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Correct option is (D) 50%

P(A) = Probability of winning a person = 50%  (given)

\(=\frac{50}{100}=\frac12\)

\(\therefore\) P(A') = Probability of loosing that person = 1 - P(A)

\(=1-\frac12=\frac12\)

Correct option is  D) 50%

Correct option is (D) 6

When a dice is rolled, then possible outcomes are S = {1, 2, 3, 4, 5, 6}.

Hence, total number of possible outcomes are 6.

Correct option is  D) 6

Correct option is: D) 1 – x

The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.

(c) \(\frac{52}{77}\)

Given, odds against Event 1 = 5 : 2

⇒ P(Event 1 not happening) = \(\frac{5}{5+2}\) = \(\frac{5}{7}\)

Odds in favour of Event 2 = 6 : 5

⇒ P(Event 2 happens) = \(\frac{6}{6+5}\) = \(\frac{6}{11}\)

⇒ P(Event 2 not happening) = 1 - \(\frac{6}{11}\) = \(\frac{5}{11}\)

∴ P(None of the events happen) = \(\frac{5}{7}\)x \(\frac{5}{11}\) = \(\frac{25}{77}\)

(∵ Both event are independent)

⇒ P(At least one event happens) = 1 - \(\frac{25}{77}\) = \(\frac{52}{77}\).

If there is a chance that an event will happen, then, its probability is between zero and 1.

The most the probability of an event occurring can be is 1which means the event has a 100% probability of happening. But 5/3 is greater than 1 so it can’t be the probability of an event.

The probability of an event of a trials is less than 1.

The formula for the probability of occurrence of at least one event out of three events A, B and C is as follows:

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

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

U is a finite sample space and A and B are any two events of U. The probability of event B, under the condition that event A is happened, is called conditional probability of the event B.

Total number of ways of selecting two persons from 5 persons

= ⁵C₂ = \(\frac{5×4}{2×1}\) = 10

The sample space for the random experiment of selecting two persons from five persons a, b, c, d, e is expressed as follows:

U = {(a, b), (a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, d), (c, e), (d, e)}

The sample space for the random experiment of tossing together a balance die and a balanced coin is expressed as follows :

U = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5,T), (6, T)}

Where, in each bracket the first term is the number on die and second term is the outcome on the coin.

A student can get 0 mark or any number of marks in the close interval [1, 100].

So, the sample space for the marks scored by a student appearing for an examination of 100 marks is expressed as follows :

U = {0, 1, 2, 3, ……………, 100}

The number of sample points in the sample space U = 101.

The sample space of a random experiment of throwing one balanced die and a balanced coin simultaneously is obtained as follows:

U = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}

Where, H = Head; T = Tail;

1, 2, 3, 4, 5, 6 = Numbers on die.

Suppose, four persons are denoted by a, b, c and d. Firstly, one minister can be selected in ⁴C₁ ways than secondly one deputy minister can be selected in ³C₁ ways.

∴ Total number of selections = ⁴C₁ × ³C₁
= 4 × 3
= 12

Hence, the sample space for randomly selecting one minister and one deputy minister from four persons is expressed as follows:

U = {(a, b), (a, c), (a, d), (b, a), (b, c), (b, d), (c, a), (c, b), (c, d), (d, a), (d, b), (d, c)}.

Where, in each bracket, first place indicates minister and second place indicates deputy minister.

A balanced coin is thrown till the first head is obtained.

The options of this random experiment are as follows :

• Head (H) is obtained in the first trial, i.e., outcome H is obtained.
• Tail (T) is obtained in the first trial and Head (H) is obtained in second trial, i.e., outcome TH is obtained.
• Tail (T) is obtained in first two trials and Head (H) is obtained in third trial, i.e., outcome TTH is obtained.
• Tail (T) is obtained in first three trials and Head (H) is obtained in fourth trial, i.e., outcome TTTH is obtained.

In this manner outcomes TTTTH, TTTTTH, … are obtained. Thus, the number of outcomes of this random experiment is indefinite.

Hence, the sample space of the random experiment of throwing a balanced coin till the first head is obtained, is expressed as follows:

U = (H, TH, TTH, TTTH, …}

The sample space of this experiment is infinite.

