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Total no. of possible outcomes = 12 {1, 2, 3,…., 12}

(i) Let E ⟶ event of pointing 10

No. favourable outcomes = 1 {10}

P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes) = 1/12

(ii) Let E ⟶ event of pointing at an odd no.

No. favourable outcomes = 6 {1, 3, 5, 7, 9, 11}

P(E) = 6/12 = 1/2

(iii) Let E ⟶ event of pointing at a no. multiple of 3

No. favourable outcomes = 4 {3, 6, 9, 12}

P(E) = 4/12 = 1/3

(iv) Let E ⟶ event of pointing at an even no.

No. favourable outcomes = 6 {2, 4, 6, 8, 10, 12}

P(E) = 6/12 = 1/2

Correct option is: C) Both (i) & (iv)

Correct option is: A) 0 ≤ P(A) ≤ 1

The probability of forming 3-digit no. with digits 0, 2, 4, 6, 8 is as follows,

Since it is a 3 digit no., first place cannot have 0 as it will make it a 2 digit no.

\(\therefore\) 1^st place can be filled in 4 ways, 2^nd place in 5 ways and 3^rd in 5 ways.

\(\therefore\) total number = 4 x 5 x 5 = 100

The three digit numbers with same digits are 222,444,666,888

\(\therefore\) P (forming three-digit no.) = \(\cfrac{4}{100}=\cfrac{1}{25}=0.04\)

Let E₁, E₂ and A be the events defined as follows: 

E₁ = Six occurs, 

E₂ = Six does not occur 

A = man reports it is a six 

Then, P(E₁) = \(\frac{1}{6}\) , P(E₂) = 1- \(\frac{1}{6}\) = \(\frac{5}{6}\)

 P(A/E₁) = Probability of man reporting it a six when six occurs = Probability of speaking truth = \(\frac{3}{4}\)

P(A/E₂) = Probability of man reporting a six when six does not occur 

= Probability of not speaking truth = 1-\(\frac{3}{4}\) = \(\frac{1}{4}\)

∴ P(Throw is actually a six) =\(\frac{P(E_1)\times P(A/E_1)}{P(E_1)\times P(A/E_1)+P(E_2)\times P(A/E_2)}\)                                 ... (Baye’s Theorem) 

= \(\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}\) = \(\frac{\frac{3}{24}}{\frac{8}{24}}\) = \(\frac{3}{8}\)

Total number of possible outcomes = 3 + 4 + 5 = 12

No. favourable outcomes = 3

P(candle is red) = 3/12 = 1/4

P(candle is not red) = 1 - P (candle is red)

= 1 - (1/4) = (4 - 1)/4 = 3/4

Total no. of pens = 144

Defective one : 20

Good ones = 744-20 = 124

Probability of purchasing pen  

= 124/144 = 31/36

Probability of not purchasing pen

= 20/144 = 5/36

A bag contains 10 balls each marked with one of the digit from 0 to 9. 

Probability that one ball is in marked 0 drawn 

P = 1/10 = 0.1 = 0.1 

Probability that ball is not marked 0 

q = 1 – p 

= 1 – 0.1 = 0.9 

Now 4 balls are drawn successively with replacement. 

∴ Probability that any of them ball is marked 0

= P(X = 0) = 4C₀p⁰q⁴

= (0.9)⁴ = (9/10)⁴

Probability that a bulb gets fuse after 150 days of its use = 0.05 Probability that the bulb will not fuse after 150 days of its use = 1 – 0.05 = 0.95

(i) Probability that no bulb will fuse after 150

days of its use = P(none) = (0.95)⁵ = 0.7738

= 0.77(approx)

(ii) P(not more than one ) = O(0) + P(1)

= (0.95)⁵ + ⁵C₁ x (0.95)⁴ x (0.05)

= (0.95)⁵[0.95 + 5 x 0.05] = (0.95)⁴ x 1.2

(iii) P(more than one)

= P(2) + P(3) + P(4) + P(5)

= [P(0) + P(1) + P(2) + P(3) + P(4) + P(5)] - [P(0) + P(1)]

= 1 - [P(0) + P(1)] = 1 - (0.95)⁴ x 1.2

Given, Sample space is the set of first 100 natural numbers. 

