The document discusses probability and experiments with random outcomes. It defines key probability concepts like sample space, events, and probability functions. It provides examples of common experiments with finite sample spaces like coin tosses, die rolls, and card draws. It also discusses experiments with infinite discrete sample spaces like repeated coin tosses until the first tail. The document establishes the basic properties and rules of probability, including that it is a function between 0 and 1, that probabilities of disjoint events add, and that probabilities of subsets are less than the original set.

1. Probability
An experiment is a process - natural or set up
deliberately - that has an observable outcome. In
the deliberate setting, the word experiment and
trial are synonymous. An experiment has a random
outcome if the result of the experiment can't be
predicted with absolute certainty. An event is a
collection of possible outcomes of an experiment. An
event is said to occur as a result of an experiment if
it contains the actual outcome of that experiment.
Individual outcomes comprising an event are said to
be favorable to that event. Events are assigned a
measure of certainty which is called probability (of
an event.)
Quite often the word experiment describes an
experimental setup, while the word trial applies to

2. actually executing the experiment and obtaining an
outcome. P(E) =
A formal theory of probability has been developed in
the 1930s by the Russian mathematician A. N.
Kolmogorov.
The starting point is the sample (or probability)
space - a set of all possible outcomes. Let's call it Ω.
For the set Ω, a probability is a real-valued function
P defined on the subsets of Ω:
(*) P: 2Ω → [0, 1]
Thus we require that the function be non-negative
and that its values never exceed 1. The subsets of Ω
for which P is defined are called events. 1-element
events are said to be elementary. The function is

3. required to be defined on the empty subset Φ and
the whole set Ω:
(1) P(Φ) = 0, P(Ω) = 1.
This says in particular that both Φ and Ω are events.
The event Φ that never happens is impossible and
has probability 0. The event Ω has probability 1 and
is certain or necessary.
If Ω is a finite set then usually the notions of an
impossible event and an event with probability 0
coincide, although it may not be so. If Ω is infinite
then the two notions practically never coincide. A
similar dichotomy exists for the notions of a certain
event and that with probability 1. Examples will be
given shortly.
The probability function (or, as it most commonly
called, measure) is required to satisfy additional

4. conditions. It must be additive: for two disjoint
events A and B, i.e. whenever A ∩ B = Φ,
(2) P(A∪B) = P(A) + P(B),
which is a consequence of a seemingly more general
rule: for any two events A and B, their union A∪B
and intersection A∩B are events and
(2') P(A∪B) = P(A) + P(B) - P(A ∩ B).
Note, however, that (2') can be derived from (2).
Indeed, assuming that all the sets involved are
events, events A - B and A ∩ B are disjoint as are B -
A and A ∩ B. In fact, all three events A - B, B - A,
and A ∩ B are disjoint and the union of the three is
exactly A∪B. We have,
P(A)+P(B)= (P(A - B) + P(A ∩ B)) + (P(B - A) +

5. P(A ∩ B))
= P(A∪B) + P(A ∩ B)
Some properties of P follow immediately from the
definition. For example, denote the complement of an
event A, A = Ω - A. It is also called a complementary
event. Then, naturally, A and A’ are disjoint, A ∩ A’ = Φ,
and
1 = P(Ω) = P(A) + P(A’).
In other words,
(4) P(A) = 1 - P(A’).
if B = A∪C for disjoint A and C, then
P(B) = P(A∪C) = P(A) + P(C) ≥ P(A).
In other words, if A is a subset of B, A B, then

6. (5) P(B) ≥ P(A).
Probability is a monotone function - the fact that
jibes with our intuition that a larger event, i.e. an
event with a greater number of favorable outcomes,
is more likely to occur than a smaller event.
Disjoint events do not share favorable outcomes and,
for this reason, are often called incompatible or,
even more assertively, mutually exclusive.
Finite Sample Spaces
Tossing a coin. The experiment is tossing a coin (or
any other object with two distinct sides.) The coin
may land and stay on the edge, but this event is so
enormously unlikely as to be considered impossible
and be disregarded. So the coin lands on either one
or the other of its two sides. One is usually called
head, the other tail. These are two possible

7. outcomes of a toss of a coin. In the case of a single
toss, the sample space has two elements that
interchangeably, may be denoted as, say,
{Head, Tail}, or {H, T}, or {0, 1}, ...
Rolling a die. The experiment is rolling a die. A
common die is a small cube whose faces shows
numbers 1, 2, 3, 4, 5, 6 one way or another. These
may be the real digits or arrangements of an
appropriate number of dots, e.g. like these
There are six possible outcomes and the sample
space consists of six elements:
{1, 2, 3, 4, 5, 6}.

