Explore topic-wise InterviewSolutions in .

Let events A,B and C are

A : Bolt is manufactured by machine A.

B : Bolt is manufactured by machine B.

C : Bolt is manufactured by machine C.

P(A) = 25% = \(\frac{25}{100}\) = \(\frac{1}{4}\)

P(B) = 35% = \(\frac{35}{100} = \frac{7}{20}\)

P(C) = 40% = \(\frac{40}{100} = \frac{2}{5}\)

let E be event that drawn bold in defective.

P(E/A) = 5% = \(\frac{5}{100} = \frac{1}{20}\)

P(E/B) = 4% = \(\frac{4}{100} = \frac{1}{25}\)

P(E/C) = 2% = \(\frac{2}{100} = \frac{1}{20}.\)

Probability that defective bolt is manufactured by C 

is P(C/E) = \(\frac{P(E/C)P(C)}{P(E/C).P(C)+P(E/B)\, P(B)+P(E/A)P(A)}\)

= \(\cfrac{\frac{1}{50} \times \frac{2}{5}}{\frac{1}{50}\times \frac{2}{5} + \frac{1}{25} \times \frac{7}{20}+ \frac{1}{20}\times\frac{1}{4}}\)

=\(\cfrac{\frac{1}{125}}{\frac{1}{125} + \frac{7}{500}+\frac{1}{80}}\)

= \(\cfrac{\frac{1}{125}}{\frac{16+28+25}{2000}}\)

= \(\frac{16}{69}\)

Thus probability that defective bolt is not  manufactured by C 

is P(C'/E) = 1-P(C/E) 

= 1 - \(\frac{16}{69}\) = \(\frac{69-16}{69}\) = \(\frac{53}{69}\).

Since, Total cards = 52

No. of kings = 4

No. of queens = 4

No. of favorable cards that are king or queen = 8

∴ Required probability = 8/52 ​= 2/13​

(i) Number of good shirts = 88

P (Ramesh buys the shirt) = 88/100 or 22/25

(ii) Number of shirts without Major defect = 96

P(Kewal buys a shirt) = 96/100 or 24/25

Given: Math students = 30% 

Chemistry Students = 20% 

Math & Chemistry both = 10% 

To Find: P(Math or Chemistry)

Now, P(Math) = 30% = \(\frac{30}{100}\) = 0.30

P(Chemistry)  30% = \(\frac{20}{100}\) = 0.20

P(Math ∩ Chemistry) = 10% =  \(\frac{10}{100}\) = 0.10

We know that,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

Therefore, 

P(Math ∪ Chemistry) = 0.30 + 0.20 – 0.10 = 0.40

Hence, number of students studying math or chemistry are 40%.

let A denot the event that Hemant passes in english and B denote the event that hemant passes in hindi . 

Given : P(A) = \(\frac{2}{3}\), P(B) = \(\frac{5}{9}\),P(A and B) = \(\frac{2}{5}\) 

To find : Probability that he will pass in at least one of these subjects. = P(A or B) 

Formula used : P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = \(\frac{2}{3}+\frac{5}{9}-\frac{2}{5}\)

P(A or B) = \(\frac{30+25-18}{45}\) = \(\frac{37}{45}\)

P(A or B) =   \(\frac{37}{45}\) 

The probability that he will pass in at least one of these subjects. = P(A or B) =  \(\frac{37}{45}\) 

On tossing a coin three times, Possible outcomes are (HHH), (HHT), (HTH), (THH), (TTT), (TTH),(THT),(HTT)

Total number of possible outcomes = 8

Winning chances fer Ramesh = 2

and his loosing chances = 8 - 2 = 6

.^. .  P(Ramesh loosing the game) = 6/8 = 3/4

Given numbers 6,7,8,......70 form an AP with a = 6 and d = 1

Let T_n  = 70. Then,

6 + (n - 1) = 70

⇒ 6 + n - 1 = 70

⇒ n = 65

Thus, total numbers of outcomes = 65.

(i) Let E₁ be the event of getting a one digit-number.

Out of these numbers, one-digit number are 6,7,8 and 9

The number of favorable outcomes = 4

Therefore P(getting a one-digit number) = P(E₁) = \(\frac{number\, of \, outcomes\,favorable\,to\,E_1}{number \,of\, all\,possible\,outcomes}\) = \(\frac{4}{65}\)

Thus, the probability that the card bears a one-digit number is  \(\frac{4}{65}\).