∴ n(S) = 100

Let A be the event of choosing the number such that it is divisible by 4 

∴ n(A) = \([\frac{100}{4}] \) = [25] = 25 {where [.] represents Greatest integer function} 

∴ P(A) =   \(\frac{n(A)}{n(S)}\) = \(\frac{25}{100} =\frac{1}{4}\) 

Let B be the event of choosing the number such that it is divisible by 6 

∴ n(B) = \([\frac{100}{6}]\)  = [16.67] = 16 {where [.] represents Greatest integer function} 

∴ P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{16}{100} =\frac{4}{25}\) 

We need to find the P(such that number chosen is divisible by 4 or 6) 

∵ P(A or B) = P(A∪B) 

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that: 

P(E∪F) = P(E) + P(F) – P(E∩F)

∴ P(A∪B) = P(A) + P(B) – P(A∩B) 

We don’t have value of P(A∩B) which represents event of choosing a number such that it is divisible by both 4 and 6 or we can say that it is divisible by 12. 

n(A∩B) = \([\frac{100}{12}] \) = [8.33] = 8 

∴ P(A∩B) = \(\frac{n(A ∩ B)}{n(S)}\) = \(\frac{8}{100} = \frac{2}{25}\) 

∴ P(A∪B) = \(\frac{1}{4}+\frac{4}{25}-\frac{2}{25}\) = \(\frac{1}{4}+\frac{2}{25}\) =\(\frac{33}{100}\) 

In a deck of 52 cards there are 2 colours. Each colour having 26 cards. 

As we need to choose 4 cards out of 52. Let S represents the sample space. ∴ 

n(S) = ⁵²C₄ 

Let A represents the event that all 4 cards drawn are black. 

∴ n(A) = ways in which 4 cards can be selected from 26 black cards. 

⇒ n(A) = ²⁶C₄ 

∴ P(A) = \(\frac{^{26}C_4}{^{52}C_4}\)  = \(\frac{26\times25\times24\times23}{52\times51\times50\times49}\)  = \(\frac{46}{833}\)

Let B represents the event that all 4 cards drawn are red. 

∴ n(B) = ways in which 4 cards can be selected from 26 red cards. 

⇒ n(B) = ²⁶C₄ 

∴ P(B) =  \(\frac{^{26}C_4}{^{52}C_4}\)  = \(\frac{26\times25\times24\times23}{52\times51\times50\times49}\)  = \(\frac{46}{833}\)

As we need to find the probability of event such that all drawn cards are from same colour. This means we need to find 

P(A∪B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that: 

P(E∪F) = P(E) + P(F) – P(E∩F) 

∴ P(A∪B) = P(A) + P(B) – P(A∩B) 

As both events A and B have no common elements or we can say that they are mutually exclusive 

∴ P(A∩B) = 0 

Hence, 

P(A∪B) = P(A) + P(B) = \(\frac{46}{833} + \frac{46}{833} = \frac{92}{833}\)  

 Let E denotes the event that student passed in first examination.

And H be the event that student passed in second exam. 

S is the sample space containing the students who appeared for the exam. 

Given, 

n(S) = 100 

n(E) = 60 

n(H) = 50 

also no of students who passed both exam = n(E∩H) = 30 

∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{60}{100}\) = \(\frac{3}{5}\) 

Similarly, P(H) =  \(\frac{n(H)}{n(S)}\) = \(\frac{50}{100}\) = \(\frac{1}{2}\)  

And, P(E∩H) = \(\frac{P(E∩H)}{n(S)}\) = \(\frac{30}{100}\) = \(\frac{3}{10}\) 

We need to find the probability of event such that a student selected at random has passed at least one examination. 