8. Many random variables may be associated with this
experiment: the square of the outcome f(x) = x2,
with values from
{1, 4, 9, 16, 25, 36},
Drawing a card. The experiment is drawing a card
from a standard deck of 52 cards. The cards are of
two colors - black (spades and clubs) and red
(diamonds and hearts), four suits (spades, clubs,
diamonds, hearts), 13 values (2, 3, 4, 5, 6, 7, 8, 9,
10, Jack, Queen, King, Ace). (Some decks use 4
colors, others use different names. For example, a
Jack may be called a Knave. We shall abbreviate the
named designations as J, Q, K, A.) There are 52
possible outcomes with the sample space
{2♠, 2♣, 2♦, 2♥, 3♠, 3♣, 3♦, 3♥, ..., A♠, A♣,

9. A♦, A♥}.
Of course, if we are only interested in the color of a
drawn card, or its suite, or perhaps the value, then
it would be as natural to consider other sample
spaces:
{b, r},
{♠, ♣, ♦, ♥} or
{2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A}.
Choosing a birthday. The experiment is to select a
single date during a given year. This can be done, for
example, by picking a random person and inquiring
for his or her birthday. Disregarding leap years for
the sake of simplicity, there are 365 possible
birthdays, which may be enumerated

10. {1, 2, 3, 4, ..., 365}.
Tossing two coins. The experiment is tossing two
coins. One may toss two coins simultaneously, or one
after the other. The difference is in that in the
second case we can easily differentiate between the
coins: one is the first, the other second. If the two
indistinguishable coins are tossed simultaneously,
there are just three possible outcomes, {H, H}, {H,
T}, and {T, T}. If the coins are different, or if they
are thrown one after the other, there are four
distinct outcomes: (H, H), (H, T), (T, H), (T, T),
which are often presented in a more concise form:
HH, HT, TH, TT. Thus, depending on the nature of
the experiment, there are 3 or 4 outcomes, with the
sample spaces

11. Indistinguishable coins
{{H, H}, {H, T}, {T, T}}.
Distinct coins
{HH, HT, TH, TT}
Rolling two dice. The experiment is rolling two dice.
If the dice are distinct or if they are rolled
successively, there are 36 possible outcomes: 11, 12,
..., 16, 21, 22, ..., 66. If they are indistinguishable,
then some outcomes, like 12 and 21, fold into one.
There are 6×5/2 = 15 such pairs giving the total
number of possible outcomes as 36 - 15 = 21. In the
first case, the sample space is
{11, 12, ..., 16, 21, 22, ..., 66}.

12. When we throw two dice we are often interested not
in individual numbers that show up, but in their sum.
The sum of the two top numbers is an example of a
random variable, say Y(ab) = a + b (where a, b range
from 1 through 6), that takes values from the set {2,
3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. It is also possible to
think of this set of a sample space of a random
experiment. However, there is a point in working
with random variables. It is often a convenience to
be able to consider several random variables related
to the same experiment, i.e., to the same sample
space. For example, besides Y, we may be
interested in the product (or some other function) of
the two numbers.
Infinite Discrete Sample Spaces
First tail. The experiment is to repeatedly toss a coin until
first tail shows up. Possible outcomes are sequences of H

13. that, if finite, end with a single T, and an infinite sequence
of H:
{T, HT, HHT, HHHT, ..., {HHH...}}.
As we shall see elsewhere, this is a remarkable space that
contains a not impossible event whose probability is 0. One
random variable is defined most naturally as the length of
an outcome. It draws values from the set of whole numbers
augmented by the symbol of infinity:
{1, 2, 3, 4, ..., ∞}.
Mutually exclusive events: In the experiment of rolling a die,
a sample space is S={1,2,3,4,5,6}. Consider A `an odd number
appears’ and B `an even number appears’ as A={1,3,5} , B
={2,4,6} A∩B=∅
Exhaustive events: Consider the experiment of rolling a die, a
sample space is S={1,2,3,4,5,6}.Let’s define the following
Events A={1,2,3}, B={3,4} C={5,6} , then AUBUC=S OR Ei ∩ Ej = ∅
for i≠j
FOR CLASS ---XI