(ii) Let E₁ be the event of getting a number divisible by 5.

Out of these numbers, number divisible by 5 are 10,15,20,......,70.

Given number 10,15,20,.......,70 form an AP with a = 10 and d = 5

Let T_n = 70. Then,

10 + (n - 1)5 = 70

⇒ 10 + 5n - 5 = 70

⇒ 5n = 65

⇒ n = 13

Thus, number of favorable outcomes = 13.

Therefore P(getting a number divisible by 5) = P(E₂) = \(\frac{number\, of \, outcomes\,favorable\,to\,E_2}{number \,of\, all\,possible\,outcomes}\) = \(\frac{13}{65}\) = \(\frac{1}{5}\)

Thus, the probability that the card bears a number divisible by 5 is \(\frac{1}{5}\).

(iii) Let E₃ be the event of getting an odd number less then 30.

Out of these numbers, odd number less then 30 are 7,9,11,......,29.

Given number 7,9,11,.......,29 form an AP with a = 7 and d = 2.

Let T_n = 29. Then,

7 + (n - 1)2 = 29

⇒ 7 + 2n - 2 = 29

⇒ 2n = 24

⇒ n = 12

Thus, number of favorable outcomes = 12.

Therefore P(getting a odd number less than 30) = P(E₃) = \(\frac{number\, of \, outcomes\,favorable\,to\,E_3}{number \,of\, all\,possible\,outcomes}\) = \(\frac{12}{65}\)

Thus, the probability that the card bears a odd number less than 30 is  \(\frac{12}{65}\).

(iv) Let E₄ be the event of getting a composite number between 50 and 70.

Out of these numbers, composite number between 50 and 70 are 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68 and 69.

Thus, number of favorable outcomes = 15.

Therefore P(getting a composite number between 50 and 70) = P(E₃) = \(\frac{number\, of \, outcomes\,favorable\,to\,E_3}{number \,of\, all\,possible\,outcomes}\) = \(\frac{15}{65}\) = \(\frac{3}{13}\) 

Thus, the probability that the card bears a composite number between 50 and 70 is \(\frac{3}{13}\).

Tickets numbered 2, 3, 4, 5, …, 100, 101

Total number of tickets = 100

(i) Even numbers from 2 to 101 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100

Total number of even number = 50

P(getting a even number) = 50/100 = 1/2.

(ii) Numbers less than 16 from 2 to 101 = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

Total number of numbers less than 16 = 14

P(getting a number less than 16) = 14/100 = 7/50.

(iii) Perfect square from 2 to 101 = 4, 9, 16, 25, 36, 49, 64, 81, 100

Total number of perfect squares = 9

P(getting a perfect square) = 9/100.

(iv) Prime numbers less than 40 from 2 to 101 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Total number of prime numbers = 12

P(getting a prime number less 40) = 12/100 = 3/25.

All possible outcomes are 2, 3, 4, 5,......, 101

Number of all possible outcomes = 100

(i) Out of these the numbers that are even = 2, 4, 6, 8,......,100

Let E₁ be the event of getting an even number.

Then, number of favorable outcomes = 50

T_n = 100 

⇒ 2 + (n - 1) x 2 = 100,

⇒ n = 50

therefore, P(getting an even number) = \(\frac{50}{100}\) = \(\frac{1}{2}\)

(ii) Out of these, number that are less than 16 = 2, 3, 4, 5,.......,15.

Let E₂ be the event of getting a number less than 16.

Then, number of favorable outcomes = 14

 therefore, P(getting a number less than 16) = \(\frac{14}{100}\) = \(\frac{7}{50}\)

(iii) Out of these, number that are perfect square = 4, 9, 16, 25, 36, 49, 64, 81 and 100

Let E₃ be the event of getting a number that is a perfect square.

Then, number of favorable outcomes = 9

 therefore, P(getting a number that is a perfect square) = \(\frac{9}{100}\)

(iv) Out of these, prime number less than 40 = 2,3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37.

Let E₄ be the event of getting a prime number less than 40.

Then, number of favorable outcomes = 12

 Therefore, P(getting a prime number less than 40) = \(\frac{12}{100}\) = \(\frac{3}{25}\)

Total number of discs in the box = 80

Perfect squares from 1 to 80 = 1, 4, 9, 16, 25, 36, 49 and 64.

