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So, let’s start the learning now.
Getting Started
This post contains three sections –
 Section A – This section contains important formulas, short tricks, concepts on Probability
 Section B – This section contains some examples that demonstrate how to solve questions on Probability using short tricks, formulas shown in section A.
 Section C – Go to this section to solve some problems on Probability
Now,
Where to start from?
We recommend go through section A, then section B, then section C. Even if you are familiar with this topic, do not just skip section A and section B.
SECTION A:
Probability in mathematics deals with calculating the likelihood of a particular event’s occurrence, which is expressed as a number between 1 and 0. An event with a probability of 1 can be considered a certainty: for example, the probability of a coin toss which results in either “head” or “tail” is 1, because there are no other options, when we assume the coin lands flat. An event can be considered to have equal odds of occurring or not occurring: for example, the probability of a coin toss in a test match will be 0.5. As it will have 2 chances either a head or a tail. Result in “head” is 0.5, and that in “tail” is 0.5. An event with a probability of “ZERO” can be considered impossible: for example, the probability that the coin will land on the ground without either of the sides facing up is 0, because there must be either “head” or “tail” must be facing up.
Probability is expressed mathematically as: The number of favourable outcomes of a targeted event divided by the number of favourable outcomes plus the number of failure outcomes:
Probability = n(a)/[n(a) + n(b)]
Where n(a) = number of favourable outcomes
n(b) = number of failure outcomes
Important Facts:
 Experiment: An operation that will produce some welldefined outcomes is called an experiment.
 Random Experiment: An experiment in which all the outcomes possible are known and the output cannot be exactly predicted in advance is called a random experiment.
Examples of Random Experiment:
 Rolling an unbiased dice.
 Tossing a fair coin.
 Picking a card from a pack of well shuffled cards.
 Details:

 A dice is a solid cube having 6 faces, marked 1, 2, and 3,4,5,6 respectively. When a dice is thrown, the outcome is the number that appears on its upper face.
 When we throw a coin, the either a Head (H) or a Tail (T) appears.
 A pack of cards has 52 cards. A pack has 13 cards of each suit namely Spades, Clubs, Hearts and Diamonds.

 Clubs and Spades are black cards.
 Hearts and Diamonds are red cards.
 There are 4 honours in each suit. These are Aces, Kings, Queens and Jacks. These are called face cards.
 Sample Space:When we perform an experiment, then the set S of all possible outcomes is called the Sample space.
 In tossing a coin, S = (H,T)
 If two coins are tossed, then S= {HH, HT, TH, TT}
 When a dice is rolled, we have S = {1,2,3,4,5,6}
 P(S) = 1
 0 ≤ P(E) ≤ 1
 If A and B represents the events, we have :
 The complement of an event A is denoted as A’ and is written as P (A’) = 1 – P (A)
Examples of Sample Space:
5. Event: A subset of a sample space is called an event.
6. Probability of Occurrence of an Event:
If S determines sample space and let E determines event.
Then, E⊂ S
So, P(E) = n(E)
n(S)
7. Results on Probability:
P(A∪B) = P(A) + P(B) – P(A∩B)
Formulae:
Rule of Addition
P(A∪B) = P(A) + P(B) – P(A∩B)
Disjoint Events
Events A and B are disjoint if
P(A∩B) = 0
Independent Events
Events A and B are independent if
P(A∩B) = P(A) ⋅ P(B)
Compound probability
P(A or B) = P(A) + P(B) – P(A and B)
Where A and B are any two events.
P(A or B) is the probability of the occurrence of at least one of the events.
P(A and B) is the probability of the occurrence of both A and B at the same time.
For the formula application, please refer to Section B.
SECTION B:
Examples:
1.What is the probability of getting number 1 or number 4 when a dice is rolled?
Solution:
Taking the individual probabilities of each number, getting a 1 is 1/6 and so is getting a 4.
Applying the formula of compound probability,
P(A or B) = P(A) + P(B) – P(A and B)
Probability of getting a 1 or a 4,
P(1 or 4) = P(1) + P(4) – P(1 and 4)
==> 1/6 + 1/6 – 0
==> 2/6 = 1/3.
2. A coin is tossed twice. On what probability will the tossed coin get two consecutive tails?
Solution:
The chances of getting a tail in one toss = 1/2
The coin is tossed twice.
P(A∩B) = P(A) ⋅ P(B)
So 1/2 * 1/2 = 1/4 is the answer.