14. Question:21(ncert) In a class of 60 students, 30 opted for NCC,
32 opted for NSS and 24 opted for both NCC and NSS. If
One of these students is selected at random, find the
probability that
(i) The student opted for NCC or NSS.
(ii) The student opted neither NCC nor NSS.
(iii) The student opted for NSS but not NCC.
Solution: P(A) =30/60, P(B) = 32/60, P(A∩B) = 24/60 , where
the event that student choose NCC be taken by A and
The student choose NSS be taken as B.
(i) P(AUB) = 19/30
(ii) P(A’∩B’) = P(AUB)’ = 1 – P(P(AUB)=11/30 (De Morgan’s
law)
(iii) P(A’∩B) = P(B) – P(A∩B) ( Only B)
= 8/15 – 6/15 = 2/15.
If A, B, C ARE THREE EVENTS ASSOCIATED WITH A RANDOM
EXPERIMENT( Addition theorem fo three events)
P(A B C) = P(A) + P(B) + P(C) – P(B∩C) – P(A∩B) –
P(A∩C) +P(A∩B∩C)
Question5 ( misc.): Out of 100 students, two sections of 40
and 60 are formed. If you and your friend are among the 100
students, what is the probability that
(a) You both enter the same section.
(b) You both enter the different section.

15. Answer: (a) required probability = = 17/33 [
C(n,r) = , n! = n(n-1)(n-2)……3.2.1 ]
(b) required probability = = 16/33 or 1 – 17/33
= 16/33.
Question 6 (misc.): Three letters are indicated to three
persons and an envelope is addressed to each of them. The
letters are inserted into the envelopes at random so that
each envelope contains exactly one letter. Find the prob.
That at least one letter is in it’s proper envelope.
Answer: Let the envelope be denoted by E1, E2, E3 are the
corresponding letters by L1, L2, L3 .
Total outcomes = 3! = 6
There are 4 favourable cases when at least one letter is in
proper envelope = 4/6 = 2/3.
E1 E2 E3
L1 L3 L2
L3 L2 L1
L2 L1 L3
L1 L2 L3
L2 L3 L1
L3 L1 L2
2 cases when no letter goes in correct
envelope
1 case when 2 letters & hence all are in correct
3 cases when 1 letter is in correct & 2 letters in wrong
envelope.

16. Question 9(misc.): If 4- digit nos. greater than or equal to 5000
are randomly formed from the digits
0,1,3,5 and 7, what is the prob. Of forming a number divisible by
5 when (i) the digits may be repeated, (ii) the repetition of digits is
not allowed.
Answer: 5 or 7 in the extreme left position i.e., in thousand place
(i) when digits may be repeated , let 4 places in a 4 digit
number be marked as below:
I II III IV
1 5 5 5 ways
If place I is filled by 5, then each of the places II,III,IV can be
filled in 5 ways and hence 5x5x5=125 ways
Similarly if place number I is filled by 7, then total no. of
ways = 125 , therefore n(S) =125+125=250
Now to find favourable cases , when the number is divisible
by 5
I II III IV
(2ways 5,7) 5 5 (0 or 5)2 ways favourable outcomes =
100 , then req. prob. = 100/ 250=2/5
(ii) No repetition
5 is fixed at I place and II,III,IV can be filled in (4)(3)(2) =
24 ways
Similarly when 7 is fixed in place at I , then total ways =
24+24=48ways
favourable outcomes = 18 ways as
I II III IV
5 - - 0
7 - - 0
7 - - 5
3 2

