# Class 11 NCERT Solutions- Chapter 16 Probability – Exercise 16.2

• Difficulty Level : Easy
• Last Updated : 19 Jan, 2021

### Question 1. A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?

Solution:

Let us consider that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

Hence, Space = (1, 2, 3, 4, 5, 6)

As per the conditions given in the question

E be the event “die shows 4”

E = (4)

F be the event “die shows even number”

F = (2, 4, 6)

E ∩ F = (4) ∩ (2, 4, 6) = 4

4 ≠ ∅          -(since there is a common element in E and F)

For that reason, we can conclude that E and F are not mutually exclusive event.

### Question 2. A die is thrown. Describe the following events:

(i) A: a number less than 7

(ii) B: a number greater than 7

(iii) C: a multiple of 3

(iv) D: a number less than 4

(v) E: an even number greater than 4

(vi) F: a number not less than 3

Also find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ FI, FI

Solution:

Let us consider that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

Hence, Space = (1, 2, 3, 4, 5, 6)

As per the conditions given in the question,

(i) A: a number less than 7:

Here, all the numbers in the die are less than 7,

Hence, A = (1, 2, 3, 4, 5, 6)

(ii) B: a number greater than 7:

There is no number present which is greater than 7 on the die

That’s why, B = (∅)

(iii) C: a multiple of 3:

Two numbers are present which are multiple of 3.

That’s why, C = (3, 6)          -(As, 3×1 = 3, 3×2 = 6)

(iv) D: a number less than 4:

There are three numbers which are less than 4.

That’s why, D= (1, 2, 3)

(v) E: an even number greater than 4:

There is only one even number which is greater than 4

That’s why, E = (6)

(vi) F: a number not less than 3:

That means, we have to find the number(s) which is/are greater than 3

There are four numbers which are greater than 3.

That’s why, F = (3, 4, 5, 6)

According to the question, we also have to solve, A B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, D – E, A – C, E ∩ F’, F’

Hence,

• A ∩ B = (1, 2, 3, 4, 5, 6) ∩ (∅) = (∅)
• B U C = (∅) ∪ (3, 6) = (3, 6)
• E ∩ F = (6) ∩ (3, 4, 5, 6) = (6)
• D ∩ E = (1, 2, 3) ∩ (6) = (∅)
• D – E = (1, 2, 3) – (6) = (1, 2, 3)
• A – C = (1, 2, 3, 4, 5, 6) – (3, 6) = (1, 2, 4, 5)
• F’ = S – F = (1, 2, 3, 4, 5, 6) – (3, 4, 5, 6) = (1, 2)
• E ∩ F’ = (6) ∩ (1, 2) = (∅)

### Question 3.  An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:

A: the sum is greater than 8,

B: 2 occurs on either die

C: the sum is at least 7 and a multiple of 3.

Which pairs of these events are mutually exclusive?

Solution:

Let us consider that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

In the question, it is given that pair of dice is thrown, so sample space will be-

{(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)}

A: the sum is greater than 8:

According to the above sample space the maximum sum will be (6+6) = 12

Hence, the possible sum greater than 8 are: 9, 10, 11, and 12

Where A= {(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}

B: 2 occurs in either die:

In this case, there are three possibilities:

(i) 2 may come in the first die: B1 = {(2, 1), (2, 3), (2, 4), (2, 5), (2, 6)}

(ii) 2 may come in the second die: B2 = {(1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}

(iii) 2 may come in both the die simultaneously: B3 = {(2, 2)}

Hence, B = {B1 ∪ B2 ∪ B3} = {(2, 1), (2, 3), (2, 4), (2, 5), (2, 6),

(1, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 2)}

C: The sum is at least 7 and multiple of 3:

According to this condition the sum can be 9 or 12 [As, 9 = (3×3) & 12 = (3×4)

both are multiple of 3 and greater than 7]

Hence, C= {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}

Now, we have to find pairs of these events are mutually exclusive or not.

A = {(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}

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

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

(i) A∩ B = ∅

We find that, there is no common element in A and B

Therefore, A & B are mutually exclusive

(ii) B ∩ C = ∅

We find that, there is no common element between B and C

Therefore, B and C are mutually exclusive.

(iii) A ∩ C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}

⇒ {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)} ≠ ∅

We find that, A and C has common elements.

Therefore, A and C are mutually exclusive.

### Question 4. Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event “two heads and one tail show”, C denotes the event” three tails show, and D denote the event ‘a head shows on the first coin”. Which events are

(i) Mutually exclusive?