The above answer can be verified with the help of sample space.
A coin when is tossed twice, the sample space is {(H,H), (H,T), (T,H), (T,T)}.
Our required event is (T,T) whose occurrence is only once out of four possible outcomes and so, our answer is 1/4.
3. A pack contains 3 blue, 1 red and 5 black pens. If a pen is drawn at random from the pack, replaced and the process is repeated 2 more times, what is the probability of drawing 2 blue pens and 1 black pen?
Solution:
Total number of pens = 9
Probability of drawing 1 blue pen = 3/9
Probability of drawing another blue pen = 3/9
Probability of drawing 1 black pen = 5/9
Probability of drawing 2 blue pens and 1 black pen = 3/9 * 3/9 * 5/9 = 45/729 = 5/81
SECTION C:
1. Tickets that are numbered from 1 to 20 are mixed in a box and then a ticket is taken at random. What is the probability that the ticket taken is a multiple of 2 or 5?
a. 3/5
b. 8/5
c. 11/20
d. 9/20
Solution: a
Explanation
Here, S = {1, 2, 3, 4, …., 19, 20}
Let E = event of getting a multiple of 2 or 5
= {2, 4 ,5,6,8,10, 12,14, 15,16, 18, 20}
∴P(E)=n(E)/n(S)=12/20=3/5
2. What is probability of drawing two hearts cards from a well shuffled pack of 52 cards?
a. 12/5
b. 1/17
c. 1/29
d. 13/18
Solution: b
Explanation
Probability =[ Favourable outcomes]/ Total
Total no. of cards = 52
No.of heart cards = 13
Probability of 1^{st} card as heart = 13/52
Probability of 2^{nd} card as heart = 12/51
So, total probability = 13/52*12/51
= 1/17
3.What is the possibility of having 53 Fridays in a nonleap year?
a. 5/7
b. 1/7
c. 1/366
d. 54/365
Solution: b
Explanation
A nonleap year has 365 days.
No. of weeks in a year = 52
So, no. of Fridays = 52
But we want 53 Fridays. So 1 day extra should be Friday.
Probability = [ Favourable outcomes]/ Total
Total no. of days (in a week)= 7
So, probability of 53 Fridays = 1/7
4. A basket has 4 white and 2 blue shirts. One shirt is picked in random and is put in another basket. The second basket has 2 white and 4 blue shirts. Now a shirt is picked from second basket. What is the probability of it being a white shirt?
a. 5/9
b. 29/72
c. 8/21
d. 3/17
Solution: c
Explanation
Probability = [ Favourable outcomes]/ Total
 Pick white shirt
Box 1 has a total of = 2+4 = 6 shirts
Probability of white from box1 = 4/6 = 2/3
Now we add 1 white shirt to box 2.
So box 2 has (2+1) = 3 white and 4 blue shirts
Total in box 2 = 3+4 = 7
Probability of white from box 2 = 3/7
So white shirt probability from box 1 and 2 = 2/3*3/7
= 2/7
 Pick blue shirt
Probability of blue from box 1 = 2/6= 1/3
Now we add 1 blue shirt to box 2.
So box 2 has (4+1) = 2 white and 5 blue shirts
Total in box 2 = 2+5 = 7
Probability of white from box2 = 2/7
So white shirt probability from box 1 and 2 = 1/3*2/7
= 2/21
Total probability = 2/7+2/21
= 6/21+2/21 (Find the LCM)
= 8/21
5. In a set of 20 game cards, 15 are black and rest are yellow. 3 black and 6 yellow are marked IMPORTANT. If a card is chosen in random from this set, find the possibility of choosing a yellow card or an ‘IMPORTANT’ card?
a. 13/20
b. 2/5
c. 17/20
d. 8/13
Solution: b
Explanation
We need yellow card or important card.