17. Total no. of ways = 6+6+6 = 18, so req. pob. = 18/48=3/8.
Question (misc.): A combination lock on a suitcase has 3
wheels each labeled with nine digits
from 1 to 9 . If an opening combination is a particular
sequence of three digits with no repeats, what is the prob. Of
a person guessing the right combination?
Answer: 1/p(9,3) , p(n,r) =
SOME USEFUL RESULTS:
1. P(A∩B’) = P(A-B) = P(A) – P(A∩ B) = P( Only A)
2. P(A’∩ B’) = P(AUB)’ = 1 – P(AUB) and P(A’UB’) = 1 –
P(A∩B)
3. P(at least one) = 1 – P(None) = 1 – P(0)
4. For any two events A and B, P(A∩B) ≤
P(A)≤P(AUB)≤P(A)+P(B)
Question: 1 A box contains 100 bolts and 50 nuts, It is given
that 50% bolt and 50% nuts are rusted. Two
Objects are selected from the box at random. Find the
probability that both are bolts or both are rusted.
Answer: A= event of getting a bolt , B= event of getting a
rusted item and A ∩B= event of getting a rusted bolt.
Required probability = P(A)+P(B)-P(A ∩B) = +
- = 260/447.
Question:2 A pair of dice is rolled. Find the probability of
getting a doublet or sum of number to be at least 10.
Solution: Total no. = 36 , A = event of getting a doublet i.e.,
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
P(A) = 1/6 , P(B) = 6/36 = 1/6 {
(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)+, P(A∩B) = 1/18

18. P(AUB) = 5/18.
**Question:3 A pair of dice is rolled. Find the probability that
the sum of the number on the two faces is
(i) neither divisible by 3 nor by 4 ? (ii) is either a multiple of
3 or 2?
Answer: A = event that sum of numbers on the two faces is
divisible by 3 = *(1,2),(2,1),(1,5),(5,1)…..(5,4),(6,6)+ = 12 , B
= sum divisible by 4 = 9 *(1,3), (2,2), (2,6)……(4,4),(6,6)+
P(A) = 1/3, P(B) = ¼, P(A∩B) = 1/36 , P(AUB) = 5/9 , then
(i) P(A’∩B’) = 1 – P(AUB) = 4/9.
(ii) A = sum of numbers on two faces multiple of 2 , B =
multiple of 3
P(A) = 18/36 = ½ , P(B) = 12/36 = 1/3 P(AUB) = 2/3.
**Question:4 Of the students attending a lecture, 50% could
not see what was written on the board and 40% could not hear
what the lecture was saying. Most unfortunate 30% fell into
both of these categories. What
Is the probability that a student picked at random was able to
see and hear satisfactorily?
Answer: Let A = event that a student can see satisfactorily , B
= student can hear satisfactorily
P(A’) =50% = ½ , P(B’) = 40% = 2/5 , P(A) =1- p(A’) = ½ ,
P(B) 1 – P(B’) = 3/5
P(A’∩B’) = 30% = 3/10 , P(AUB)’ = 1 – P(AUB) = 3/10 ⇨
P(AUB)=7/10.
P(A∩B) = 2/5.
Question:5 Two dice are tossed once. Find the prob. Of getting
“an even number on the first die or a total of 2”.
[Hint A = even number on first die , B =total of 2
Find P(AUB) = P(A) + P(B) = 19/36.]
Question:6 Three squares of chess board are selected at
random. Find the porb. Of getting 2 squares of one
Color and other of a different color .

19. Answer Total 64 squares in chess board , 32 are of white and
32 of black colour ,
selecting as 1w & 2b or 1b &2w
Req. prob. = = 16/21.
Question:7 Three of the six vertices of a regular hexagon are
chosen at random. What is the prob.
That the triangle with these vertices is equilateral?
Answer: Total no. of triangles c(6,3) = 20 ( since no three
points are collinear). Of these only ∆ ACE;
∆BDF are equilateral ∆’s. so req. prob. = 2/20=1/10.
B C
A D
F E
Question:8 If A, B, C are three mutually exclusive and
exhaustive
Events of an experiment such that 3P(A) = 2P(B) = P(C).
Then find P(A).
Answer: use P(A)+P(B)+P(C)=1 as mutually exclusive and
exhaustive events, p(A)=2/11
Question: 9 Two unbiased dice are thrown. Find the prob. That
neither a doublet nor a total of 10 will appear.
Answer: (5,5) is only common event , P(neither a doublet
nor a total of 10) = 1-P(doublet or total 10)
= 1 – 2/9 = 7/9.
Question:10 In a class of 100 students, 60 drink tea, 50 drink
coffee and 30 drink both. A student from this class is selected
at random. Find the prob. That the student takes (i) at least
one of 2 drinks (ii) only one of two drinks.
Answer: You can use venn diagram also , (i) P(TUC)=4/5 ,