Here, U = {1, 2, 3, …….., 20}

(1) A = Event that the selected number is odd number.

∴ A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

(2) B = Event that the selected number is divisible by 3

∴ B = {3, 6, 9, 12, 15, 18}

(3) C = Event that the selected number is divisible by 2 or 3

∴ C = (2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20}

First five natural numbers are 1, 2, 3, 4, 5. The number of ways of selecting three numbers from these five numbers is ⁵C₃ =\( \frac{5×4×3}{3×2×1}\) = 10.

Hence, the sample space for the experiment of randomly selecting three numbers from the first five natural numbers is expressed as follows:

U = {(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5), (1, 4, 5), (2, 3, 4), (2, 3, 5), (2, 4, 5), (3, 4, 5)}.

Given: There are three coins tossed once. 

To Find: Describe the events according to the subparts? 

Explanation: when three coins are tossed, then the sample spaces are: 

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

According to the question, 

A = {HHH} 

B = {HHT, HTH, THH} 

C = {TTT} 

D = {HHH, HHT, HTH, HTT} 

Now, A∩B = ϕ,A∩C = ϕ,A∩D = {HHH}

B∩C = ϕ,B∩D = {HHT,HTH},C∩D = ϕ

Since, If the intersection of two sets are null or empty it means both the sets are Mutually Exclusive. 

(i) Events A and B, Events A and C, Events B and C and events C and D are mutually exclusive. 

(ii) Here, We know, If an event has only one sample point of a sample space, then it is called elementary events. 

So, A and C are elementary events. 

(iii) If There is an event that has more than one sample point of a sample space, it is called a compound event,

Since, B∩D = {HHT,HTH}

So, B and D are compound events.

Given: A dice is thrown twice. And each time number appearing on it is recorded.

When the dice is thrown twice then the number of sample spaces are 6² = 36

Now,

The possibility both odd numbers are:

A = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}

Since, possibility of both even numbers is:

B = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}

And, possible outcome of sum of the numbers is less than 6.

C = {(1, 1)(1, 2)(1, 3)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)}

Hence,

(AՍB) = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5) (2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6)}

(AՌB) = {Փ}

(AUC) = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5) (1, 2)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)}

(AՌC) = {(1, 1), (1, 3), (3, 1)}

∴ (AՌB) = Փ and (AՌC) ≠ Փ, A and B are mutually exclusive, but A and C are not.

Here, U = {1, 2, 3, …, 100}

One number is selected at random.

∴ Total number of primary outcomes

n = ¹⁰⁰C₁ = 100A = Event that the number selected is divisible by 7.

A = {7, 14, 21, 28, …, 91, 98}

∴ Favourable outcomes for the event A is m = 14.

Hence, P(A) = \(\frac{m}{n} = \frac{14}{100} = \frac{7}{50}\)

Given: A dice is thrown twice. And each time number appearing on it is recorded. 

To Find: Describe the given events. 

Explanation: when the dice is thrown twice then the number of sample spaces are 6 2 = 36 

Now, 

The possibility both odd numbers are: 

A = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)} 

Since, Possibility of both even numbers are: 

B = {(2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6)} 

And, Possible outcome of sum of the numbers is less than 6 

C = {(1, 1)(1, 2)(1, 3)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)} 

Therefore, 

(AՍB) = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5) (2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6)} 

(AՌB) = {Փ} 

(AUC) = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5) (1, 2)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)} 

(AՌC) = {(1, 1)(1, 3)(3, 1)} 

Hence, (AՌB) = Փ and (AՌC)≠Փ, A and B are mutually exclusive, but A and C are not.

Given: A dice is thrown once. 

To Find: Find the given events, and also find the Also, find A ∪ B, A ∩ B, B ∩ C, E ∩ F, D ∩ F and \(\bar{F}\)

Explanation: In a single throw of a die, the possible events are: 

S = {1, 2, 3, 4, 5, 6} 

Now, According to the subparts of the question, we have certain events as: 

(i) A = getting a number below 7 

So, The sample spaces for A are: 

A = {1, 2, 3, 4, 5, 6} 

(ii) B = Getting a number greater than 7 

So, the sample spaces for B are: 

B = {Փ} 

(iii) C = Getting multiple of 3 

So, The Sample space of C is 

C = {3, 6} 

(iv) D = Getting a number less than 4 

So, The sample space for D is 

D = {1, 2, 3}

(v) E = Getting an even number greater than 4. 