No. of yellow cards = 20 – 15 = 5
No. of important cards = 6+3 = 9
Total cards = 20
No. of yellow important cards = 6
Probability = [ Favourable outcomes]/ Total
So probability = 5/20+9/206/20
= 8/20 = 2/5
6. What is the probability of getting two numbers whose product is odd if we throw a dice two times simultaneously?
a. 1/4
b. 3/5
c. 3/8
d. 4/2
Solution: a
Explanation
When the two dice are thrown simultaneously, we have n(S) = (6 x 6) = 36
Then, E = {(1,1),(1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
∴n(E)=9
∴P(E)=n(E)/n(S)
=9/36 = 1/4
7. In a lucky draw, there are 15 prizes and 20 blank cards. What is the probability of getting a prize if a card is taken at random?
a. 1/5
b. 3/7
c. 2/9
d. 5/4
Solution: b
Explanation
Probability = [ Favourable outcomes]/ Total
= 15/20+15
= 15/35
= 3/7
8. A man and his friend applied for the same post in a company. The probability of the man’s selection is 1/5 and the probability of the friend’s selection is 1/8. What is the probability that only one of them will be selected?
a. 4/3
b. 4/9
c. 11/40
d. None of these
Solution: c
Explanation
Let A = event that the man is selected and
I = event that the friend is selected.
Then,
P(A)=1/5
And
P(I)=1/8
P (Ᾱ) = (1−1/5) = 4/5
And
P(Ī)=(1−1/8)= 7/8
∴ Required probability = P [( A and not I) or (I and not A)]
= P [(A∩Ī)or(I∩Ᾱ)]
= P (A∩Ī) + P(I∩Ᾱ)
= P(A). P(Ī)+P(I).P(Ᾱ)
= 1/5*7/8 + 1/8*4/5
= 7/40+ 1/10
= 11/40
9. 70 bulbs are selected at random from a lot and the lifetime (in hrs) is recorded in the form of a frequency table given below:
Lifetime(in hrs)  400  300  700  800  1000 
Frequency  11  13  20  24  9 
1 bulb is selected at random from the lot. The probability that the life is 1250 hrs is:
a. 1/70
b. 7/18
c. 0
d. 2
Solution: c
Explanation
There is no bulb with lifetime of 1250 hrs. So, the probability is 0.
Refer to the above table and answer the question:
10. The probability that bulbs selected at random has life less than 800 hrs is:
a. 10/30
b. 4/16
c. 7/18
d. 44/70
Solution: d
Explanation
Total no. of bulbs = 70
Let E be the event that the bulbs have life less than 800 hrs.
No. of favourable outcomes E = 11+13+20
= 44
So probability P(E)= n(E)
n(S)
= 44/70
11.If a number is selected at random from the set {1, 2, 3, …., 100}, then the probability that the selected number is a perfect cube is .
a. 1/25
b. 2/30
c. 5/18
d. 4/19
Solution: a
Explanation
When counting from 1 to 100, we have 1, 8, 27 and 64 as perfect cubes
So, the probability of getting a perfect cube is
4/100 = 1/25.
12. X speaks truth in 80% of cases and Y in 90% of cases. What is the percentage of cases that X and Y are likely to contradict each other, narrating the same incident?
a. 26%
b. 45%
c. 50%
d. 55%
Solution: a
Explanation
Let X = Chances of event that X will speak truth
Y = Event that Y speaks the truth
Then P(X) = 80/100 = 4/5
P(Y) = 90/100= 9/10
P(Xlie) = 1−4/5 = 1/5
P(Ylie) = 1−9/10= 1/10
Now, X and Y contradict each other = [X lies and Y true] or [Y true and Xlies]
= P(X).P(Ylie) + P(Xlie).P(Y)
[When we are adding at the place of OR]
= (4/5*1/10) + (1/5*9/10)
= 2/25 + 9/50
= 13/50
In percentage = 13/50 * 100%
= 26%
13. Two brothers A and B appeared for an exam. Let X be the event that A is selected and Y is the event that B is selected.
The probability of X is 1/8 and that of Y is 2/5. What will be the probability that both of them are selected.
a. 3/20
b. 1/20
c. 4/15
d. 5/3
Solution: b
Explanation
Given, X be the event that A is selected and Y is the event that B is selected.
P(X)=1/8
P(Y)=2/5.
Assuming C to be the event in which both are selected.
P(C)=P(X)×P(Y) where X and Y are independent events
=(1/8)×(2/5)
= 1/20
14. A bag contains 25 apples numbered 1 to 25. An apple is drawn and then another apple is drawn without replacement.
Find the probability that both apples will show odd numbers.
a. 11/50
b. 13/20
c. 4/17
d. 5/20
Solution: a
Explanation
The probability that first apple shows the odd number =12/25
Since, the apple is not replaced there are now 11 odd numbered apples and total 24 apples left.