(ii) Simple?

(iii) Compound?

Solution:

When a coin is thrown, the possible outcomes are either Head(H) or Tail(T).

Now according to the question, three coins are tossed once so the possible sample space contains,

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

Now,

A = (HHH)

B: “two heads and one tail”

B = (HHT, THH, HTH)

C: ‘three tails’

C = (TTT)

D: a head shows on the primary coin

D = (HHH, HHT, HTH, HTT)

(i) Mutually exclusive:

A ∩ B = (HHH) ∩ (HHT, THH, HTH) = ∅

Therefore, A and B are mutually exclusive.

A ∩ C = (HHH) ∩ (TTT) = ∅

Therefore, A and C are mutually exclusive.

A ∩ D = (HHH) ∩ (HHH, HHT, HTH, HTT) = (HHH)

A ∩ D ≠ ∅

So they aren’t mutually exclusive

B ∩ C = (HHT, HTH, THH) ∩ (TTT) = ∅

Since there’s no common element in B & C, in order that they are mutually exclusive.

B ∩ D = (HHT, THH, HTH) ∩ (HHH, HHT, HTH, HTT) = (HHT, HTH)

B ∩ D ≠ ∅

We find that, there are common elements in B & D,

So, they not mutually exclusive.

C ∩ D = (TTT) ∩ (HHH, HHT, HTH, HTT) = ∅

Since there’s no common element in C & D,

So they are mutually exclusive.

(ii) Simple event

If an occasion has just one sample point of a sample space,

it’s called an easy (or elementary) event.

A = (HHH)

C = (TTT)

Both A & C have just one element,

So, they are simple events.

(iii) Compound events

If an occasion has quite one sample point, it’s called a Compound event

B = (HHT, HTH, THH)

D = (HHH, HHT, HTH, HTT)

Both B & D have quite one element,

So, they’re compound events.

### Question 5. Three coins are tossed. Describe

(i) Two events which are mutually exclusive.

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

(iii) Two events, which are not mutually exclusive.

(iv) Two events which are mutually exclusive but not exhaustive.

(v) Three events which are mutually exclusive but not exhaustive.

Solution:

When a coin is thrown, the possible outcomes are either Head(H) or Tail(T).

Now according to the question, three coins are tossed once so the possible sample space contains,

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

(i) Two events which are mutually exclusive.

Let us consider A be the event of getting only head

A = (HHH)

And also let us consider B be the event of getting only Tail

B = (TTT)

So, A ∩ B =∅

We find that, there is no common element in A & B so these two are mutually exclusive.

(ii) We find that, three events which are mutually exclusive and exhaustive

Now,

Let us consider P be the event of getting exactly two tails

P = (HTT, TTH, THT)

Let us consider Q be the event of getting at least two heads

Q = (HHT, HTH, THH, HHH)

Let us consider R be the event of getting only one tail

C= (TTT)

P ∩ Q = (HTT, TTH, THT) ∩ (HHT, HTH, THH, HHH) = ∅

We find that, there is no common element in P and Q,

Therefore, they are mutually exclusive

Q ∩ R = (HHT, HTH, THH, HHH) ∩ (TTT) = ∅

We find that, there is no common element in Q and R

Hence, they are mutually exclusive.

P ∩ R = (HTT, TTH, THT) ∩ (TTT)= ∅

We find that, there is no common element in P and R,

So they are mutually exclusive.

Hereby, P and Q, Q and R, and P and R are mutually exclusive

∴ P, Q, and R are mutually exclusive.

And also,

P ∪ Q ∪ R = (HTT, TTH, THT, HHT, HTH, THH, HHH, TTT) = Space

Hence, P, Q and R are exhaustive events.

(iii) Two events, which are not mutually exclusive

Let us consider ‘A’ be the event of getting at least two heads

A = (HHH, HHT, THH, HTH)

Let us consider ‘B’ be the event of getting only head

B = (HHH)

Now A ∩ B = (HHH, HHT, THH, HTH) ∩ (HHH) = (HHH)

A ∩ B ≠ ∅

We find that, there is a common element in A and B,

So, they are not mutually exclusive.

(iv) Here, two events which are mutually exclusive but not exhaustive

Let us consider ‘P’ be the event of getting only Head

P = (HHH)

Let us consider ‘Q’ be the event of getting only tail

Q = (TTT)

P ∩ Q = (HHH) ∩ (TTT) = ∅

We find that, there is no common element in P and Q,

Hence, these are mutually exclusive events.