20. (ii) P(event of taking only tea or only coffee)=
P(EUF)=P(E)+P(F)=1/2 = exactly one drink.
FOR CLASS --XII
conditional Probability
If E and F are two events associated with the same
sample space of a random experiment, the conditional
Probability of the event E given that F has occurred,
i.e., P(E/F) is given by P(E/F) = provided
P(F)≠0.
PROPERTIES:
(I) Let E & F be events of a sample space S of an
experiment, then we have P(S/F) = P(F/F) = 1
(II) If A & B are any two events of a sample space S
and F is an event of S such that P(F)≠0, then
P((AUB)/F) = P(A/F) + P(B/F) – P((A∩B)/F) ,
for disjoint events ⇨ P((A∩B)/F) =0.
(IIII) P(E’/F) = 1 – P(E/F).
EXAMPLE(ncert): Consider the experiment of tossing a coin. If
the coin shows head, toss it again but if it shows tail, then throw
a die. Find the conditional prob. Of the event that `the die shows
a number greater than 4’ given that `there is at least one tail’.

21. SOLUTION: S = {(H,H), (H,T), (T,1), (T,2), (T,3), (T,4), (T,5),
(T,6)}
E ={(T,5), (T,6)} , F ={(H,T), (T,1), (T,2), (T,3), (T,4), (T,5),
(T,6)+ & E∩F =*(T,5), (T,6)+
P(F) = ¼+1/12+1/12+1/12+1/12+1/12+1/12 = ¾ [ P(H,T)
= ½.1/2 =1/4 , P(T,1) = ½ .1/6 =1/12]
P(E∩F) = 1/12+1/12= 1/6 , then P(E/F) = =
(1/6)/(3/4) = 2/9.
Question13(ncert): An instructor has a question bank
consisting of 300 easy true/false questions, 200 difficult
true/false
Questions , 500 easy multiple choice questions and 400 difficult
multiple choice questions. If a question is selected at random
from the question bank, what is the prob. That it will be an easy
question give it is a multiple choice question?
Answer: E denotes easy questions, D = difficult , T = true/false
questions and M =multiple
Total no. of questions = 1400, no. of easy multiple choice
questions = 500
P(E∩M) = 500/1400, no. of multiple choice questions
500+400=900
P(M) = 900/1400 , P(E/M) = P(E∩M)/P(M) = 5/9.
Question: A can solve 90% of the problems given in a book and
B can solve 70%. What is the prob. That at least one of them will
solve the problem, selected at random from the book?

22. Answer: Use P(AUB)= 1 – P(A’)P(B’), then answer is .97, events
are indep.]
Question: A can hit a target 4 times in 5 shots, B 3 times in 4
shots, and C 2 times in 3 shots. Calculate the prob. That
(i) A, B, C all may hit (ii) B, C may hit and A may lose (iii)
any two of A, B and C will hit the target
(iv) non of them will hit the target.
Answer: Hint: A, B, C are indep. (i) P(A∩B∩∩C) =
P(A).P(B).P(C) =4/5.3/4.2/3, (ii) find P(A’∩B∩C)= 1/10
(iii) P(A∩B∩C’)U P(A’∩B∩C)U P(A∩B’∩C)= 13/30 (IV)
P(A’∩B’∩C’)= 1/60, same as part (i)-
Question: Three persons A, B, C throw a die in succession till one
gets a `six’ and wins the game. Find their respective prob.
Of winning , if A begins.
Answer: A wins if gets `six’ in Ist or 4th or 7th …. Throw of
unbiased die , then P(A)=1/6 and P(A’)=5/6
A will get 4th throw if he fails in 1st , B fails in in second and C
fails in 3rd throw.
Prob. Of winning of A in 4th throw= P(A’∩A’∩A’∩A) =
(5/6)3 .1/6
Similarly, prob. Of winning of A in 7th throw = (5/6)6 . 1/6
and so on…
Prob. Of winning of A 1/6+(5/6)3 .1/6 + (5/6)6. 1/6 +…….=

