Samacheer Class 11 Maths - Introduction to Probability Theory

Two coins are tossed simultaneously = one coin is tossed two times.
The sample space S = { H, T } × { H, T }
S = {HH, HT, TH, TT }
n(S) = 4
Let A be the event of getting one head and one tail, B be the event of getting atmost two tails.
A = {HT, TH}
n(A) = 2
B = {HH, HT, TH, TT}
n (B) = 4
(i) P (getting one head and one tail)
(ii) P (getting atmost two tails)

(i) One is a mango and the other is an apple:
Let S be the sample space, A be the event of taking one mango and one apple.
n(S) = 9C 2
n (A) = 5C 1 × 4C 1
= 5 × 4 = 20
(ii) Both are of the same variety:
Let S be the sample space, A be the event of taking 2 mangoes and B be the event of taking 2 apples
∴ n(s) = 9C 2
P(taking 2 fruits are of the same colour)
= P(A or B)
= P(A ∪ B)
= P(A) + P(B)

Non-leap year
No of days = 365
= \(\frac{365}{7}\) weeks = 52 weeks + 1 day
In 52 weeks we have 52 Sundays. So we have to find the probability of getting the remaining one day as Sunday. The remaining 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
(i.e.,) n(S) = 7
In the n (Sunday) = A {Saturday to Sunday or Sunday to Monday}
(i.e.,) n(A) = 1
So, P(A) = 1.
∴ Probability of getting 53 Sundays \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{1}{7}\)
(ii) Leap year should have fifty-three Sundays?
In 52 weeks we have 52 Sundays. We have to find the probability of getting one Sunday to form the remaining 2 days can be a combination of the following S = {Saturday to Sunday, Sunday to Monday, Monday, to Tuesday, Tuesday to wednes¬day, Wednesday to Thursday, Thursday to Friday, Friday and Saturday}.
(i.e) n(s) =7
In this = A {Saturday to Sunday, Sunday to Monday}
(i.e) n(A) = 2
So, P(A) = \(\frac{2}{7}\)

Eight coins are tossed simultaneously one time = one coin is tossed eight times.
Let S be the sample space.
S = {H, T} × {H, T} × ………….. × {H, T} 8 times
Let A be the event of getting exactly two heads,
B be the event of getting atleast two tails and
C be the event of getting atmost two tails.
When eight coins are tossed, the number of elements in the sample space
n(S) = 2 8 = 256
n(A) = 8C 2
= \(\frac{8 \times 7}{1 \times 2}\) = 28
n (B) = 8C 2 + 8C 3 + 8C 4 + 8C 5 + 8C 6 + 8C 7 + 8C 8
= n (S) – (8C 8 + 8C 1 )
= n (S) – {n (Event of getting all heads) + n (Event of getting one head)}
= n(S) – (1 + 8)
= 256 – 9 = 247
n (C) = 8C 0 + 8 C 1 + 8 C 2
= 1 + 8 + \(\frac{8 \times 7}{1 \times 2}\)
= 1 + 8 + 28 = 37
(i) P {getting exactly two tails) =
(ii) P (getting atleast two tails ) =
(iii) P (getting atmost two tails) =

Let S be the sample space
S = { 1, 2, 3, …………., 100 }
Let A be the event of choosing a prime number
A = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
n(A) = 25
Let B be the event of choosing an integer a multiple of 8.
B = { 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96}
n(B) = 12
P(Choosing a prime number) = P(A)
P (Choosing an integer a multiple of 8) = P (B)
P ( Choosing an integer a prime or multiple of 8)
= P(A or B)
= P(A ∪ B)
P(A) + P(B)
(since A and B are mutually exclusive that is A ∩ B = Φ)

Let S be the sample space, A be the event of taking 3 red balls and B be the event of taking one red and
2 black balls.
A bag contains 7 red balls and 3 black balls.
3 balls are drawn at random.
∴ The number of outcomes n(S) = 11C 3
= \(\frac{11 \times 10 \times 9}{1 \times 2 \times 3}\)
= 11 × 5 × 3 = 165
n (A) = 7C 3
n(B) = 7C 1 × 4C 2
= 7 × \(\frac{4 \times 3}{1 \times 2}\) = 7 × 6 = 42
(i) p (getting 3 red balls) = p(A)
(ii) p (getting one red and 2 blacks) = p(B)

