Probability - 1

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Session
Probability - 1
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Session Objectives

• Experiment

• Sample Space

• Event

• Types of Events

• Probability of an Event

• Class Exercise

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Experiment
Experiment An operation, which results in some
well-defined outcomes is called an experiment.
Random Experiment If we conduct an experiment
and we do not know which of the possible outcome
will occur this time, the experiment is called a
random experiment.

• For example
• Tossing a coin
• Throwing a die
• 3. Drawing a card from a well shuffled pack of
  cards

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Sample Space
Sample Space The sample space of an random
experiment is the set of all possible elementary
outcomes. It is denoted by S.

• For example
• When we toss a coin, the sample space
• S H, T
• It is a random experiment, because when we
  toss a coin this time, we do
• not know whether we shall get head or tail.

• When we throw a die, the sample space
• S 1, 2, 3, 4, 5, 6

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Event
Event A subset E of a sample space is called
an event. An event is a combination of one or
more of the possible outcomes of an experiment.
For example In a single throw of a die, the
event of getting a prime number is given by E
2, 3, 5 and the sample space S 1, 2, 3, 4,
5, 6.
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Algebra of Events
For any three events A, B and C with sample space
S.
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Algebra of Events
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Types of Events (Sure Event)
Sure Event In the throw of a die, sample space S
1, 2, 3, 4, 5, 6
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Types of Events (Impossible Event)
Impossible Event In the throw of a die, sample
space S 1, 2, 3, 4, 5, 6.
Let E be the event of getting an 8 on the die.
Clearly, no outcome can be a number 8.
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Simple and Compound Event
Simple or Elementary Event An event that contains
only one element of the sample space is called a
simple or an elementary event.
Compound Event A subset of sample space which
contains more than one element is called
compound event or mixed event or composite event.
For example In a simultaneous toss of two coins,
the sample space is S HT, TH, HH, TT
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Equally Likely Outcomes
Equally Likely Outcomes The outcomes are said to
be equally likely, if none of them is expected to
occur in preference to the other or the chances
of occurrence of all of them are same.
For example In throwing of a single die, each
outcome is equally likely to happen.
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Mutually Exclusive Events
Mutually Exclusive Events Two or more events are
said to be mutually exclusive if no two or
more of them can occur simultaneously in the same
trial.

• Facts
• Elementary events related to an experiment are
  always
• mutually exclusive.
• 2. Compound events may or may not be mutually
  exclusive.

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Example
In throwing a die, sample space S 1, 2, 3, 4,
5, 6
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Exhaustive Events
Exhaustive Events In a random experiment, two or
more events are exhaustive if their union is the
sample space.
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Example
In throwing a die, sample space S 1, 2, 3, 4,
5, 6
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Example-1
Solution A (1,1), (1,3), (1,5), (3,1), (3,3),
(3,5), (5,1), (5,3), (5,5) B (1,1), (1,2),
(1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2),
(4,1) C (2,2), (2,4), (2,6), (4,2),
(4,4), (4,6), (6,2), (6,4), (6,6)
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Solution Cont.
(1,1), (1,2), (1,3), (1,4), (1,5),
(2,1), (2,2), (2,3), (3,1) (3,2),
(3,3), (3,5), (4,1), (5,1), (5,3), (5,5)
A and C are mutually exclusive
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Example 2

• Three coins are tossed . Describe
• two mutually exclusive events A and B.
• three mutually exclusive exhaustive events A, B
  and C.
• two events E and F which arent mutually
  exclusive.

