.6) = 1 – 0.216 = 0.784

Que10: If A and B are two independent events such that P(A )> 0,and P(B) ≠ 1, then  is equal to
(a) 1 – P(A/B)
(b)
(c)
(d)

Ans: (b)