S be the sample space one card is drawn from a pack of 52 cards.
∴ n(S) = 52C 1
n(S) = 52
(i) The card is an ace or a king
Let A be the event of getting an ace.
n(A) = 4C 1 = 4
Let B be the event of getting a king.
n(B) = 4C 1 = 4
P (getting an ace or a king)
= P(A or B)
= P(A ∪ B)
= P(A) + P(B)
(since A and B are mutually exclusive events, A ∩ B = Φ)
(ii) The card will be 6 or smaller:
Let A be the event of getting a number 6.
∴ n(A) = 4C 1 = 4
Let B be the event of getting numbers less than 6.
n(B) = 16C 1 = 16
P (the card will be 6 or less than 6)
= P(A or B)
= P(A ∪ B)
= P(A) + P(B)
(since A and B are mutually exclusive events A ∩ B = Φ)
(iii) The card is either a queen or 9?
Let A be the event of getting a Queen.
∴ n(A) = 4C 1 = 4
Let B be the event of getting a number 9.
n(B) = 4C 1 = 4
P (the card is either a Queen or 9)
= P(A or B)
= P(A ∪ B
= P(A) + P(B)
(since A and B are mutually exclusive events)

Number of members in the cricket club = 16
Number of bowlers = 5
Number of batters = 16 – 5 = 11
The probability that a team of 11 members consisting of atleast 3 bowlers = (Probability of selecting 3 bowlers and 8 batters) + ( Probability of selecting 4 bowlers and 7 batters) + (Probability of selecting 5 bowlers and 6 batters)
The probability that a team of 11 members consisting of atleast 3 bowlers
[Selection procedure:
First out of total 16 members selecting 11 members in 16C 11 ways.
Selection of 11 members consisting minimum of 3 bowlers.
∴ Selection of 11 members as follows
(1) 3 bowlers from 5 bowlers and 8 batters from 11 batters.
(2) 4 bowlers from 5 bowlers and 7 batters from 11 batters.
(3) 5 bowlers from 5 bowlers and 6 batters from 11 batters.]

Given the odds, that event A occurs is 5 to 7.
(ii) Suppose p (B) = \(\frac{2}{5}\). Express the odds that the event B occurs.
∴ The odds that the event B occurs is 2 to 3.

(i) P(A̅)
P(A̅) = 1 – P(A)
= 1 – \(\frac{3}{8}\)
P(A̅) = \(\frac{8-3}{8}\) = \(\frac{5}{8}\)
(ii) P(A ∪ B)
P(A ∪ B) = P(A) + P(B)
since A and B are mutually exclusive events.
(iii) P(A̅ ∩ B)
P(A̅ ∩ B) = P(B) – P(A ∩ B)
Since A and B are mutually exclusive we have A ∩ B = Φ
∴ P(A ∩ B) = 0
∴ P(A̅ ∩ B) = p(B)
= \(\frac{1}{8}\)
(iv) P(A̅ ∪ B̅)
P(A̅ ∪ B̅) = P\((\overline{A \cup B})\)
= 1 – P(A ∩ B)
Since A and B are mutually exclusive we have A ∩ B = Φ
∴ P(A ∩ B) = 0
P(A̅ ∪ B̅) = 1 – 0 = 1

Given P (A) = 0.35,
P (A and B) = P(A ∩ B) = 0.15
P(A or B) = P(A ∪ B) = 0.85
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.85 = 0.35 + P (B) – 0.15
0.85 + 0.15 – 0.35 = P(B)
P(B) = 1 – 0.35
P(B) = 0.65
(i) P (only B)
P (only B) = P(A̅ ∩ B)
= P(B) – P(A ∩ B)
= 0.65 – 0.15
P (only B) = 0.50
(ii) P (B̅)
P (B̅) = 1 – P(B)
= 1 – 0.65
P (B̅) = 0.35
(iii) P (only A)
P (only A) = P(A ∩ B̅)
= P(A) – P(A ∩ B)
= 0.35 – 0.15
P (only A) = 0.20

A die is thrown twice. Let S be the sample space
s = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3,2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2),(5, 3 ),(5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }
n(S) = 36
Let A be the event ‘First die shows 5’
A = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5,0)}
n(A) = 6
Let B be the event ‘Second die shows 5’
B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
n(B) = 6