23. = 36/91
B wins if gets a six in 2nd or 5th or 8th or……..throw.
Prob. Of winning of B = (5/6)1/6+
(5/6)4.1/6+(5/6)7.1/6+…… 30/91, for prob. Of C = 1 –
[P(A)+P(B)]= 25/91.
The probability of 7 when rolling two die is 1/6 (= 6/36)
because the sample space consists of 36 equiprobable
elementary outcomes of which 6 are favorable to the event
of getting 7 as the sum of two die.
Denote this event A: P(A) = 1/6.Consider another event B
which is having at least one 2. There are still 36 elementary
outcomes of which 11 are favorable to B; therefore, P(B) =
11/36. We do not know whether B happens or not, but this is
a legitimate question to inquire as to what happens if it
does. More specifically, what happens to the probability of A
under the assumption that B took place?
The assumption that B took place reduces the set of possible
outcomes to 11. Of these, only two - 25 and 52 - are
favorable to A. Since this is reasonable to assume that the 11
elementary outcomes are equiprobable, the probability of A
under the assumption that B took place equals 2/11. This
probability is denoted P(A|B) - the probability of A assuming
B: P(A|B) = 2/11. More explicitly P(A|B) is called the
conditional probability of A assuming B. Of course, for any
event A, P(A) = P(A|Ω), where, by convention, Ω is the
universal event - the whole of the sample space - for which

24. all available elementary outcomes are favorable.
We see that, in our example, P(A|B) ≠ P(A). In general, this
may or may not be so.
Retracing the steps in the example,
P(A|B) = P(A∩B) / P(B),
and this is the common definition of the conditional
probability. The formula confirms to the earlier definitions:
P(A|Ω) = P(A∩Ω) / P(Ω) = P(A)
since P(Ω) = 1 and A∩Ω = A (because A⊂Ω.)
For another example, let's look at the events associated with
rolling a dice in the form of octahedron, a shape with eight
faces. The sample space naturally consists of 8 equiprobable
outcomes:
Ω = {1, 2, 3, 4, 5, 6, 7, 8}.
Let A be the event of getting an odd number, B is the event
getting at least 7. Then P(A) = 4/8 = 1/2, P(B) = 2/8 = 1/4.
A∩B = {7}, so that
P(A|B) = P(A∩B)/P(B) = 1/2 = P(A).
Observe also that
P(B|A) = P(A∩B)/P(A) = 1/4 = P(B).
On the other hand, define
A+ = {1, 2, 3, 5, 7} and

25. A- = {3, 5, 7}.
Then
P(B|A+) = P(A+∩B)/P(A+) = 1/5 < 1/4 = P(B),
whilst
P(B|A-) = P(A-∩B)/P(A-) = 1/3 > 1/4 = P(B).
So we see that, in general, there is no definite relationship
between the probability P(B) and the conditional probability
P(B|A). They may be equal, or one of them may be greater
than the other. In the former case the events are said to be
independent.
Bayes' Theorem
Bayes' theorem (or Bayes' Law and sometimes Bayes' Rule) is
a direct application of conditional probabilities. The
probability P(A|B) of "A assuming B" is given by the formula
P(A|B) = P(A∩B) / P(B)
[it is known as multiplication rule of prob.] which for our

26. purpose is better written as
P(A∩B) = P(A|B)·P(B).
The left hand side P(A∩B) depends on A and B in a
symmetric manner and would be the same if we started with
P(B|A) instead:
P(B|A)·P(A) = P(A∩B) = P(A|B)·P(B).
This is actually what Bayes' theorem is about:
(1) P(B|A) = P(A|B)·P(B) / P(A).
Most often, however, the theorem appears in a somewhat
different form
(1') P(B|A) = P(A|B)·P(B) / (P(A|B)P(B) + P(A|B)P(B)),
where B is an event complementary to B: B∪B = Ω, the
universal event. (Of course also B∩B = Φ, an empty event.)
This is because

27. A = A ∩ (B ∪ B)
= (A∩B) ∪( A∩B)
and, since A∩B and A∩B are mutually exclusive,
P(A) = P(A∩B ∪ A∩B)
= P(A∩B) + P(A∩B)
= P(A|B)P(B) + P(A|B)P(B).
More generally, for a finite number of mutually exclusive and
exhaustive events Hi (i = 1, ..., n), i.e. events that satisfy
Hk ∩ Hm = Φ, for k ≠ m and
H1 ∪ H2 ∪ ... ∪ Hn = Ω,
Bayes' theorem states that
P(Hk|A) = P(A|Hk) P(Hk) / ∑i P(A|Hi) P(Hi),
where the sum is taken over all i = 1, ..., n.
Question: (ncert) A labrotary blood test is 99% effective in