Solution Sample space for the toss of three
coins is S (H, H, H), (H, H, T), (H, T, H),
(T, H, H), (H, T, T), (T, H, T), (T, T,
H), (T, T, T)
Let AEvent of getting three heads (H, H, H)
and BEvent of getting three tails(T, T,
T) A and B are mutually exclusive
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Solution (Cont.)
Let CEvent of getting one or two head
(H, H, T),(H, T, H),(T, H, H),(H, T, T),(T,
H,T),(T,T,H) A, B and C are mutually exclusive
exhaustive events.
Let E Event of getting two heads
(H, H, T),(H, T, H),(T, H, H) and F
Event of getting at least one tail (H,
H,T), (H, T, H),(T, H, H), (H, T, T), (T, H, T),
(T, T, H), (T,T,T) E and F arent mutually
exclusive.
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Probability of an Event
In a random experiment, the probability of
happening of the event A with sample space S is
defined as
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Probability of an Event (Cont.)
If m is the number of outcomes favourable to an
event and n is the total number of possible
outcomes. Then,
Hence, the probability of an event always lies
between 0 and 1.
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Probability of an Event (Cont.)
Sure Event An event A is said to be sure or
certain if P(A) 1
Impossible Event An event A is said to be
impossible if P(A) 0
Odds in favour of occurrence of an event A are
defined as m n - m, i.e. ratio of favorable
outcomes to unfavorable ones.
Odds against occurrence of event A are defined as
n - m m , i.e. ratio of unfavourable outcomes
to favourable ones.
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Example-3
What is the probability of getting at least two
heads in a simultaneous throw of three coins?
Solution If three coins are tossed together
possible outcomes are
S HHH, HHT, HTH, THH, TTH, THT, HTT, TTT
Number of these exhaustive outcomes, n(S) 8
At least two heads can be obtained in the
following ways
E HHH, HHT, HTH, THH
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Solution Cont.
Number of favourable outcomes, n(E) 4
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Example-4
In a single throw of two dice what is the
probability of getting (i) 8 as the sum
(ii) a total of 9 or 11
Solution In throwing of a pair of dice, total
number of outcomes in sample space 6 6 36
(i) To get 8 as the sum favourable outcomes are
(2, 6), (3, 5), (4, 4), (5, 3) and (6, 2).
Thus, the number of such outcomes 5
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Solution Cont.
(ii) Favourable outcomes to the event of getting
the sum as 9 or 11 are (3, 6), (4, 5),
(5, 4), (6, 3), (5, 6) and (6, 5).
Thus, the number of such outcomes 6
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Example-5
The letters of the word SOCIETY are placed at
random in a row. What is the probability that
three vowels come together.
Solution There are 7 letters in the word
SOCIETY.
These 7 letters can be arranged in a row in 7!
ways.
O, I, E are three vowels in the word SOCIETY.
Assuming these three vowels as one letter, we get
5 letters which can be arranged in a row in 5!
ways.
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Solution Cont.
But three vowels O, I, E can be arranged in 3!
ways.
The total number of arrangements in which three
vowels come together is 5! 3!.
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Example-6
A five-digit number is formed by the digits 1, 2,
3, 4, 5 without repetition. Find the
probability that the number is divisible by 4.
Solution Total number of ways in which a five
digit number can be formed by digits 1, 2, 3,
4, 5 5!
A number is divisible by 4 if the numbers formed
by last two digits are divisible by 4.
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Solution Cont.
Thus for an outcome to be favorable, the last two
digits can be (1, 2), (2, 4), (3, 2), (5, 2).
The last two digits can have only these 4
arrangements. But the rest of the three digits
can be arranged in 3! ways.
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Example-7
A bag contains 8 red and 5 white balls. Three
balls are drawn at random. Find the probability
that one ball is red and two balls are white.
Solution Total number of balls 8 5 13
n(S) number of ways of selecting 3 out of 13
balls
Let A be the event of selecting one red and 2
white balls out of 8 red and 5 white balls.
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Solution Cont.
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Example-8
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Solution Cont.
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Example-9
Out of 9 outstanding students in a college, there
are 4 boys and 5 girls. A team of four students
is to be selected for a quiz programme. Find
the probability that two are boys and two are
girls.
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Solution Cont.
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Example-10

• Four cards are drawn at random from a pack 52
  playing cards.
• Find the probability of getting
• all the four cards of the same suit
  (CBSE 1993)
• (ii) all the four cards of the same number
  (CBSE 1993)

Solution (i) There are four suits club,
spade, heart and diamond, each of 13 cards.
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Solution Cont.
(ii) Four cards of the same number (1, 1,
1, 1), (2, 2, 2, 2), (3, 3, 3, 3), (13, 13, 13,
13).
Favourable number of events 13
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Example 11
Two dice are thrown. Find the odds in favour of
getting the sum to be (i) 4 (ii) 5 (iii) what
are the odds against getting the sum to be six.
Solution The sample space when two dice are
thrown is S (1, 1), (1, 2), ... (1, 6), (2,
1), (2, 2), ... (2, 6), (3, 1), (3, 2),
... (3, 6), (4, 1), (4, 2), ... (4, 6),
(5, 1), (5, 2), ... (5, 6), (6, 1), (6, 2), ...
(6, 6)
n(S) 36
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Solution (Cont.)

• Let A be the event of getting the sum on the pair
  of dice to be 4
• A (1, 3), (2, 2), (3, 1)

• Let B be the event of getting the sum on the pair
  of dice to be 5.
• B (1, 4), (2, 3) (3, 2) (4, 1)

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Solution (Cont.)

• Let C be the event of getting the sum to be six
  on the pair of dice
• C (1, 5), (2, 4) (3, 3) (4, 2), (5, 1)

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