P(A) = 0.5, P(B) = 0.3
Here A and B are mutually exclusive.
(i) P(A ∪ B) = P(A) + P(B)
= 0.5 + 0.3 = 0.8
(ii) P(A ∩ B) = P(A) + P(B) – P(A ∪ B) = 0.5 + 0.3 – 0.8
P(A ∩ B) = 0
P(A ∩ \(\overline{B}\)) = P(A) – P(A ∩ B) = 0.5 – 0 = 0.5
(iii) P(\(\overline{A}\) ∩ B) = P(B) – P(A ∩ B) = 0.3 – 0 = 0.3

A be the event of availability of a fire B be the event of a fire engine when needed. Availability of a second fire engine when needed.
Given P(A) = 0.96, P(B) = 0.96
Then A̅ is the event of non-availability of the first fire engine and B̅ is the event of non-availability of second fire engine when needed.
P (A̅) = 1 – P(A)
= 1 – 0.96
= 0.04
Also P(B) = 0.04
(i) P(atleast one engine is available) = (1 – probability of no engine available)
= 1 – P(A’ ∩ B’)
= 1 – P (A’) P(B’)
= 1 – (0.04) (0.04)
= 1 – 0.0016
= 0.9984
(ii) P (A’ ∩ B’) = P (A’) P(B’)
= 0.04 × 0.04
= 0.0016

(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.48 + 0.36 – 0.2
= 0.64.
(ii) P (Getting only one award)
= P(A) – P(A ∩ B) + P(B) – P(A ∩ B)
= (0.48 – 0.2) + (0.36 – 0.2)
= 0.28 + 0.16
= 0.44.

Given A and B are twp events such that
P(A ∪ B) = 0.7, P(A ∩ B) = 0.2 and P(B) = 0.5
To prove A and B are independent it is enough to prove
P(A ∩ B) = P(A). P(B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.7 = P (A) + 0.5 – 0.2
0.7 = P(A) + 0.3
P(A) = 0.7 – 0.3 = 0.4
P(A). P(B) = 0.4 × 0.5 = 0.20
= P(A ∩ B)
∴ P(A∩B) = P(A). P(B)
∴ A and B are independent.

(i) What is the probability that the problem is solved?
Let A 1 denote the event of that first student solves the problem.
A 2 denote the event that second student solves the problem.
A 3 denote the event that third student solves the problem.
Given P(A 1 ) = \(\frac{1}{3}\), P(A 2 ) = \(\frac{1}{4}\) P(A 3 ) = \(\frac{1}{5}\)
We note that A 1, A 2, A 3 are independent events.
The problem will be solved if atleast one of them
solves it we have to find P(A 1 ∪ A 2 ∪ A 3 )
Probability of at least one solves the problem = 1 – Probability of no one solving it
P(A 1 ∪ A 2 ∪ A 3 ) = 1 – P(A̅ 1 ∪ A̅ 2 ∪ A̅ 3 )
= 1 – P(A̅ 1 ). P(A 2 ). P(A 3 )
A 1, A 2, A 3 are independent then A̅ 1, A̅ 2, A̅ 3 are also independent.
= 1 – [1 – p(A 1 )] [1 – P(A 2 )] [1 – P(A 3 )]
(ii) What is the probability that exactly one of them will solve it?
p(A̅ 1 ) = 1 – P(A 1 ) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
p(A̅ 1 ) = 1 – P(A 2 ) = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
p(A̅ 1 ) = 1 – P(A 3 ) = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Probability of Exactly one student solving the problem
[Probability of exactly one student solving the problem = Probability [(A 1 solving the problem and A 2, A 3 non solving the problem) or (A 1, A 3 non solving and A 2 solving) or (A 1, A 2 solving and A 3 non solving)]

Let A be the event of changing oil, B be the event of changing oil filter.
Given P(A) = 0.30, P(B) = 0.40, P(A ∩ B) = 0.15
(i) Probability of new oil filter B needed when the oil A changed is
(ii) Probability of oil A changed when new oil filter B is changed is

