Question 1: A coin is tossed. Find the total number of elementary events and also the total number events associated with the random experiment.

Answer:

Note: If there an n elements in a set, then the number of its subset is 2^n .

When a coin is tossed, there will be two possible outcomes, Head (H) and Tail (T).

Sample space \displaystyle S =  \{ H, T \}

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Question 2: List all events associated with the random experiment of tossing of two coins. How many of them are elementary events?

Answer:

We know, when two coins are tossed then the no. of possible outcomes are \displaystyle 2^2 = 4

So, the Sample spaces are \displaystyle S = \{ HH, HT, TT, TH \}

Therefore there are total 4 events associated with the given experiment.

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Question 3: Three coins are tossed once. Describe the following events associated with this random experiment.

A = Getting three heads,      B = Getting two heads and one tail,

C = Getting three tails,         D = Getting a head on the first coin.

(i) Which pairs of events are mutually exclusive?

(ii) Which events are elementary events?

(iii) Which events are compound events?

Answer:

When three coins are tossed, then the sample spaces are: S = \{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \}

So, according to the question,

A = \{ HHH \}

B = \{ HHT, HTH, THH \}

C = \{  TTT \}

D = \{ HHH, HHT, HTH, HTT \}

Now,

A \cap B = \varnothing

A \cap C = \varnothing

A \cap D = \{ HHH \}

B \cap C = \varnothing

B \cap D = \{ HHT, HTH \}

C \cap D = \varnothing

We know that, if the intersection of two sets are null or empty it means both the sets are Mutually Exclusive.

We know that, 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 \cap D = \{ HHT, HTH \} So, B and D are compound events.

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Question 4: In a single throw of a die describe the following events:

(i) A – Getting a number less than 7          (ii) B – Getting a number greater than 7

(iii) C – Getting a multiple of 3          (iv) D – Getting a number less than 4

(v) E=Getting an even number greater than 4 (vi) F=Getting a number not less than 3.

Also find A \cup B, A \cap B, B \cap C, E \cap F, D \cap F \text{  and  }  \overline{F}.

Answer:

A dice is thrown once. Therefore S = \{ 1, 2, 3, 4, 5, 6 \}

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 = \{ \varnothing \}

(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. So, the sample space for E is E = \{ 6 \}

(vi) F = Getting a number not less than 3. So, the sample space for F is F = \{ 3, 4, 5, 6 \}

Now,

A = \{1, 2, 3, 4, 5, 6\} \text{ and }  B = \{ \varnothing \} \Rightarrow A \cup B = \{1, 2, 3, 4, 5, 6\}

A = \{1, 2, 3, 4, 5, 6\} \text{ and }  B = \{ \varnothing \} \Rightarrow A \cap B = \{ \varnothing \}

B = \{ \varnothing \} \text{ and }  C = \{3, 6\} \Rightarrow B \cap C = \{ \varnothing \}

F = \{3, 4, 5, 6\} \text{ and }  E = \{6\} \Rightarrow E \cap F = \{6\}

E = \{6\} \text{ and }  D = \{1, 2, 3\} \Rightarrow D \cap F = \{3\}

\overline{F} = S - F = \{1, 2, 3, 4, 5, 6\} - \{ 3, 4, 5, 6 \} = \{ 1, 2 \}

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Question 5: Three coins are tossed. Describe:

(i) two events A and B which are mutually exclusive.

(ii) three events A, B and C which are mutually exclusive and exhaustive.

(iii) two events A and B which are not mutually exclusive.

(iv) two events A and B which are mutually exclusive but not exhaustive.

Answer:

When three coins are tossed, then the sample space is

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

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

A: getting no tails

B: getting no heads

Then, A = \{HHH\} \text{ 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\} \text{ and }  C = \{HHH, HHT, HTH, THH\} 

Since, A \cup B = B \cap C = C \cap A = \varnothing \text{ and }  A \cup B \cup 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 \cap B = \{HHH\} = \varnothing

(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\} \text{ and }  B = \{HHT, HTH, THH\}

It is because A \cap B = \varnothing \text{ but } A \cup B \neq S  

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Question 6: A die is thrown twice. Each time the number appearing on it is recorded. Describe the following events:

(i) A = Both numbers are odd.

(ii) B – Both numbers are even.

(iii) C =sum of the numbers is less than 6

Also, find A \cup B, A \cap B , A \cup C, A \cap C. Which pairs of events are mutually exclusive?

Answer:

When the dice is thrown twice then the number of sample spaces are 6^2 = 36

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)\}

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)\}

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)\}

\text{Hence, } (A \cup B) = \{(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)\}

(A \cap B) = \{ \varnothing \}

(A \cup C) = \{(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)\}

(A \cap C) = \{(1, 1), (1, 3), (3, 1)\}

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Question 7: Two dice are thrown. The events A , B,C,D,E and F are described as follows:

A = Getting an even number on the first die.

