Probability is the one of the most important topics in engineering because it helps us to understand some aspects of the future of an event. Probability is not only used in mathematics but also is various domains of engineering.

1. Probability
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2. Definition
 Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted
with total certainty. We can predict only the chance of an event to occur i.e., how likely they are
going to happen, using it. Probability can range from 0 to 1, where 0 means the event to be an
impossible one and 1 indicates a certain event.
 Note: The probability of all the events in a sample space adds up to 1.
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3. For example
 when we toss a coin, either we get Head OR Tail, only two possible outcomes are possible (H,
T). But when two coins are tossed then there will be four possible outcomes, i.e
 {(H, H), (H, T), (T, H), (T, T)}.
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4. Terminology of Probability Theory
 The following terms in probability help in a better understanding of the concepts
of probability.
 Experiment: A trial or an operation conducted to produce an outcome is called an
experiment.
 Sample Space: All the possible outcomes of an experiment together constitute a
sample space. For example, the sample space of tossing a coin is head and tail.
 Favorable Outcome: An event that has produced the desired result or expected
event is called a favorable outcome. For example, when we roll two dice, the
possible/favorable outcomes of getting the sum of numbers on the two dice as 4
are (1,3), (2,2), and (3,1).
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5.  Trial: A trial denotes doing a random experiment.
 Random Experiment: An experiment that has a well-defined set of outcomes
is called a random experiment. For example, when we toss a coin, we know
that we would get ahead or tail, but we are not sure which one will appear.
 Event: The total number of outcomes of a random experiment is called an
event.
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6.  Equally Likely Events: Events that have the same chances or probability of
occurring are called equally likely events. The outcome of one event is
independent of the other. For example, when we toss a coin, there are equal
chances of getting a head or a tail.
 Exhaustive Events: When the set of all outcomes of an experiment is equal to
the sample space, we call it an exhaustive event.
 Mutually Exclusive Events: Events that cannot happen simultaneously are
called mutually exclusive events. For example, the climate can be either hot
or cold. We cannot experience the same weather simultaneously.
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7. Probability Formula
 P(A)=Number of favorable cases/Total number of possible cases
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8. Probability formula with addition:
 Whenever an event is the union of two other events, say A and B, then
 P(A or B) = P(A) + P(B) - P(A∩B)
 P(A ∪ B) = P(A) + P(B) - P(A∩B)
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9. Probability formula with the
complementary rule:
 Whenever an event is the complement of another event, specifically, if A is an
event, then
 P(not A) = 1 - P(A) or P(A') = 1 - P(A).
P(A) + P(A′) = 1.
OR
 Complementary events occur when there are just two events, and one event
is exactly opposite to the other event. Hence, A∪A¯= setofsamplespace
 For an event with probability P(A), its complement is P(A¯) such that
 P(A¯)+P(A)=1
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10. Probability formula with the conditional
rule:
 When event A is already known to have occurred and the probability of event
B is desired, then P(B, given A) = P(A and B), P(A, given B). It can be vice
versa in the case of event B.
 P(B∣A) = P(A∩B)/P(A)
 Conditional Probability: This probability of the event E is called the
conditional probability of E given that F has already occurred, and is denoted
by P (E|F).
 P(E|F) = Number of elementary events favorable to E∩F/Number of
elementary events which are favorable to F
 P(E|F) = n(E∩F)/n(F)
 Or
 P(E|F) = P(E∩F)/P(F) where P(F) ≠ 0
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11. Example:
 If P(A) = 7/13, P(B) = 9/13 and P(A∩B) = 4/13, evaluate P(A|B).
 Solution: P(A|B) = P(A∩B)/P(B)
 = (4/13)/(9/13)
 = 4/9.
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12. Example:
 A coin is tossed 2 times. The toss resulted in one head and one tail. What is
the probability that the first throw resulted in a tail?
 Solution: The sample space of a coin tossed two times is given as
 S = {HH, HT, TH, TT}
 Let Event A be the first throw resulting in a tail.
 Event B be that one tail and one head occurred.
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13.  P(A)=P(TH,TT)/P(HH,HT,TH,TT)
 =2/4
 =1/2
 P(A∩B)=P(TH)/P(HH,HT,TH,TT)
 =1/4
 So P(A/B)=P(A∩B)P(A)=1/4/1/2
 =1/2
 = 0.5
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14. Probability formula with multiplication
rule:
 Whenever an event is the intersection of two other events, that is, events A
and B need to occur simultaneously. Then P(A and B) = P(A)⋅P(B).
 P(A∩B) = P(A)⋅P(B∣A)
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15. Example:
 A coin is tossed. What is the probability of getting a head?
 Total number of equally likely outcomes (n) = 2 (i.e. head or tail)
 Number of outcomes favorable to head (m) = 1
 P(head)=1/2
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16. For Mutually Exclusive Events
 The additive theorem of probability states if A and B are two mutually
exclusive events then the probability of either A or B is given by
 P(A or B)=P(A)+P(B)
 P(A∪B)=P(A)+P(B)
 In the same way:
 P(A∪B∪C) = P(A)+P(B)+P(C)
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17. Example:
 A card is drawn from a pack of 52, what is the probability that it is a king or a
queen?
 Let Event (A) = Draw of a card of king
 Event (B) Draw of a card of queen
 P (card draw is king or queen) = P (card is king) + P (card is queen)
 P(A∪B)=P(A)+P(B)=4/52+4/52=1/13+1/13=2/13
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18. For Non-Mutually Exclusive Events
 In case there is a possibility of both events to occur then the additive
theorem is written as:
 P(A or B)= P(A)+P(B)−P(A and B)
 P(A∪B) = P(A)+P(B)−P(AB)
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19. Bayes' Theorem
 Bayes' Theorem states that the conditional probability of an event, based on
the occurrence of another event, is equal to the likelihood of the second
event given the first event multiplied by the probability of the first event.
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P(A∣B)= P(B)/P(A⋂B)
=P(B)/P(A)⋅P(B∣A)
where:
P(A)= The probability of A occurring
P(B)= The probability of B occurring
P(A∣B)=The probability of A given B
P(B∣A)= The probability of B given A
P(A⋂B))= The probability of both A and B occurring
Formula for Bays theorem