First Bag contains 5 white and 3 black balls Total number of balls in the first bag 8 Second Bag contains 4 white and 6 black halls Total number of balls in the second bag = 10
One ball is drawn from each bag.
(i) Probability of both are white:
P (getting both are white) = P (getting white ball from the first bag) × P (getting the white ball from the second bag)
(ii) Probability of both are black:
P (getting both are black) = P (getting black ball from the first bag) × P (getting the ball from the second bag)
(iii) Probability of one white and one black.
P (getting one white and one black) = P ( getting one white from the first bag or one white from the second bag) + P (getting one black from the first bag or one black from the second bag)

Let G be the event of choosing a boy and G be the event of choosing a Girl.
Given P(B) = \(\frac{2}{3}\), P(G) = \(\frac{1}{3}\)
Let B 1 be the event of a boy getting first grade
P(B 1 ) = 0.70
Let G 1 be the event of a girl getting first grade
P(G 1 ) = 0.85
Probability of a. student getting a first grade = Probability of a boy getting first grade or Probability
of a Girl getting first grade
= P(B) × P(B 1 ) + P(G) × P(G 1 )
= \(\frac{2}{3}\) × 0.70 + \(\frac{1}{3}\) × 0.85
= \(\frac{1.4+0.85}{3}\)
= \(\frac{2.25}{3}\) = 0.75

P(A) = 0.4, P(A ∪ B) = 0.7
(i) When A and B are mutually exclusive
P(A ∪ B) = P(A) P(B)
(i.e.,) 0.7 = 0.4 + P(B)
0.7 – 0.4 = P(B)
(i.e.,) P(B) = 0.3
(ii) Given A and B are independent
⇒ P(A ∩ B) = P(A). P(B)
Now, P(A ∪ B) = P(A) + P(B) – P (A ∩ B)
(i.e.,) 0.7 = 0.4 + P(B) – (0.4) (P(B))
(i.e.,) 0.7 – 0.4 = P(B) (1 – 0.4)
0.3 = P (B) 0.6
⇒ P(B) = \(\frac{0 \cdot 3}{0 \cdot 6}=\frac{3}{6}\) = 0.5
(iii) P(A/B) = 0.4
(i.e.,) \(\frac{P(A \cap B)}{P(B)}\) = 0.4
⇒ P(A ∩ B) = 0.4 [P(B)] …………. (i)
But We know P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
⇒ P(A ∩ B) = 0.4 + P(B) – 0.7
= P(B) – 0.3 …………. (ii)
from (i) and (ii) (equating R.H.S) We get
0.4 [P(B)] = P(B) – 0.3
0.3 = P(B) (1 – 0.4)
0.6 (P(B)) = 0.3 ⇒ P(B) = \(\frac{0.3}{06}=\frac{3}{6}\) = 0.5
(iv) P(B/A) = 0.5
(i.e.,) \(\frac{P(A \cap B)}{P(A)}\) = 0.5
(i.e.,) P(A ∩ B) = 0.5 × P(A)
= 0.5 × 0.4 = 0.2
Now P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.7 = 0.4 + P(B) – 0.2
⇒ 0.7 = P(B) + 0.2
⇒ P(B) = 0.7 – 0.2 = 0.5

Probability of year being a leap year = \(\frac{1}{4}\)
Probability of year being non – leap year = \(\frac{3}{4}\)
(i) It contains 53 Sundays:
A non – leap year has 365 days.
365 days = 52 weeks + 1 day.
52 weeks contain 52 Sundays.
In order to get 53 Sundays in a non – leap year the remaining I day must be a Sunday.
Remaining one day may be Sunday or Monday or Tuesday or Wednesday or Thursday or Friday or Saturday.
Probability of getting Sunday from the remaining one day = \(\frac{1}{7}\)
A leap year has 366 days.
366 days = 52 weeks + 2 odd days
52 weeks contain 52 Sundays.
In order to get 53 Sundays in a leap year the remaining 2 days must contain a Sunday. Remaining Two days may be
S = { (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday ), ( Friday, Saturday), (Saturday, Sunday) }
n(S) = 7
Let A be the event set of getting a Sunday then
A = { (Sunday, Monday), ( Saturday, Sunday) }
n(A) = 2
P (getting a Sunday from the remaining 2 days)
P(getting 53 Sundays in a year) = P(getting a leap year) × P(getting a Sunday from the remaining 2 days) + P(getting a non-leap year) × P(getting a Sunday from the remaining 1 day)
∴ Probability of getting 53 Sundays in a year = \(\frac{5}{28}\)
(ii) A leap year has 366 days
\(\frac{366}{7}\) = 52 weeks + 2 days
In 52 weeks, we get 52 Sundays.
From the remaining two days we should get one Sunday, the remaining two days can be any one of the following combinations.
Saturday and Sunday, Sunday and Monday, Monday and Tuesday, Tuesday and Wednes¬day, Wednesday and Thursday, Thursday and Friday, Friday and Saturday of the seven combinations two have Sundays.
∴ (Probability of getting a Sunday = \(\frac{2}{7}\)
Selecting a leap year = \(\frac{1}{4}\)
{∴ In every four consecutive years we get one leap year}
∴ Probability of getting 53 Sundays = \(\frac{2}{7} \times \frac{1}{4} \times \frac{1}{14}\)