Possible outcomes = 8

P(getting a number which is a perfect square) = 8/80 = 1/10.

The total number of outcomes = 80.

Let E₁ be the event of getting a perfect square number.

Out of these number perfect square number are 1, 4, 9, 16, 25, 36, 49, and 64.

Thus, the number of favorable outcomes = 8

Therefore, P(getting a perfect square number) = P(E₁) = \(\frac{number\, of \,outcomes\,favorable\,to\,E_1}{number\,of \,all\,possible\,outcomes} \) = \(\frac{8}{80}\) = \(\frac{1}{10}\).

Thus, the probability  tht the disc bears a perfect square number is \(\frac{1}{10}\).

It is given that.

P(getting a red ball) = \(\frac{1}{4}\) and P(getting a blue ball) = \(\frac{1}{3}\)

Let P(getting an orange ball) be x.

since, there are only 3 types of balls in the jar, the sum of probability of all the balls must be 1.

Therefore, \(\frac{1}{4}\) + \(\frac{1}{3}\) + x = 1

⇒ x = 1 - \(\frac{1}{4}\) - \(\frac{1}{3}\)

⇒ x = \(\frac{12-3-4}{12}\)

⇒ x = \(\frac{5}{12}\)

P(getting an orange ball) = \(\frac{5}{12}\).

 Let the total number of balls in the jar be n.

Therefore,  P(getting an orange ball) = \(\frac{10}{n}\)

⇒ \(\frac{10}{n}\) = \(\frac{5}{12}\)

⇒ n = 24

Thus, the total number of balls in the jar is 24.

Number of 50-p coins = 100.

Number of 1 coins = 70.

Number of 2 coins = 50.

Number of 5 coins = 30.

Thus, total number of outcomes = 250.

(i) Let E₁ be the event of getting a Rs. 1 coin.

The number of favorable outcomes = 70.

Therefore, P(getting a Rs. 1 coin) = P(E₁) = \(\frac{number\, of \,outcomes\,favorable\,to\,E_1}{number\,of \,all\,possible\,outcomes} \) = \(\frac{70}{250}\) = \(\frac{7}{25}\)

Thus, the probability that the coin will be a Rs. 1 coin is \(\frac{7}{25}\)

(ii) Let E₂ be the event of not getting a Rs. 5

Number of favorable outcomes = 250 - 30 = 220

Therefore, P( not getting a Rs. 5 coin) = P(E₂) = \(\frac{number\, of \,outcomes\,favorable\,to\,E_2}{number\,of \,all\,possible\,outcomes} \) = \(\frac{220}{250}\) = \(\frac{22}{25}\)

Thus, probability that the coin will not be a Rs. 5 coin is \(\frac{22}{25}\).

(iii) Let E₃ be the event of getting a 50-p or a Rs 2 coins.

Number of favorable outcomes = 100 + 50 = 150

Therefore, P(getting a a 50-p or a Rs 2 coin)  P(E₃) = \(\frac{number\, of \,outcomes\,favorable\,to\,E_3}{number\,of \,all\,possible\,outcomes} \) = \(\frac{150}{250}\) = \(\frac{3}{5}\)

Thus, probability that the coin will be a 50-p or a Rs 2 coin is \(\frac{3}{5}\).

Total number of discs in the box = 90

(a) Two digit numbers from 1 to 90 = 81

P(getting a two-digit number) = 81/90 = 9/10

(b) Numbers divisible by 5 from 1 to 90 = 5, 10, …. , 90,

Total numbers = 18

P(getting a number which is divisible by 5) = 18/90 = 1/5

Correct option is: B) 91%

We have 

P (E) = 0.09

Then P (not E) = 1 - P (E) = 1-0.09 = 0.91 = \((\frac {91}{100} \times 100)\%\) = 91 %

Hence, P (not E) in percentage is 91 %.

Correct option is: B) 91%

Correct option is: D) \(1.\overline 3 \)

Probability of any event always lie between 0 and 1.

\(\because\) \(1.\overline 3 > 1\) 

\(\therefore\) \(1.\overline 3 \) can not be the probability of any event.

Correct option is: D) \(1.\overline{3}\)

Correct option is: C) 0.35

\(\because\) E and \(\overline E\) are complementary events.