This can be given as – P(E or H) = P(E∪H) 

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

∴ P(E∪H) = P(E) + P(H) – P(E∩H) 

⇒ P(E∪H) = \(\frac{3}{5}+\frac{1}{2}-\frac{3}{10}\) = \(\frac{11}{10}-\frac{3}{10}\) 

= \(\frac{8}{10} = \frac{4}{5} \) 

∴ P(E∪H) = \(\frac{4}{5}\)

Total numbers of elementary events are: 15 

Let E be the event of getting a multiple of 4 

Favorable outcomes are: 4, 8, 12 

Numbers of favorable outcomes are = 3 

P (multiple of 4) = P (E) = \(\frac{3}{15}\) = \(\frac{1}{5}\)

Total number of elementary events are = 7 

Let E be the event of getting a multiple of 4 

Favorable outcome are: 4, 8 

Numbers of favorable outcome are= 2 

P (multiple of 4) = P (E) = \(\frac{2}7\)

Apply general term formula of an AP, to find the total numbers of cards.

Here a (first term) = 6, d (common difference) = 1 and l( last term) = 70

l = a + (n-1)d

70 = 6 + (n-1)(1)

n = 65

Total number of cards = 65

(i) Favorable outcome = a one-digit number = 6, 7, 8, 9

Favorable number of outcomes = 4

P(getting a one-digit number) = 4/65

(ii) Favorable outcome = a number divisible by 5 = 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70

Favorable number of outcomes = 13

P(getting a number divisible by 5) = 13/65 = 1/5

(iii) Favorable outcome = an odd number less than 30 = 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 and 29

Favorable number of outcomes = 12

P (getting a odd number less than 30) = 12/65

(iv) Favorable outcome = a composite number between 50 and 70 = 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69

The number of favorable outcomes = 15

P(getting a composite number between 50 and 70) = 15/65 = 3/13

The total number of outcomes = 30.

(i) Let E₁ be the event of getting a number not divisible by 3.

out of these numbers, numbers divisible by 3 are 3,6,9,12,15,18,21,245,27 and 30.

number of favorable outcomes  = 30 - 10 = 20

Therefore P(getting a number not divisible by 3) = P(E₁) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_1}{number\,of\,all\,possible\,outcomes}\) = \(\frac{20}{30}\) = \(\frac{2}{3}\).

Thus, the probability that the number on the card is not divisible by 3 is \(\frac{2}{3}\).

(ii) Let E₁ be the event of getting a prime number greater than 7.

out of these numbers, prime numbers greater than 7 are 11,13,17,19,23 and 29.

number of favorable outcomes = 6

Therefore P(getting a prime number greater than 7) = P(E₂) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_2}{number\,of\,all\,possible\,outcomes}\) = \(\frac{6}{30}\) = \(\frac{1}{5}\).

Thus, the probability that the number on the card is a prime number greater than 7 is \(\frac{1}{5}\).

(iii) Let E₃ be the event of getting a number which is not a perfect square number.

out of these numbers, perfect square numbers are 1,4,9,16 and 25.

number of favorable outcomes = 30 - 5 = 25

Therefore P(getting non-perfect square number) = P(E₃) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_3}{number\,of\,all\,possible\,outcomes}\) = \(\frac{25}{30}\) = \(\frac{5}{6}\).

Thus, the probability that the number on the card is not a perfect square number is \(\frac{5}{6}\).

Given numbers 1,3,5.....35 form an AP with a = 1 and d = 2

Let T_n = 35 Then,

1 + (n - 1)2 = 35

⇒ 1 + 2n - 2 = 35

 ⇒ 2n = 36 

⇒ n = 18

Thus, Total number of favorable outcomes = 18

(i) Let E₁ be the event of getting a prime number less than 15.

out of these numbers, prime numbers less than 15 are 3,5,7,11 and 13.