Given
Probability X hitting the targent P (X) = \(\frac{3}{4}\)
Probability Y hitting the targent P (Y) = \(\frac{4}{5}\)
Probability Z hitting the targent P (Z) = \(\frac{2}{3}\)
P(X̅) = 1 – P(X) = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
P(Y̅) = 1 – P(Y) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
p(Z̅) = 1 – P(Z) = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
Probability hitting the target exactly by 2 hits

Let A 1 be the event that items are produced by machine – I, A 2 be the event that items are produced by machine – II.
Let B be the event of drawing a defective item. We have to find the total probability of event B. That is P(B) clearly A 1 and A 2 are mutually exclusive and exhaustive events.
∴ P(B) = P(A 1 ). P(B/A 1 ) + P(A 2 ). P(B/A 2 )
P(B) = P(A 1 ). P(B/A 1 ) + P(A 2 ). P(B/A 2 )
= 0.60 × 0.02 + 0.40 × 0.04
= 0.012 + 0.016
= 0.028

(i) Let A 1 be the event of selecting Urn – I and A 2 be the event of selecting Urn – II.
Let B be the event of selecting one black ball.
We have to find the total probability of event B. That is P(B).
Clearly, A 1 and A 2 are mutually exclusive and exhaustive events.
Probability of selecting Urn – I
P(A 1 ) = \(\frac{1}{2}\)
Conditional Probability of B, given A 1
Probability of selecting Urn – II
P(A 2 ) = \(\frac{1}{2}\)
Conditional Probability of B, given A 2
(ii) The conditional Probability of A 1 given B is P(A 1 /B)
By Bayes’ theorem

Let A 1 be the daily volume of production by plant X, A 2 be the daily volume of production by plant Y, A 3 be the daily volume of production by plant Z.
Let B be the defective output we have to find P (B).
(i) Find the probability that the selected pipe is a defective one:
Clearly, A 1, A 2, and A 3 are mutually exclusive and exhaustive events.
Probability that the selected pipe is a defective one = \(\frac{7}{250}\)
(ii) If the selected pipe is defective, then what is the probability that it was produced by plant Y?
Probability that the defective pipe produced by plant Y = \(\frac{3}{7}\)

Let A 1, A 2, and A 3 be the events of A, B, and C becoming managers of the company respectively.
Let B be the event that the office canteen will be improved.
We have to find the conditional probability P (A 2 /B).
Since A 1, A 2 and A 3 are mutually exclusive and exhaustive events, applying Bayes theorem.
If the office canteen is improved than the probability of that B was appointed as the manager is \(\frac{15}{41}\)

(4) \(\frac{10}{21}\)
Explaination:
Number of Men = 3
Number of Women = 2
Number of Children = 4
Number of ways of choosing 2 children out of 4 children = 4C 2
= \(\frac{4 \times 3}{1 \times 2}\) = 6
Number of ways of choosing 4 persons from total 9 persons = 9C 4
= 9 × 2 × 7
= 126
Number of ways of choosing 2 persons (other than 2 children) from the remaining 5 persons (excluding children) = 5C 2
Probability that exactly two of them are children

(3) \(\frac{2}{5}\)
Explaination:
Sample space S is
S = {1, 2, 3, ………., 20}
n(S) = 20
Let A be the event of selecting a number divisible by 3
A = {3, 6, 9, 12, 15, 18}
n(A) = 6
Let B be the event of selecting a number divisible by 4
B = {4, 8, 12, 16, 20}
n(B) = 5
A ∩ B = {12}
n(A ∩ B) = 1
P (Number divisible by 3 or 4) = P (A ∪ B)
= P(A) + P(B) – P(A ∩ B)

