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If P(A) = \(\frac{3}{5}\) then P(A’) = 1 - P(A) = 1 - \(\frac{3}{5}\) = \(\frac{2}{5}\)

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}

P(odds in favour of 3 : 5) = \(\frac{3}{8}\)

In the English language there  are 26 alphabets.Consonant are 21. The probability of chosen a consonant = 21/26. 

4 cards can be drawn from 52 in ⁵²C₄ ways. 

P(3 diamond and one spade)

= (C₃ x ¹³C₁)/⁵²C₄ = 286/20825

∵ 13 diamond cards and 13 spade cards in a pack of [1] cards.

Total number of marbles = 10 + 20 + 30 = 60
Number of ways of drawing 5 marbles from 60 marbles = ⁶⁰C₅

(i) All the drawn marbles will be blue if we draw 5 marbles out of 20 blue marbles.
5 blue marbles can be drawn from 20 blue marbles in ²⁰C₅ ways.
∴ Probability that all marbles will be blue =²⁰C_{5 / ⁶⁰}C₅

(ii) Number of ways in which the drawn marble is not green =^(20+10)C₅

∴ Probability that no marble is green = 30C5 / 60C5
Probability that at least one marble is green = 1 – (30C5 / 60C5)

There are 25 cards from which one card is drawn.

Total number of elementary events = n(S) = 25

(i) From number 1 to 25, there are 8 number which are multiple of 3 i.e. {3, 6, 9, 12, 15, 18, 21, 24}, 

Favorable number of events = n(E) = 8

Probability of selecting a card with a multiple of 3 = n(E)/n(S) = 8/25

(ii) From number 1 to 25, there are 5 number which are multiple of 5 i.e. {5, 10, 15, 20, 25}

Favorable number of events = n(E) = 5

Probability of selecting a card with a multiple of 5 = n(E)/n(S) = 5/25 = 1/5

(iii) From number 1 to 25, there is one number which are multiple of 3 and 5 i.e. {15}

Favorable number of events = n(E) = 1

Probability of selecting a card with a multiple of 3 and 5 = n(E)/n(S) = 1/25 

(iv) From number 1 to 25, there are 12 numbers which are multiple of 3 or 5 i.e. {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25}

Favorable number of events = n(E) = 12

Probability of selecting a card with a multiple of 3 or 5 = n(E)/n(S) = 12/25

(odds against of 2 : 7) = \(\frac{7}{9}\)

Event A: Card is a queen 

=> n(A) = 4 & n(S) = 52. 

P(A) = \(\frac{n(A)}{n(S)} = \frac{4}{52}\)

Two heads

Possible number of favourable outcomes = 1 (i.e. HH)

Total number of possible outcomes = 4

P(E) = 1/4

S = {HH, HT, TH, TT}, n(S) = 4 

Event A = Two heads n(A) = 1. 

∴ P(A) = \(\frac{n(A)}{n(S)} = \frac{1}{4}\)

Let E be the event that the doctor visits the patient late and let T₁, T₂, T₃, T₄ be the events that the doctor comes by train, bus, scooter, and other means of transport respectively.

Then P(T₁) = 3/10, P(T₂) = 1/5, P(T₃) = 1/10 and P(T₄) = 2/5

P(E|T₁) = Probability that the doctor arriving late comes by train = 1/4

Similarly, P(E|T₂) = 1/3, P(E|T₃) = 1/12 and P(E|T₄) = 0, since he is not late if he comes by other  means of transport.

Therefore, by Bayes' Theorem, we have 

P(T₁|E) = Probability that the doctor arriving late comes by train

P(T₁/E) = (P(T₁)P(E/T₁))/(P(T₁)P(E/T₁) + P(T₂)P(E/T₂) + P(T₃)P(E/T₃) + P(T₄)P(R/T₄))

= (3/10 x 1/4)/(3/10 x 1/4 + 1/5 x 1/3 + 1/10 x 1/12 + 2/5 x 0) = 3/40 x 120/18 = 1/2

Hence, the required probability is 1/2

Required sample space = S 

= {HH,HT,TH,TT}. 