28. detecting a certain disease when it is in fact , present.
However , the test also yields a false positive result for
0.5% of the healthy person tested(i.e., if a healthy person is
tested , then, with prob. .005, the test will imply he has the
disease). If 0.1% of the population actually has the disease,
what is the prob. That a person has the disease given that his
test result is positive?
Answer: E , E’ are events that a person has disease and does
not have, A = blood test is positive
P(E) = 0.1% , P(E’) = .999, P(A/E) = .99 and P(A/E’)=
.005
P(E/A) = 22/133.
Question(NCERT) A card from a pack of 52 cards is lost.
From the remaining cards of the pack, two cards are drawn
and found to be both dimond. F ind the prob. Of the lost
card being a diamond.
Answer: E1 , E2 are events that lost card is diamond and

29. not, A denotes a diamond (lost card)
P(A/E1)=C(12,2)/C(51,2) = (12 /(51 ) , P(A/E2)
= C(13,2)/C(51,2)=13 /51 , P(E1/A) = 11/50
Conditional Probability and Independent Events
P(A|B) = P(A∩B) / P(B),
P(B|A) = P(A∩B) / P(A).
The events A and B are said to be independent provided
P(A|B) = P(A), or, which is the same
P(B|A) = P(B).
Neither the probability of A or B is affected by the
occurrence (or a occurrence) of the other event. A
symmetric way of expressing the same fact is this
P(A∩B) = P(A) P(B).

30. Multiplication rule of prob. For more than two events:
P(E∩F∩G) = P(E)P(F/E)P(G/(E∩F)= P(E)P(F/E)P(G/EF)
Independent Events and Independent
Experiments
1. Independent experiments: When two events E & F are
such that the prob. Of occurrence of one of them is not
affected by occurrence of the other. The trials - the
experiments - may or may not affect the outcomes of later
trials. If they do, the experiments are called dependent;
otherwise, they are independent.
2. Events may or may not be independent; two events, A and
B, are independent iff P(A∩B) = P(A) P(B).
3. Some times there a confusion b/w indep. Events &
mutually exclusive events. Indep. Is related to prob. Of
events
Whereas mutually exclusive is defined in terms of events
(subset of sample space).Indep. events, may have common

31. outcome but mutually never have outcome common, so
they are different in meaning
Nonzero prob. In two indep. Events cannot be nonzero
prob. In mutually exclusive events , converse is also not
true.
If A & B are two indep. Events, then the prob. Of occurrence
of at least one of A & B [P(AUB)] is given by 1 – p(A’)P(B’)
Consider tossing a coin three times in a row. Since each of the
throws is independent of the other two, we consider all 8 (= 23)
possible outcomes as equiprobable and assign each the
probability of 1/8. Here is the sample space of a sequence of
three tosses:
{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
There are 28 possible events, but we are presently interested in,
say, two:
A = {HHH, HTH, THH, TTH} and
B = {HHH, HHT, THH, THT}.

32. A is the sequence of tosses in which the third one came up
heads. B is the event in which heads came up on the second
toss. Since each contains 4 outcomes out of the equiprobable 8,
P(A) = P(B) = 4/8 = 1/2.
The result might have been expected: 1/2 is the probability of
the heads on a single toss. Are events A and B independent
according to the definition? Indeed they are. To see that,
observe that
A ∩ B = {HHH, THH},
the event of having heads on the second and third tosses. P(A ∩
B) = 2/8 = 1/4. Further, let's find the conditional probability
P(A|B):
P(A|B) = P(A ∩ B) / P(B)
= 1/4 / 1/2
= 1/2
= P(A).