(1) \(\frac{21}{64}\)
Explaination:
Given Probability of hitting the target by A is P(A) = \(\frac{3}{4}\)
Probability of hitting the target by B is P(B) = \(\frac{1}{2}\)
Probability of hitting the target by C is P(C) = \(\frac{5}{8}\)
Given A, B, C are Independent.
∴ P(A ∩ B ∩C) = P(A) P(B). P(C)
Probability of the target hit by A or B but not by C is
P(A ∩ B ∩ C̅) = P(A ∪ B). P(C̅)
= [P(A) + P(B) – P (A ∩ B)] [1 – P(C)]
= [P(A) + P(B) – P(A) P(B)] [1 – P(C)]

(2) Independent but not equally likely
Explaination:
If P(A ∩ B) = p(A). p(B), then A and B are independent.
we have (A ∩ B) = \(\frac{1}{4}\) ………… (1)
P(A). P(B) = \(\frac{3}{4} \times \frac{1}{3}=\frac{1}{4}\) ………… (2)
From equations (1) and (2) we get
P(A ∩ B) = p(A). p(B)
∴ A and B are independent
Since P(A) ≠ P(B), not equally likely.
∴ A and B are independent but not equally likely.

(1) \(\frac{19}{33}\)
Explaination:
Total number of items = 12
Number of ways of choosing two items from 12 items is = 12C 2
Number of defective items = 4
Number of non defective items = 8
Probability of getting atleast one defective items

(1) 1: 12
Explaination:
Let S be the sample space and A be the event of taking 2 hundred rupee notes.
n(S) = 13C 2 = \(\frac{13 \times 12}{1 \times 2}\) = 13 × 6
n(S) = 78
n(A) = 4C 2 = \(\frac{4 \times 3}{1 \times 2}\) = 2 × 3
n(A) = 6
n(A̅) = n(S) – n(A)
= 78 – 6 = 72
∴ Odds in favour of A is 6: 72
That is 1: 12

(4) \(\frac{19}{90}\)
Explaination:
Given words ‘ASSISTANT’ ‘STATISTICS
Sample space S = {(A, S, S, I, S, T, A, N, T)
(S,T, A, T, I, S, T, I, C, S)}
n(S) = {(A, S), (A, T), (A, A), (A, T), (A, I), (A, S), (A, T), (A, I), (A, C), (A, S) }
n(S) = 9 × 10 = 90
Let A be the event of se}ecting equal letters.
A = {(A, A), (A, A), (S, S), (S, S), (S, S), (S, S), (S, S), (S, S), (S, S), (S, S), (S, S), (1, 1), (1, 1), (T, T), (T,T), (T, T), (T, T), (T, T), (T, T) }
n(A) = 19
Probability of getting equal letters = \(\frac{19}{90}\)

(2) \(\frac{3}{8}\)
Explaination:
Sample space S = Set of all 2 × 2 matrices with elements 0 or 1 only
Number of elements in S is
n(S) = 2 4 = 16
Let A be the event of getting 2 × 2 matrices with elements 0 or 1 only whose determinant is non zero.

(3) \(\frac{1}{14}\)
Explaination:
Number of White balls (W) = 5
Number of Black balls (B) = 3
Five balls are drawn successively without replacement.
Probability that they are alternately of different
colours = P (W B W B W) + P (BW BW B)
= P(W). P(B). P(W). P(B). P(W) + P(B). P(W). P(B). P(W). P(B)

(1) \(\frac{68}{105}\)
Explaination:
Number of green balls (G) = 6
Number of white balls (W) = 2
Number of black balls (B) = 7
Two balls are drawn simultaneously
P (Balls are of different colours)
= P[(GW or WG) or (WB or BW) or (BG or GB)]
= P(GW) + P(WG) + P(WB) + P(BW) + P(BG) + P(GB)
= P(G) P(W) + P(W) P(G) + P(W) P(B) + P(B) P(W) + P(B) P(G) + P(G) P(B)
= 2 [P(G) P(W) + P(W) P(B) + P(B) P(G)]