Hence, probability that second apple shows the odd number = 11/24
Required probability = (12/25) × (11/24)
= 11/50
15. A 5digit number is formed by using digits 1, 2, 3, 4 and 5 without repetition. On what probability can the number formed be divisible by 4?
a. 4/3
b. 2/7
c. 1/5
d. None of these
Solution: c
Explanation
The formed number using the digits 1, 2, 3, 4 and 5 if it has to be divisible by 4, has to have the last two digits 12 or 24 or 32 or 52.
In this case, The five digits number can be formed using the remaining 3 digits in 3!=3*2*1=6 ways.
Hence, a number divisible by 4 can be formed in 6×4=24 ways.
The Complete number that can be formed using the digits 1, 2, 3, 4 and 5 without repetition =5!= 5*4*3*2*1=120
Required probability =24/120
=1/5
16.A lucky draw is held to select a student who will live in the only luxury room in a hostel. There are 200 YearIII, 250 YearII and 100 YearI students who applied. Each YearIII’s name is placed in the lucky draw 4 times; each YearII’s name 3 times and YearI’s name, 2 times. What is the probability that a YearIII’s name will be selected?
a. 2/5
b. 3/7
c. 16/35
d. 3/8
Solution: c
Explanation
Total names in the lucky draw= 4×200+3×250+2×100
=800+750+200
=1750
Number of YearIII’s names =4×200=800
Required probability =800/1750
=16/35
17. You toss a coin AND roll a dice. What is the probability of getting a head and a 5 on the dice?
a. 1/4
b. 1/8
c. 1/12
d. 1/6
Solution: c
Explanation
The Probability of chances of getting a head when a single coin is tossed =1/2
Probability of getting 5 when a dice is thrown=1/6
Required probability =1/2×1/6
=1/12
18.A card is taken from a pack of 52 cards. The card is drawn at random. What is the probability that it is neither a club nor a king?
a. 3/13
b. 4/19
c. 5/13
d. 9/13
Solution: d
Explanation
There are 13 club and 3 more king
Probability of getting club or a king =13+3
52
= 4/13
So probability of getting neither club nor a king = 1−4/13
=9/13
19.If a is chosen at random from the set {1,2,3,4} and b is to be chosen at random from the set {5,6,7}, what is the probability that ab will be odd?
a. 1/3
b. 1/6
c. 5/4
d. 3/8
Solution: a
Explanation
S ={(1,5),(1,6),(1,7),(2,5),(2,6),(2,7),(3,5),(3,6),(3,7),(4,5),(4,6),(4,7)}
Total element n(S)=12
ab will be odd when a or b or both will be odd.
Events of a,b,ab being odd is E.
E ={(1,5),(3,5),(1,7),(3,7)}
n(E)=4
So, Probability
P=n(E)/n(S)
P=4/12
P=1/3
20.Find the probability that in a random arrangement of the letters of the word ‘VETERINARY’ the two E’s come together.
a. 1/4
b. 1/5
c. 2/9
d. 3/7
Solution: b
Explanation
Total no. of letters in ‘VETERINARY’ = 10
The total number of words which can be formed by permuting the letters of the word ”VETERINARY ‘ is 10!/2! As there is two E’s.
Hence n(S)=10!/2!
Taking two E’s as one letter, number of ways of arrangement in which both E’s are together =9!
So n(X)=9!
Hence required probability=n(X)/n(S)
=9!/10!//2!
=9*8*7*6*5*4*3*2*1
10*9*8*7*6*5*4*3*2*1/2*1
= 2/10
= 1/5
21. In a race where 10 cars are participating, the chance that car A will win is 1/5, that Bwill win is 1/8 and that Cwill win is 1/10. Assume that a dead heat is not possible. Find the chance that one of the cars will win.
a. 1/450
b. 46/250
c. 1/480
d. 17/40
Solution: d
Explanation
Required probability =P(A)+P(B)+P(C) (all the events are mutually exclusive)
=1/5+1/8+1/10
=17/40
22. What is the probability of getting a sum 9 from two throws of a dice?
a. 1/8
b. 1/6
c. 1/9
d. 1/12
Solution: c
Explanation
In two throws of a dice, n(S) = (6×6) = 36
Let E= event of getting a sum 9 = {(3,6), (4,5), (5,4), (6,3)}
So, P(E) = n(E)= 4
n(S) 36
= 1/9
23. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card?
a. 1/13
b. 4/13
c. 1/4
d. 9/52
Solution: b
Explanation
Clearly, there are 52 cards, out of which there are 16 face cards.
So, P(getting a face card) = 16/52
= 4/13