\(\therefore\) P(E) + P(\(\overline E\)) = 1

\(\Rightarrow\) P(E) =  1- P(\(\overline E\))

= 1- 0.65 (\(\because\) P (E) = 0.65)

= 0.35

Correct option is: C) 0.35

A = Event that in the first trial tall is obtained.
B = Event that in the second trial tail is obtained on the coin.

∴ P(A) = \(\frac{1}{2}\) P(B) = \(\frac{1}{2}\)

C = Event that in the first two trials tail is obtained on the coin.

∴ P(C) = P(A) ∙ P(B)

= \(\frac{1}{2}×\frac{1}{2}\)

= \(\frac{1}{4}\)

D|C = Event that tall appears on the in all three trials if the first two trials resulted in tail

∴ P(D|C) = \(\frac{P(D∩C)}{P(C)}\)

D ∩ C = Event that in the first two trails and In the third trial tail is obtained on the coin

∴ P(D ∩ C) = P(D) ∙ P(C)

= \(\frac{1}{2}×\frac{1}{4}\)

= \(\frac{1}{8}\)

Now, P(D|C) = \(\frac{P(D∩C)}{P(C)} = \frac{\frac{1}{8}}{\frac{1}{4}}\)

= \(\frac{1}{8}×\frac{4}{1}=\frac{1}{2}\)

Events A, B and C are independent events.

P(A) = P(B) = P(C) = p

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P (A ∩ B) – P (A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)

= P(A) + P(B) + P(C) – P(A) ∙ P(B) – P(A) ∙ P(C) – P(B) ∙ P(C) + P(A) ∙ P(B) ∙ P(C)

= p + p + p – (p × p) – (p × p) – (p × p) + (p × p × p)

= 3p – p² – p² – p² + p³

= 3p – 3p² + p³

= p (3 – 3p + p²)

Correct option is: C) 3

Total balls is 12 in which x are black.

\(\therefore\) Probability of drawing a black ball = \(\frac x{12}\) 

After putting 6 more black balls.

Total balls = 12 + 6 = 18

and total black balls = x + 6

\(\therefore\) Probability of drawing a black balls after putting 6 more black balls = \(\frac {x+6}{18}\) 

According to given conditions, we have

\(\frac {x+6}{18}\) = \(\frac {2x}{12}\)

\(\Rightarrow\) 12 (x + 6) = 36x (By cross multiplication)

\(\Rightarrow\) 12x + 72 = 36x

\(\Rightarrow\) 36x -12x = 72

\(\Rightarrow\) 24x = 72

\(\Rightarrow\) x = \(\frac {72}{24}= 3\)

Hence, the value of x is 3.

Correct option is: C) 3

Correct option is: A) 1

\(\because\) P is probability of any event of random experiment.

\(\therefore\) 0 \(\leq\) P \(\leq\) 1 

But given that \(K_1\) \(\leq\) P \(\leq\) \(K_2\)

\(\therefore\) 0 \(\leq\) \(K_1\) \(\leq\) P \(\leq\)  \(K_2\) \(\leq\) 1

\(\therefore\) Possible maximum value of \(K_2\) is 1 

and possible least value of \(K_1\) is 0.

\(\therefore\) Possible maximum \(K_2\) + possible least value of \(K_1\)

= 1 + 0 = 1

Correct option is: A) 1

A = Event that the tax-limit for income of males increases.

B = Event that the tax-limit for income of females increases.

A ∩ B = Event that the tax-limit for income of male and female increases.

Here, P(A) = 0.66, P(B) = 0.72 and P(A ∩ B) =0.47 are given.

(i) C = Event that the tax-limit increases only for one of the two male and female.
Event C occurs in following two ways:

A ∩ B’ = Event that the tax-limit increases for the income of male only and not of female OR

A’ ∩ B = Event that the tax-limit increases for the income of female only and not of male Events A ∩ B’ and A’ ∩ B are mutually exclusive,

∴ P(C) = P(A ∩ B’) + P(A’ ∩ B)

= [P(A) – P(A ∩ B)l + (P(B) – P(A ∩ B)]

= [0.66 – 0.47] + [0.72 – 0.47]

= 0.19 + 0.25 = 0.44

(ii) A’ ∩ B’ = Event that the tax-limit does not increase for income of either male or female.

Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.66 + 0.72 – 0.47 = 0.91

∴ P(A’ ∩ B’) = P(A ∪ B)’ = 1 – P(A ∪ B)

= 1 – 0.91 = 0.09

The event which depends on chance is called a random event. For example, head is obtained on tossing a balanced coin, a card drawn from a pack of 52 cards is king of spade.