Let A: ‘at least one tail occurs’A = {HT,TH,TT} 

We have, n(A) = 3, n(S) = 4 = 2² 

Required probability  = P(A) = n(A)/n(s) = 3/4

Required sample space associated with an experiment is S = {1, 2, 3,4, 5, 6} 

∴ n(S) = 6 

(i) Let A : ‘a prime number will appear’ 

⇒ A = {2,3,5} ∴ n(A) = 3 

∴ Required Probability

P(A) = n(A)/n(S) = 3/6 = 1/2

(ii) Let B: ‘a number ≥ 3 will appear’ 

⇒ B = {3,4,5,6} ∴ n(B) = 4 

∴ Required Probability

P(B) = n(B)/n(S) = 4/6 = 2/3

(iii) Let C: ‘a number ≤ 1 will appear’ 

⇒ C = (1) 

∴ Required Probability

P(C) = n(C)/n(S) = 1/6

(iv) Let D: ‘a number >6 will apper’ ⇒ D = { } = ϕ

= P(D) = 0/6 = 0

(v) E : ‘ a number < 6 will appear’ 

⇒ E= {1, 2, 3, 4, 5} ∴ n(E) = 5 

∴ Required probability

We have, number of possible outcomes = 52 n(S) = 52= ⁵²C₁ 

(a) There are 52 points in the sample – space. 

(b) Let A: ‘a selected card is an ace of spades’ 

Clearly the number of elements in the set A is 1. i.e., n(A) =¹C₁ 

∴ Required probability

= P(A) = n(A)/n(S) = 1/52

(c) (i) Let B: ‘a selected card is an ace’ we know that, in a deck of 52 cards, there are 4 different aces. 

∴ n(B) = 4= ⁴C₁ 

∴ Required probability

= P(B) = n(B)/n(S) = 4/52 = 1/13

(ii) Let C: ‘a selected card is black’ There are 26 black cards in a pack o f 52 cards 

∴ n(C) =  ²⁶C₁ = 26 

∴ Required probability

= P(C) = n(C)/n(S) = 26/52 = 1/2

(B) is the correct answer. Since a leap year has 366 days and hence 52 weeks and 2 days. The 2 days can be SM, MT, TW, WTh, ThF, FSt, StS.

Therefore, P (53 Sundays or 53 Mondays) =3/7 .

Let B be the event that a truck stopped at the roadblock will have faulty brakes and T be the event that it will have badly worn tires. We have P (B) = 0.23, P (T) = 0.24 and P (B∪T) = 0.38

and P (B ∪ T) = P (B) + P (T) – P (B ∩ T)

So 0.38 = 0.23 + 0.24 – P (B ∩ T)

⇒ P (B ∩ T) = 0.23 + 0.24 – 0.38 = 0.09

(c) \(\frac{1}{10}\)

Let S be the sample space.

Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon

= ⁶C₃ = \(\frac{6\times5\times4}{3\times2}\) = 20.

Let A : Event that the selected three vertices form an equilateral triangle. 

Then n(A) = 2 

(As only two equilateral triangles are formed from the vertices of a regular hexagon)

∴ Required probability = \(\frac{n(A)}{n(S)}\) = \(\frac{2}{20}\) = \(\frac{1}{10}\).

Given:

⇒ P(MA) = P(getting A in mathematics)

⇒ P(MA) = 0.2

⇒ P(MN) = P(not getting A in mathematics)

⇒ P(MN) = 1 - 0.2

⇒ P(MN) = 0.8

⇒ P(P_A) = P(getting A in physics)

⇒ P(P_A) = 0.3

⇒ P(P_N) = P(not getting A in physics)

⇒ P(P_N) = 1 - 0.7

⇒ P(P_N) = 0.3

⇒ P(C_A) = P(getting A in Chemistry)

⇒ P(C_A) = 0.5

⇒ P(C_N) = P(not getting A in chemistry)

⇒ P(C_N) = 1 - 0.5

⇒ P(C_N) = 0.5

We need to find the probability that:

i. X gets A in all subjects

ii. X gets A in no subjects

iii. X gets A in two subjects

⇒ P(X_all) = P(getting A in all subjects)

Since getting A in different subjects is an independent event, their probabilities multiply each other

⇒ P(X_all) = (P(M_A)P(P_A)P(C_A)

⇒ P(X_all) = 0.2 × 0.3 × 0.5

⇒ P(X_all) = 0.03

⇒ P(X_none ) = P(getting A in no subjects)

Since getting A in different subjects is an independent event, their probabilities multiply each other

⇒ P(X_none ) = (P(MN)P(P_N)P(C_N))

⇒ P(X_none ) = 0.8 × 0.7 × 0.5

⇒ P(X_none ) = 0.28

⇒ P(X_two ) = P(getting A in any two subjects)

Since getting A in different subjects is an independent event, their probabilities multiply each other

⇒ P(X_two ) = (P(M_A)P(P_A)P(C_N)) + (P(M_A)P(P_N)P(C_A)) + (P(M_N)P(P_A)P(C_A))

⇒ P(X_two ) = (0.2 × 0.3 × 0.5) + (0.2 × 0.7 × 0.5) + (0.8 × 0.3 × 0.5)

⇒ P(X_two ) = 0.03 + 0.07 + 0.12

⇒ P(X_two ) = 0.22

∴ The required probabilities are 0.03, 0.28, 0.22.