B = Getting an odd number on the first die.

C = Getting at most 5 as sum of the numbers on the two dice.

D = Getting the sum of the numbers on the dice greater than 5 but less than 10.

E = Getting at least 10 as the sum of the numbers on the dice.

F = Getting an odd number on one of the dice.

(i) Describe the following events: A and B, B or C, B and C, A and E, A or F, A and F

(ii) State true or false:

(a) A and B are mutually exclusive,

(b) A and B are mutually exclusive and exhaustive events.

(c) A and C are mutually exclusive events.

(d) C and D are mutually exclusive and exhaustive events.

(e) C, D and E are mutually exclusive and exhaustive events.

(f) A’ and B’ are mutually exclusive events.

(g) A, B, F are mutually exclusive and exhaustive events:

Answer:

When two dice are thrown then the no. of possible outcomes are 6^2 = 36

A = Getting an even number of the first die. 

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

B = Getting an odd number on the first dice 

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

C = Getting at most 5 as the sum of numbers on the two dices. 

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

D = Getting a sum greater than 5 but less than 10 

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

E = Getting at least 10 as the sum of numbers on the dices

E = \{(4, 6),(5, 5),(5, 6),(6, 4),(6, 5),(6, 6)\}

F = Getting an odd number on one of the dices 

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

Now, 

(i) A \text{ and } B = A \cap B = \varnothing 

Since, There is no common events in Both A and B so the intersection is Null ( \varnothing ).

\Rightarrow B \text{ or } C = B \cup C

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

\Rightarrow B \text{ and } C = B \cap C

B \cap C = \{ (1, 1), (1, 2),(1, 3),(1, 4),(3, 1),(3, 2) \}

\Rightarrow A \text{ and } E = A \cap E

A \cap E = \{ (4, 6), (6, 4),(6, 5),(5, 6),(6, 6) \}

\Rightarrow A \text{ or } F = A \cup F

A \cup F = \{ (2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6),(4, 1),(4, 2), \\ (4, 3),(4, 4),(6, 1), (6, 2), (6, 3), (6, 4),(6, 5),(6, 6),(1, 2),(1, 4),(1, 6), \\ (3, 2),(3, 4),(3, 6),(4, 5),(5, 2),(5, 4),(5, 6) \}

(ii)  (a) True, because A \cap B = \varnothing

(b) True, because A \cap B = \varnothing \text{ and } A \cup B = S

(c) False, because A \cap C = \varnothing

(d) False, because C \cap D = \varnothing \text{ but } C \cup D \neq S

(e) True, because C \cap D \cap E = \varnothing \text{ and } C \cup D \cup E = S

(f) True, because A' \cap B' = \varnothing

(g) False, because A \cap B \cap F = \varnothing \text{ and } A \cup B \cup F = S

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Question 8: The numbers 1,2,3 and 4 are written separately on four slips of paper. The slips are then put in a box and mixed thoroughly. A person draws two slips from the box one after the other, without replacement. Describe the following events:

A = The number on the first slip is larger than the one on the second slip.

B = The number on the second slip is greater than 2

C = The sum of the numbers on the two slips is 6 or 7

D = The number on the second slips is twice that on the first slip.

Which pair(s) of events is (are) mutually exclusive?

Answer:

Here, Four slips of paper 1, 2, 3, and 4 are put in a box. 

If Two slips are drawn from it one after the other without replacement. 

Then The sample space for the experiment is:  S = \{(1, 2)(1, 3), (1, 4)(2, 1)(2, 3)(2, 4)(3, 1)(3, 2)(3, 4)(4, 1)(4, 2)(4, 3)\}

(i) A = number on the first slip is larger than the one on the second slip, 

So, The sample space for A is:  \{(2, 1)(3, 1)(3, 2)(4, 1)(4, 2)(4, 3)\}

(ii) B = number on the second slip is greater than 2 

So, The sample space for B is  \{(1, 3)(2, 3)(1, 4)(2, 4)(3, 4)(4, 3)\}

(iii) C = sum of the numbers on the two slips in 6 or 7 

So, The sample space for C is  \{(2, 4)(3, 4)(4, 2)(4, 3)\}

(iv) D = number on the second slip is two times the number on the first slip

So, The sample space for D is  \{(1, 2)(2, 4)\}   

Now,  We can see, A \cap D = \varnothing

Therefore, A and D are mutually exclusive events.

Hence, A and D are mutually exclusive events.

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Question 9: A card is picked up from a deck of 52 playing cards.

(i) What is the sample space of the experiment?

(ii) What is the event that the chosen card is black faced card?

Answer:

(i) If a card is picked up from a deck of 52 playing cards, then the sample space of this experiment S = 52 cards in a deck

(ii) Let E is the event, and a black face card is chosen, 

Then The possible outcomes in E are jack, queen, and king of clubs and the jack, queen and king of spades.