(2) \(\frac{1}{2}\)
Explaination:
Number of Red balls n (R) = 5
Number of Black halls n (B) = 5
Number of elements in the sample space n (S) = 5 + 5 = 10
Case (i)
P (drawing a red ball first) = \(\frac{5}{10}\)
Let P(E 1 ) = \(\frac{1}{2}\)
Now two red balls are added.
P (drawing a red ball after adding) = \(\frac{7}{12}\)
P(A/E 1 ) = \(\frac{7}{12}\)
Case (ii)
P (drawing a black ball first) = \(\frac{5}{10}\)
Let P(E 2 ) = \(\frac{1}{2}\)
Now two black halls are added.
P (drawing a black ball after adding) = \(\frac{5}{12}\)
P(A/E 2 ) = \(\frac{5}{12}\)
P(A) = P(A/E 1 ). P(E 1 ) + P(A/E 2 ). P(E 2 )

(3) 0.71
Explaination:
Given x is choosen from the first 100 natural numbers.
n(S) = 100
Satifies when x takes the values 31 to 100 and also at x = 10
∴ A = { 10, 31, 32, 33, ……………, 100}
n(A) = 71

(1) \(\frac{5}{13}\)
Explaination:
Given A and B are independent events
P(A ∩ B) = P(A) P(B)
Also given P(A) = 0.35 and P(A ∪ B) = 0.6
P(A ∪ B) = P(A) – P(B) – P(A ∩ B)
P(A ∪ B) = P(A) + P(B) – P(A). P(B)
0.6 = 0.35 + P(B) – 0.35 P(B)
0.6 = 0.35 + (1 – 0.35) P (B)
0.6 = 0.35 + 0.65 P (B)
0.65 P (B) = 0.6 – 0.35
P(B) = \(\frac{0.25}{0.65}\)
P(B) = \(\frac{25}{65}\) = \(\frac{5}{13}\)

(3) 0.56
Explaination:
Given A and B are two events.
P(A) = 0.4, P(B) = 0.8 and P(B/A) = 0.6
P (A ∩ B) = 0.6 × 0.4 = 0.24
P(A̅ ∩ B) = P(B) – P(A ∩ B)
= 0.8 – 0.24
P(A̅ ∩ B) = 0.56

(3) \(\frac{3}{16}\)
Explaination:
x 2 + ax + b = 0 ⇒ x = \(\frac{-a \pm \sqrt{a^{2}-4 b}}{2}\)
Given that the roots are real ⇒ a 2 – 4b ≥ 0 or a 2 > 4b
When a = 1, b = 1 or 2 or 3 or 4 a 2 – 4b < 0
When a = 2, b = 1 a 2 – 4b = 0
When a = 3, b = 1 or 2 for which a 2 – 4b ≥ 0
When a = 4, b = 1 or 2, 3 or 4 for which a 2 – 4b ≥ 0
So, Selecting from the 4 number 4 2 = 16 ways.
(i.e.,) n(s) = 16
n(A) = (2 or 3 or 4) = 3
n(B) = (1 or 2 or 3 or 4) = 4
P(A) + P(B) = \(\frac{3}{16}+\frac{4}{16}=\frac{7}{16}\)

(4) \(\frac{7}{128}\)
Explaination:
Favourable events for atleast 8 heads
n(A) = 10C 8 + 10C 9 + 10C 10
= 10C 2 + 10C 1 + 10C 0
= \(\frac{10 \times 9}{1 \times 2}\) + 10 + 1
= 5 × 9 + 11 = 45 + 11
n(A) = 56
Ten coins are tossed
∴ n(S) = 2 10 = 1024

(4) 0.28
Explaination:
P(A) = 0.3, P(B) = 0.6
P(A ∩ B) = 0.18
So P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.3 + 0.6 – 0.18
= 0.9 – 0.18 = 0.72
P(A’ ∩ B’) = P[(A ∪ B)’] = 1 – P(A ∪B)
= 1 – 0.72 = 0.28

(3) \(\frac{3}{5}\)
Explaination:
2x 2 + 2mx + m + 1 = 0
roots are real ⇒ m 2 – 2m – 2 ≥ 0
Here m ≤ 5 ⇒ n(S) = 5
When m= 1,m 2 – 2m – 2
When m = 2, m 2 – 2m- 2
When m = 3, m 2 – 2m – 2
When m = 4, m 2 – 2n- 2
When m = 5, m 2 – 2m – 2
⇒ n{A) = 3 and so P(A) = \(\frac{3}{5}\)