The correct answer is (D). From the given data P (A) + P (B) = P (A∪B).

This shows that P (A∩B) = 0. Thus P (A | B) = P(A∩B)/P(B) = 0

(a) All possible genders are expressed as :

S = {BBB, BBG, BGB, BGG, GBB,GBG, GGB, GGG}

(b) (i)Let A denote the event : ‘exactly one child is a girl’

A = {BBG, BGB, GBB}

(ii) Let B denote the event that at least two children are girls.

B = {GGB, GBG, BGG, GGG}, P (B) =4/8

(iii) Let C denote the event : ‘no child is a girl’.

C = {BBB}

∴ P (C) =1/8

The correct answer is (A).

Answer is True

Total number of possible outcomes, n(S) = 52 – 8 = 44 

(i) Number of favorable outcomes, 

n(E) = 2

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{2}{44}\) = \(\frac{1}{22}\)

(ii) Number of favorable outcomes, n(E) = 26 – 4 = 22

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{22}{44}\) = \(\frac{1}{2}\)

Total number of possible outcomes, n(S) = 52 – 2 – 2 = 48 

(i) Number of favorable outcomes, 

n(E) = 4

∴ P(E) \(\frac{n(E)}{n(S)}\) = \(\frac{4}{48}\) = \(\frac{1}{12}\)

(ii) Number of favorable outcomes, 

n(E) = 24

∴ P(E) \(\frac{n(E)}{n(S)}\) = \(\frac{24}{48}\) = \(\frac{1}{2}\)

(iii) Number of favorable outcomes, n(E) = 12 – 4 = 8

 ∴ P(E) \(\frac{n(E)}{n(S)}\) = \(\frac{8}{48}\) = \(\frac{1}{6}\)

(iv) Number of favorable outcomes, 

n(E) = 2

∴ P(E) = \(\frac{n(E)}{n(S)}\)  = \(\frac{2}{48}\) = \(\frac{1}{24}\)

Total number of possible outcomes, n(S) = 20 

(i) Number of favorable outcomes, n(E) = 10 + 3 = 13 {(2, 4, 6, 8, 10, 12, 14, 16, 18, 20) (3, 9, 15)}

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{13}{20}\)

(ii) Number of favorable outcomes, 

n(E) = 8

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{8}{20}\) = \(\frac{2}{5}\)

Number of all possible outcomes = 36

Let E be the event of getting all those numbers whose product is 12.

These number are (2,6),(3,4),(4,3) and (6,2)

Therefore, p(getting all those numbers whose product is 12) = P(E) = \(\frac{number\,of\,outcomes\,favorable\,to\,E}{number\,of\,all\,possible\,outcomes}\) = \(\frac{4}{36}\) = \(\frac{1}{9}\)

Thus, the probability of getting all those numbers whose product is 12 is \(\frac{1}{9}\).

Total number of possible outcomes, n(S) = 44 

(i) Number of favorable outcomes, 

n(E) = 20

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{20}{44}\) = \(\frac{5}{11}\)

(ii) Number of favorable outcomes, n(E) = (24 – 9) + (20 – 11) = 24

∴ P(E) = \(\frac{n(E)}{n(S)}\) =  \(\frac{24}{44}\) = \(\frac{6}{11}\)

(iii) Number of favorable outcomes, 

n(E) = 9

∴ P(E) = \(\frac{n(E)}{n(S)}\) =  \(\frac{9}{44}\)

(iv) Number of favorable outcomes, 

n(E) = 9

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) =  \(\frac{9}{44}\)

All possible outcomes are 5 ,6, 7, 8.........50.

Number of all possible outcomes = 46

(i) Out of the given numbers, the prime numbers less than 10 are 5 and 7.

Let E₁ be the event of getting a prime number less than 10.

Then, number of favorable outcomes = 2

Therefore, p(getting a prime number less than 10) = P(E) = \(\frac{2}{46}\) = \(\frac{1}{23}\)

(ii) out of the given numbers, the perfect squares are 9, 16, 25, 36 and 49.

Let E₂ be the event of getting a perfect square.

Then, number of favorable outcomes = 5

Therefore, P(getting a perfect square) = P(E) = \(\frac{5}{46}\)

Total number of possible outcomes, n(S) = 52 – 6 = 46 

(i) Number of favorable outcomes, n(E) = 26 – 6 = 20

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{20}{46}\) = \(\frac{10}{23}\)

(ii) Number of favorable outcomes, n(E) = 12 – 6 = 6

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{6}{46}\) = \(\frac{3}{23}\)

(iii) Number of favorable outcomes, n(E) = 13 – 3 = 10

 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{10}{46}\) = \(\frac{5}{23}\)