The sample space for E is 

E = {6} 

(vi) F = Getting a number not less than 3. 

The sample space for F is 

⇒ F = {3, 4, 5, 6} 

Now, 

A = {1, 2, 3, 4, 5, 6} and B = {Փ}

⇒ A∪B = {1, 2, 3, 4, 5, 6}

A = {1, 2, 3, 4, 5, 6} and B = {Փ}

⇒ A∩B = {ϕ}

B = {Փ} and C = {3, 6}

⇒ B∩D = {ϕ}

F = {3, 4, 5, 6}and E = {6}

⇒ E∩F = {6}

E = {6} and D = {1, 2, 3}

⇒ D∩F = {3}

And, For \(\bar{F}\) = S - F

S = {1, 2, 3, 4, 5, 6} and F = {3, 4, 5, 6}

⇒ \(\bar{F}\) = {1,2}

Hence, These are the events for given ecperiment

Let 

(i) 

A: ‘atleast two heads appear’ 

B: ‘atleast two tails appear’ 

Here, A = {HHT, HTH, THH, HHH} 

B = {TTH, THT, HTT, TTT} 

∴ A ∩ B = ϕ , ⇒ A, B mutually exclusive. 

(Note: Give any suitable example (events) associated with experiment) 

(ii) Let A : ‘No tail appears’ 

B: ‘Exactly one tail appears’ 

C: ‘Atleast two tails appear’ 

Here A = {HHH] 

B = {THH, HTH, HHT} 

C = {TTH, THT, HTT, TTT} 

Clearly, A ∪ B ∪ C = sample space = S and A ∩ B = ϕ , B ∩ C = ϕ, A ∩ C 

∴ A, B, C are mutually exclusive and mutually exclusive and exhaustive events.

(iii) Let 

A: ‘almost two tails appear’ 

B: ‘exactly two tails appear’ 

Here, 

A = {HHH, THH, HTH, HHT, HTT, THT, TTH} 

B = {HTT,THT,TTH} 

Clearly, A ∩ B ≠ ϕ

∴ A and B are not mutually exclusive 

(iv) ‘Getting exactly one head’ and ‘Getting exactly two heads’. 

(v) ‘Getting exactly one tail’, ‘Getting exactly two tails’ and ‘Getting exactly three tails’.

Given: Three coins are tossed.

When three coins are tossed, then the sample space is

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

Now, the subparts are:

(i) The two events which are mutually exclusive are when,

A: getting no tails

B: getting no heads

Then, A = {HHH} and B = {TTT}

So, the intersection of this set will be null. Or, the sets are disjoint.

(ii) Three events which are mutually exclusive and exhaustive are:

A: getting no heads

B: getting exactly one head

C: getting at least two head

So, A = {TTT} B = {TTH, THT, HTT} and C = {HHH, HHT, HTH, THH}

Since, A ⋃ B = B ⋂ C = C ⋂ A = Փ and

A⋃ B⋃ C = S  

(iii) The two events that are not mutually exclusive are:

A: getting three heads

B: getting at least 2 heads

So, A = {HHH} B = {HHH, HHT, HTH, THH}

Hence, A ⋂ B = {HHH} = Փ

(iv) The two events which are mutually exclusive but not exhaustive are:

A: getting exactly one head

B: getting exactly one tail

So, A = {HTT, THT, TTH} and B = {HHT, HTH, THH}

It is because A ⋂ B = Փ but A⋃ B ≠ S 

We take, B = Boy, G = Girl

∴ The sample space for the families having two children is expressed as follows:

U = {BB, BG, GB, GG}

Now, the total number of primary outcomes of the sample space of selecting a family at random is n = ⁴C₁ = 4.

(1) A = Event that one child is a girl and one child is a boy. = {BG, GB}

∴ Favourable outcomes for the event A is m = 2.

Hence, P(A) = \(\frac{m}{n} = \frac{2}{4} = \frac{1}{2}\)

(2) B = Event that at least one child is a girl among two children of the selected family.
= {GB, BG, GG}

∴ Favourable outcomes for the event B is m = 3.