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Bayes’ Theorem Formula Derivation
From conditional probability, the Bayes theorem can be given as follows.
P(A|B) = P(A⋂B)/ P(B)
Where, P(B) ≠ 0
P(B|A) = P(B⋂A)/ P(A)
Where, P(A) ≠ 0
Here, the joint probability P(A ⋂ B) of both events A and B being true such that,
P(B ⋂ A) = P(A⋂ B)
P(A⋂ B) = P(A | B) P(B) = P(B | A) P(A)
P(A|B) = [P(B|A) P(A)]/ P(B)
Where, P(B) ≠ 0

22. Types of Probability
 There can be different perspectives or types of probabilities based on the
nature of the outcome or the approach followed while finding the probability of
an event happening. The four types of probabilities are,
• Classical Probability
• Empirical Probability
• Subjective Probability
• Axiomatic Probability
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23.  . Theoretical Probability
 Theoretical probability is based on the possible chances of something happening.
It is based on what is expected to happen in an experiment without conducting it.
It is the ratio of the number of favorable outcomes to the total number of
outcomes.
 2. Experimental Probability
 Experimental probability is a probability that is determined based on a series of
experiments. Therefore, it is based on the data which is obtained after an
experiment is carried out. It is the ratio of the number of times an event occurs to
the total number of experiments that are conducted.
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24.  3. Axiomatic Probability
 In axiomatic probability, a set of rules or axioms are set, which applies to all
types. In this probability, the chances of occurrence and non-occurrence of
the events can be quantified. It is the likelihood of an event or outcome
occurring based on the occurrence of a previous event or outcome.
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All Probability Formulas
Let A and B are two events. The probability formulas are listed below:
Range of Probability 0⩽P(A)⩽1
Addition rule of Probability P(A∪B)=P(A)+P(B)−P(A∩B)
Complementary event P(A¯)=1–P(A)
Mutually exclusive events P(A∩B)=0
Independent events P(A∩B)=P(A)×P(B)
Bayes formula P(AB)=P(B/A)×P(A)/P(B)