A = Event that the price of petrol rises

∴ P(A) = \(\frac{80}{100}\)

B = Event that the price of diesel rises

∴ P(B) = \(\frac{77}{100}\)

A ∩ B = Event that the prices of both petrol and diesel rises

∴ P(A ∩ B) = \(\frac{68}{100}\)

B|A = Event that the price of diesel rises knowing that the price of petrol rises

∴ P(B|A) = \(\frac{P(A∩B)}{P(A)}\)

\(=\cfrac{\frac{68}{100}}{\frac{80}{100}}\)

\(= \frac{68}{100}×\frac{100}{80}=\frac{17}{20}\)

The experiment which can be independently repeated under identical conditions and all its possible outcomes are known but it cannot be predicted with certainty which of the outcome will appear is called a random experiment.

A = Event that it rains on Thursday
B = Event that it rains on Friday
C = Event that it rains on Saturday

Here, P(A) = 0.8, P(B) = 0.7 and P(C) = 0.6 are given.
A, B and C are independent events.

∴ P(A ∩B) = P(A) ∙ P(B) = 0.8 × 0.7 = 0.56

P(A ∩ C) = P(A) ∙ P(C) = 0.8 × 0.6 = 0.48

P(B ∩ C) = P(B) ∙ P(C) = 0.7 × 0.6 = 0.42

P(A ∩ B ∩ C) = P(A) ∙ P(B) ∙ P(C)

= 0.8 × 0.7 × 0.6 = 0.336

Now, A ∪ B ∪ C = Event that it rains on at least one of the three days in the next week

According to the law of addition of probability,

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) –

P(B ∩ C) + P(A ∩ B ∩ C)

= 0.8 + 0.7 + 0.6 – 0.56 – 0.48 – 0.42 + 0.336

= 0.976

• It can be repeated under identical conditions.
• Its all possible outcomes are known.
• It cannot be predicted which of the outcome will appear at the end of experiment.
• It results into a certain outcome.

The set of all possible outcomes of a random experiment is called a sample space of that random experiment. It is denoted by ‘U’ or ‘S’.

• Finite Sample Space: A sample space which has countable or finite number of elements is called a finite sample space. “For example, sample space obtained by throwing a balanced die.
• Infinite Sample Space: A sample space which does not have countable elements, i. e., if sample space is not finite then that sample space is called an infinite sample space. For example, the sample space of an experiment of drawing a card from the pack of 52 cards until the ace of heart is not obtained.

Any subset of the sample space of a random experiment is called an event. It is denoted by letters A, B, C,…

1. Impossible Event:

• A special subset 0 or { } of the sample space U of any random experiment is called an impossible event.
• For example, an event of getting both H and T in a single toss of an unbiased coin is an impossible event.

2. Certain Event:

• A special subset U of the sample space U of any random experiment is called a certain event.
• For example, an event of getting H or T in a single toss of an unbiased coin is a certain event.

3. Intersection of two events A and B: If A and B be any two events of the finite sample space U, then the event that ‘event A and event B both occur simultaneously’ is called the intersection of events A and B. It is denoted by the symbol A ∩ B.
Thus, A ∩ B = {x; x ∈ A and x ∈ B}

4. Union of two events A and B: If A and B be any two events of the finite sample space U, then the event that ‘either event A or B or both occur together is called the union of two events A and B. It is denoted by the symbol A ∪ B. Thus, A ∪ B = {x; x ∈ A or x ∈ B or x ∈ A ∩ B}

5. Complementary Event: If A be an event of the finite sample space U, then, the event that A does not occur is defined as the set of those elements (or outcomes) of sample space U, which are not in A is called the complementary event of A. It is denoted by the symbol A’, A̅ or Ac.
Thus, A’ = {x; x ∉ A, x ∉ U}

6. Mutually Exclusive Events: If A and B be any two events of a finite sample space; U, then the event that ‘events A and B; cannot occur together, i.e., if A ∩ B = Φ, the events A and B are said to be mutually exclusive events.