The number of favorable outcomes = 5

Therefore P(getting a prime number less than 15) = P(E₁) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_1}{number\,of\,all\,possible\,outcomes}\) = \(\frac{5}{18}\)

Thus, the probability of getting card bearing a prime number less than 5 is \(\frac{5}{18}\)

(ii) Let E₂ be the event of getting a number divisible by 3 and 5

out of these numbers, the number divisible by 3 and 5 means number divisible by 15 is 15

The number of favorable outcomes = 1

Therefore P(getting a number divisible by 3 and 5) = P(E₂) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_2}{number\,of\,all\,possible\,outcomes}\)= \(\frac{1}{18}\)

Thus, the probability of getting card bearing a number divisible by 3 and 5 is \(\frac{1}{18}\)

Cards numbered 1 to 30 are put in a bag.

So, total number of cards = 30

(i) Numbers divisible by 3 form 1 to 30 = {3,6,9,12,15,18,21,24,27,30}

Total numbers divisible by 3 = 10

Numbers not divisible by 3 = 30 – 10 = 20

P(drawn card is not divisible by 3) = 20/30 = 2/3

(ii) A prime numbers greater than 7 form 1 to 30 = {11, 13, 17, 19, 23, 29}

Total prime number greater than 7 = 6

P(drawn card is a prime number greater than 7) = 6/30 = 1/5

(iii)

Perfect square numbers form 1 to 30 = {1,4,9,16,25}

Number of Perfect square = 5

Non perfect square numbers = 30-5 = 25

P(drawn card is not a perfect square number) = 25/30 = 5/6

It is given that

Total number of salt packets = 12

Number of packets having more than 2kg of salt = 5

We know that

Probability that the chosen packet contains more than 2kg of salt = number of packets having more than 2kg of salt/ total number of salt packets

By substituting the values

Probability that the chosen packet contains more than 2kg of salt = 5/12

It is given that

Total number of students = 30

(i) We know that

Probability that the marks of the chosen students are 30 or less = (7 + 10 + 6)/ 30

So we get

Probability that the marks of the chosen students are 30 or less = 23/30

(ii) We know that

Probability that the marks of the chosen students are 31 or more = (4 + 3)/30

So we get

Probability that the marks of the chosen students are 31 or more = 7/30

(iii) We know that

Probability that the marks of the chosen students lie in the interval 21 – 30 = 6/30

So we get

Probability that the marks of the chosen students lie in the interval 21 – 30 = 1/5

No. of all possible outcomes = 7

No. of favourable outcomes = -1,0,1 = 3

.^. .  Required probability = 3/7

All possible outcomes are : (HHH), (THH), (HTH), (HHT), (TTT), (TTH), (THT), (HTT).

No. of favourable outcomes = 3

P(getting two heads) = 3/8

The law of addition of probability for two events A and B is as follows :

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

If A and B are mutually exclusive events, A ∩ B = 0 and P (A ∩ B) = 0. Hence, the law of addition of probability for two events A and B is written as follows:

P(A ∪ B) = P(A) + P(B)

For two independent events A and B, the results are obtained as follows:

1. P(A ∩ B) = P(A) × P(B)

2. P(A’ ∩ B’) = P(A’) × P(B’)

3. P(A ∩ B’) = P(A) × P(B’)

4. P(A’ ∩ B) = P(A’) × P(B) .

Interpretation of P(A|B): The conditional probability of the event A under the condition that the event B is occurred.

Interpretation of P(B|A): The conditional probability of the event B under the condition that the event A has occurred.

The law of multiplication of probability for two events A and B is as follows:

P(A ∩ B) = P(A|B) ∙ P(B) OR
P(A ∩ B) = P(B|A) ∙ P(A)

If two events A and B are independent, P(A|B) = P(A) and P(B|A)= P(B).

So the law of multiplication of probability for two events A and B is written as follows:

P(A ∩ B) = P(A) ∙ P(B)

U is a finite sample space and A and B are any two events of U. If the probability of occurrence of the event A does not depend on the occurrence or non-occurrence of the event B, then the events A and B are called independent events.