33. So that P(A|B) = P(A) and, according to the definition, events A
and B are independent, as expected.
Bernoulli trial is an experiment with only two possible outcomes
that have positive probabilities p and q such that p + q = 1. The
outcomes are said to be "success" and "failure", and are
commonly denoted as "S" and "F" or, say, 1 and 0.
For example, when rolling a die, we may be only interested
whether 1 shows up, in which case, naturally, P(S) = 1/6 and
P(F) = 5/6. If, when rolling two dice, we are only interested
whether the sum on two dice is 11, P(S) = 1/18, P(F) = 17/18.
The Bernoulli process is a succession of independent Bernoulli
trials with the same probability of success. One important
question about a succession of n Bernoulli trials is the
probability of k success.
Since the individual trials are independent, we are talking of the
product of probabilities of successes and failures. Such a

34. product is independent of the order in individual successes and
failures come about. For example,
P(SSFSF) = P(SFFSS) = P(FFSSS) = p3q2.
In general, the probability of k successes in n trials is denoted
b(k; n, p) and is equal to
b(k; n, p) = C(n, k)pkqn - k,
where C(n, k) is the binomial coefficient n choose k. Observe
that, by the binomial formula, ∑b(k; n, p) over k from 0 to n is
exactly 1:
∑b(k; n, p) = ∑C(n, k)pkqn - k = (p + q)n = 1.
As a function of k, b(k; n, p) is known as the binomial
distribution and plays an important role the theory of
probabilities.
Binomial distribution is the distribution of a total number of
successes in a given number of Bernoulli trials. The common
notation is b(k; n, p), where k is the number of successes, n is

35. the number of trials, p is the probability of success. We know
that b(k; n, p) = C(n, k) pk(1 - p)n - k.
The result for the expected value np = ∑ k b(k; n, p) might have
been anticipated given the interpretation of the probability as a
relative frequency.
Note that the expected value np is always different from the
most likely value (n + 1) p, provided of course p ≠ 0.
ASSIGNMENT CLASS XI PROBABILITY
1. A pair of dice is rolled. Consider the following events: A : the
sum is greater than 8, B : 2 occurs on either
die , C : the sum is atleast 7 and a multiple of 3. Which pair of
events are mutually exclusive?
2. A card is drawn at random from a deck of 52 playing cards. Find
the probability that it is:
(i) an ace (ii) a jack of hearts (iii) a three of clubs or a six of
diamonds (iv) a heart

36. (v) any suit except heart (vi) a ten or a spade (vii) neither a four
nor a club
(viii) an honours card (ix) a face card (x) a spade or a face card
3. Four cards are drawn at random from a pack of 52 cards. Find
the probability of getting:
(i) all the four cards of the same suit (ii) all the four cards of the
same number
(iii) one card from each suit (iv) all four face cards
(v) two red cards and two black cards (vi) all cards of the same
color (vii) getting four aces
4. An urn contains 9 red, 7 white and 4 black balls. If two balls are
drawn at random, find probability that:
(i) both the balls are red (ii) one ball is white
(iii) the balls are of the same color (iv) one is white and other is
red
5. 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.
6. One number is chosen from numbers 1 to 200. Find the
probability that it is divisible by 4 or 6.
7. The letters of word “SOCIETY” are placed at random in a row.
What is the probability that three vowels
come together?
8. Fine the probability that in a random arrangement of the letters
of the word “UNIVERSITY” the two I’s
come together.
9. The letters of the word “MUMMY” are placed at random in a
row. What is the chance that letters at the
extreme are both M ?
** 10. A bag contains 50 tickets numbered 1, 2, 3… ,50 of which
five are drawn at random and arranged in
ascending order of magnitude x1 < x2 < x3< x4 <x5 . Find the
probability that x3= 30 .

37. ANSWERS: 1. A, B; B, C are mutually exclusive; but A, C are not.
2.
(i ) 1/13 (ii ) 1/52 (iii ) 1/26 (iv ) 1/4 (v ) 3/4 (vi ) 4/13 (vii ) 9/13
(viii ) 4/13 (ix ) 3/13 (x ) 11/26
3. (i ) 44/4165 (ii ) 1/20825 (iii ) 2197/20825 (iv ) 99/54145 (v )
325/833 (vi ) 92/833 (vii ) 1/270725
4.
(i ) 18/95 ( ii) 91/190 (iii ) 63/190 (iv ) 63/190
5. 1/5 , 6. P(A)=50/200, P(B)= 33/200 & P(A∩B)=16/200,
P(AUB)= 67/200 , 7. (5!.3!)/7! = 1/7 , 8. 1/5 , 9. 3/10 , 10.
X1,x2 < 30 , they come from tickets 1 to 29 & x4,x5>30 come
from 20 (31to 50)
REQ. PROB. =( 29C2.20C2)/50C5 = 551/15134.