7. Difference Events: If A and B be any: two events of the finite sample space U, then the set of all those elements of U, which belong to event A but do not belong to event B is called the difference event of A and B. It is denoted by the symbol A – B or A ∩ B’. Similarly, the set of all those elements of U which belong to event B but do not belong to event A is called the difference event of B and A. It is denoted by the symbol B – A or B ∩ A’.
Thus, A – B = {x; x ∈ A and x ∉ B}
B – A = {x; x ∈ B and x ∉ A}

8. Exhaustive Events: If U is a sample space and A and B are any two events and A ∪ B = U, then events A and B are said to be exhaustive events.

9. Mutually Exclusive and Exhaustive Events: If A and B be any two events of a finite sample space U such that A ∪ B = U and A ∩ B = Φ then A and B are said to be mutually exclusive and exhaustive events.

10. Elementary Events: The events consisting of only a single element of a sample space U are called elementary events. The elementary events are mutually exclusive and exhaustive events.

11. Equi-probable Events: If there is no apparent reason to believe that out of one or more events of a random experiment, any one event is more or less likely to occur them the other events, then the events are called equi-probable events.

12. Favourable Outcomes:

If some elementary outcomes out of all the elementary outcomes of a random experiment indicate the occurence of an event A, then these outcomes are said to be favourable to the occurence of the event A.

The characteristics of random experiment are as follows :

• It can be repeated under identical conditions,
• Its all possible outcomes are known,
• It cannot be predicted with certainty which outcome will appear and
• It results into a certain outcome.

Suppose U is a sample space of a random experiment and the total number of exhaustive, mutually exclusive and equiprobable elementary outcomes of it are n. If m (0 ≤ m ≤ n) outcomes out of n outcomes are favourable to the occurrence of an event A then, the probability of the event A is defined as m/n. If we denote the probability of event A by the symbol P (A) then,

\(P (A) = \frac{m}n. Thus, P (A) = = \frac{Number \,of\, favourable \,outcomes\, of \,event A}{ Total \,number \,of \,mutually exclusive, exhaustive \,and equiprobable \,outcomes \,of \,sample \,space }\)

A and B are any two events of a finite sample space U. The probability that at least one of the two events A and B will occur is called the law of addition of probability. This rule is written as under :
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

• If A ∩ B = Φ, then law of addition of probability is written as under:
  P(A ∪ B) = P(A) + P(B)
• If A and B are mutually exclusive and exhaustive events, then
  P(A ∪ B) = P(A) + P(B) = 1.
• The law of addition of probability for three events A, B and C of a sample space is written as under:
  P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)
• If A, B, C are mutually exclusive events, then P (A ∪ B ∪ C) = P(A) + P(B) + P(C).
• If A, B, C are mutually exclusive and exhaustive events, then
  P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = 1.

• The number of outcomes of the sample space is finite.
• The number of all possible outcomes of the sample space is known.
• The outcomes of the sample space are equiprobable.

The assumptions of mathematical definition of probability are as follows:

• The number of outcomes in the sample space is finite.
• The number of all possible outcomes of the sample space is known.
• The outcomes of the sample space are equi-probable.

True

Probability of each student getting distinction in their final examination is less than or equal to

1, sum of the probabilities of two may be 1.2.

Hence, it is true statement.

False

A ∩B⊆ A, B

P(A ∩B) ≤ P(A), P(B)

But given that P(B) = 0.3 and P(A ∩B) = 0.4, which is not possible.

True

We know that A ∩ B ⊂ A

P(A ∩ B) ≤ P(A)

Hence, it is a true statement.

False

Given that, P(A ) = 0.5, P(A ∩B)< 0.3

Now, P(A) x P(B) ≤ 0.3

=>0.5 x P(B) ≤0.3

=> P(B) ≤0.6

Hence, it is false statement

False

Sum of these probabilities must be equal to 1.

P(0) + P( 1) + P( 2) + P(3) + P{ 4) + P(5)

= 0.12 + 0.25 + 0.36 + 0.14 + 0.08 + 0.11 = 1.06 which is greater than 1,

So, it is false statement.

False

Let A = Student will pass examination

B = Student will getting compartment

P(A) = 0.73, P(B) = 0.13 and P(A or B) = 0.96

P(A or B) = P(A) + P(B) = 0.73 + 0.13 = 0.86

But P(A or B) = 0.96

Hence, given statement is false.

False

P(to see giraffe or bear) = P (giraffe) + P (bear) – P(giraffe and bear)

=0.72 + 0.84-0.52= 1.04 which is not possible.

Answer is (B)

If M and N are any two events.

.'. P(M ∪N) = P(M) + P(N) – P(M ∩ N) .