26.  What is the probability of a sure event?
 Ans: The probability of a sure event is one.
 P(E)=1
 Can a probability be negative?
 Ans: The probability value of the event can not be negative. It is a positive
value between 0 and 1.
 What is the probability of an impossible event?
 Ans: The probability of an impossible event is 0.
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27. Practice questions: 1)
 Two coins are tossed 500 times, and we get:
 Two heads: 105 times
 One head: 275 times
 No head: 120 times
 Find the probability of each event to occur.
 Solution: Let us say the events of getting two heads, one head and no head
by E1, E2 and E3, respectively.
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28.  P(E1) = 105/500 = 0.21
 P(E2) = 275/500 = 0.55
 P(E3) = 120/500 = 0.24
 The Sum of probabilities of all elementary events of a random experiment is 1.
 P(E1)+P(E2)+P(E3) = 0.21+0.55+0.24 = 1
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29. 2)
 One card is drawn from a deck of 52 cards, well-shuffled. Calculate the probability that the card will
 (i) be an ace,
 (ii) not be an ace.
 Solution: Well-shuffling ensures equally likely outcomes.
 (i) There are 4 aces in a deck.
 Let E be the event the card drawn is ace.
 The number of favourable outcomes to the event E = 4
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30.  The number of possible outcomes = 52
 Therefore, P(E) = 4/52 = 1/13
 (ii) Let F is the event of ‘card is not an ace’
 The number of favorable outcomes to F = 52 – 4 = 48
 The number of possible outcomes = 52
 Therefore, P(F) = 48/52 = 12/13
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31. 3)
 Two players, Sangeet and Rashmi, play a tennis match. The probability of Sangeet
winning the match is 0.62. What is the probability that Rashmi will win the match?
 Solution: Let S and R denote the events that Sangeeta wins the match and Reshma
wins the match, respectively.
 The probability of Sangeet to win = P(S) = 0.62
 The probability of Rashmi to win = P(R) = 1 – P(S)
 = 1 – 0.62 = 0.38
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32. 4)
 Consider the experiment of rolling a die. Let A be the event ‘getting a prime
number’, B be the event ‘getting an odd number’. Write the sets representing
the events
 (i) Aor B
 (ii) A and B
 (iii) A but not B
 (iv) ‘not A’.
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33. Solutions:
 S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}
 (i) A or B = A ∪ B = {1, 2, 3, 5}
 (ii) A and B = A ∩ B = {3,5}
 (iii) A but not B = A – B = {2}
 (iv) not A = A′ = {1,4,6}
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34. 5)
A coin is tossed three times, consider the following events.
 P: ‘No head appears’,
 Q: ‘Exactly one head appears’ and
 R: ‘At Least two heads appear’.
 Check whether they form a set of mutually exclusive and exhaustive events
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35.  Solution: The sample space of the experiment is:
 S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and
 P = {TTT},
 Q = {HTT, THT, TTH},
 R = {HHT, HTH, THH, HHH}
 P ∪ Q ∪ R = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH} = S
 Therefore, P
, Q and R are exhaustive events.
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36.  And
 P ∩ Q = φ,
 P ∩ R = φ and
 Q ∩ R = φ
 Therefore, the events are mutually exclusive.
 Hence, P, Q and R form a set of mutually exclusive and exhaustive events.
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37. Practice problems:1)
 5 cards are drawn successively from a well-shuffled pack of 52 cards with
replacement. Determine the probability that (i) all the five cards should be
spades? (ii) only 3 cards should be spades? (iii) none of the cards is a spade?
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38. 2)
 A single card is drawn at random from a standard deck of 52 playing cards.
Find the Probability:
 1) The card is either a red or an ace
 2) The card is not a king
 3) The card is a king or queen
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39. Practice questions:
 A bag contains 2 yellow,3 green and 2 blue balls. Two balls are drawn at
random. What is the probability that none of the balls drawn is blue?
 Solutions: Total number of balls =2+3+2=7=2+3+2=7
 Let S be the sample space.

n(S) = Total number of ways of drawing 2 balls out of 7 =
7C2
 Let E = Event of drawing 2 balls, none of them is blue.
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40.  n(E) = Number of ways of drawing 2 balls , none of them is blue= Number of
ways of drawing 22 balls from the total 5 (=7−2) balls = 5C2
 (∵ There are two blue balls in the total seven balls. Total number of non-blue
balls =7−2=5=7−2=5)
 P(E)=n(E)/n(S) = 5C2 /
7C2
 (5×4/2×1) / (7×6/2×1)=10/21
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41. Question:
 1) A letter is randomly taken from English alphabets. What is the probability
that the letter selected is not a vowel?
 Ans:
21/26
 2) The probability A getting a job is 1/5 and that of B is 1/7 What is the
probability that only one of them gets a job?
Ans: 2/7
 3) A letter is chosen at random from the word 'ASSASSINATION'. What is the
probability that it is a vowel?
 Ans: 6/13
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42.  4) Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random.
What is the probability that the ticket drawn has a number which is a multiple
of 3 or 5?
 Ans: 9/20
 Explanation:
 Total number of tickets, n(S)=20
To get a multiple of 3 or 5 ticket drawn must be 3,5,6,9,10,12,15,18 or 20
Therefore, number of ways in which we get a multiple of 3 or 5 = 9
P(multiple of 3 or 5) =9/20
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