Hence, P(B) = \(\frac{m}{n} = \frac{3}{4}\)

When three coins are tossed, then the sample space is 

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

Now, The subparts are: 

(i) The two events which are mutually exclusive are when, 

A: getting no tails 

B: getting no heads 

Then, A = {HHH} and B = {TTT} 

SO, The intersection of this set will be null. 

Or, The sets are disjoint. 

(ii) Three events which are mutually exclusive and exhaustive are: 

A: getting no heads 

B: getting exactly one head 

C:getting at least two head 

So, A = {TTT} B = {TTH, THT, HTT} and C = {HHH, HHT, HTH, THH} 

Since, A∪B = B∩C = C∩A = Փ and A∪B∪C = S

(iii) The two events that are not mutually exclusive : 

A:getting three heads 

B:getting at least 2 heads 

So, A = {HHH} B = {HHH, HHT, HTH, THH} 

Since A∩B = {HHH} ≠ Փ

(iv) The two events which are mutually exclusive but not exhaustive are: 

A:getting exactly one head

B: getting exactly one tail 

So, A = {HTT, THT, TTH} and B = {HHT, HTH, THH} 

It is because A∩B = Փ but A∩B ≠ s

Two balanced dice are thrown simultaneously. So the sample space for this random experiment Is expressed as follows:

U ((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)}

∴ Total number of primary outcomes of U is
n = 36

(1) A = Event that the sum of numbers on the dice is 6.
A = {( 1, 5), (2. 4), (3, 3), (4, 2). (5. 1))

∴ Favourable outcomes for the event A is m = 5.

Hence, P(A) = \(\frac{m}{n} = \frac{5}{36}\)

(2) B = Event that the sum of numbers on the dice is more than 10.

∴ B’ = Event that the sum of numbers on the dice is not more than 10.

B = {(5, 6), (6, 5), (6, 6)}

∴ Favourable outcomes for the event B is m = 3.

Hence P(B) = \(\frac{m}{n} = \frac{3}{36} = \frac{1}{12}\)

∴ P(B’) = 1 – P(B)

= 1 – \(\frac{1}{12} = \frac{11}{12}\)

(3) C = Event that the sum of numbers on the dice is a multiple of 3, i.e., 3, 6, 9 or 12

C = {(1, 2), (2, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}

∴ Favourable outcomes for the event C is m = 12

Hence, P(C) =\( \frac{m}{n} = \frac{12}{36} = \frac{1}{3}\)

(4) D = Event that the product of the numbers on the dice is 12
D = {(2, 6), (3, 4), (4, 3), (6, 2)}

∴ Favourable outcomes for the event D is m = 4

Hence, P(D) =\( \frac{m}{n} = \frac{4}{36} = \frac{1}{9}\)

When three coins are tossed, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(i) Two events that are mutually exclusive can be
A: getting no heads and B: getting no tails
This is because sets
A = {TTT} and B = {HHH} are disjoint.
(ii) Three events that are mutually exclusive and exhaustive can be
A: getting no heads
B: getting exactly one head C: getting at least two heads
i.e.,
A = {TTT}

B = {HTT, THT, TTH}
C = {HHH, HHT, HTH, THH}
This is because A ∩ B = B ∩ C = C ∩ A = Φ and A U B U C = S
(iii) Two events that are not mutually exclusive can be
A: getting three heads B: getting at least 2 heads
i.e.,
A = {HHH}
B = {HHH, HHT, HTH, THH}
This is because A ∩ B = {HHH} ≠ Φ
(iv) Two events which are mutually exclusive but not exhaustive can be
A: getting exactly one head
B: getting exactly one tail That is
A = {HTT, THT, TTH}
B = {HHT, HTH, THH}
It is because, A ∩ B =Φ, but A B ≠ S
(v) Three events that are mutually exclusive but not exhaustive can be
A: getting exactly three heads
B: getting one head and two tails
C: getting one tail and two heads
i.e.,
A = {HHH}
B = {HTT, THT, TTH}
C = {HHT, HTH, THH}
This is because A ∩ B = B ∩ C = C ∩ A = Φ, but A U B U C ≠ S