But,

P ∪ Q = (HHH) ∪ (TTT)

= {HHH, TTT}

P ∪ Q ≠ Space

We find that, P ∪ Q ≠ Space, these are not exhaustive events.

(v) We find that, three events which are mutually exclusive but not exhaustive

Let us consider ‘X’ be the event of getting only head

X = (HHH)

Let us consider ‘Y’ be the event of getting only tail

Y = (TTT)

Let us consider ‘Z’ be the event of getting exactly two heads

Z = (HHT, THH, HTH)

Now,

X ∩ Y = (HHH) ∩ (TTT) = ∅

X ∩ Z = (HHH) ∩ (HHT, THH, HTH) = ∅

Y ∩ Z = (TTT) ∩ (HHT, THH, HTH) = ∅

Therefore, they are mutually exclusive

Also

X ∪ Y ∪ Z = (HHH TTT, HHT, THH, HTH)

X ∪ Y ∪ Z ≠ Space

So, X, Y and Z are not exhaustive.

Thus, it is proved that X, Y and X are mutually exclusive but not exhaustive.

### Question 6. Two dice are thrown. The events A, B, and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice ≤ 5.

Describe the events

(i) A’

(ii) not B

(iii) A or B

(iv) A and B

(v) A but not C

(vi) B or C

(vii) B and C

(viii) A ∩ B’ ∩ C’

Solution:

Let us consider that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

In the question, it is given that pair of dice is thrown, so sample space will be-

Space = {(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)}

There are several conditions in the question and according to those conditions,

we have to solve proper solutions.

A: getting an even number on the first die:

So, we have to make the sample space in which the first die is even-

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

(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

B: getting an odd number on the first die:

So, we have to make the sample space in which the first die is odd-

Hence, 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 the sum of the numbers on the dice ≤ 5:

According to the above sample space the minimum sum will be (1+1)=2

So, the sum can be 2, 3, 4 & 5

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

(i) A’ = {(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)} = B

(ii) B’ = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),

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

(iii) A or B = A ∪ B = {(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)} = Space

(iv) A and B = A ∩ B = ∅

(v) A but not C = A – C = {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5),

(4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

(vi) B or C =B ∪ 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), (4, 1), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

(vii) B and C = B ∩ C = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}

(viii) A ∩ B’ ∩ C’ = A ∩ A ∩ C’ = A ∩ C’

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

(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

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

C’ = {(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),

(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)}

A ∩ B’ ∩ C’ = A ∩ C’ = {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5),

(4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

### Question 7.  Refer to question 6 above, state true or false: (give reason for your answer)

(i) A and B are mutually exclusive

(ii) A and B are mutually exclusive and exhaustive

(iii) A = B’

(iv) A and C are mutually exclusive

(v) A and BI are mutually exclusive.

(vi) A’, B’, C are mutually exclusive and exhaustive.

Solution:

Let us consider that 1, 2, 3, 4, 5 and 6 are the possible outcomes when the die is thrown.

In the question, it is given that pair of dice is thrown, so sample space will be-

Space = {(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)}

There are several conditions in the question and according to those

conditions, we have to solve proper solutions.

A: getting an even number on the first die:

So, we have to make the sample space in which the first die is even-

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

(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

B: getting an odd number on the first die:

So, we have to make the sample space in which the first die is odd-

Hence, 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 the sum of the numbers on the dice ≤ 5:

According to the above sample space the minimum sum will be (1+1) = 2

So, the sum can be 2, 3, 4 & 5

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

(i) A and B are mutually exclusive:

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

(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

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)}

There are no common elements between A & B

That’s why, (A ∩ B) = ∅

So, A & B are mutually exclusive.

Hence, the given statement is true.

(ii) A and B are mutually exclusive and exhaustive:

A ∪ B = {(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)} = Space

⇒ A ∪ B =Space

Hence, A and B are mutually exhaustive.

We already know from (i) that A and   B are mutually exclusive.

Thus, A and B are  mutually exclusive and exhaustive.

Hence, the given statement is true.

(iii) A = B’:

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

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

Hence, the given statement is true.

(iv) A and C are mutually exclusive:

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

(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

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

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

As, A ∩ C ≠ ∅ , A and C are not mutually exclusive.

Hence, the given statement is false.

(v) A and B’ are mutually exclusive:

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

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

A ∩ B’ =  A ∩ A = ∅

That’s why, A and B’ are not mutually exclusive.

Hence, the given statement is false.

(vi) A’, B’, C are mutually exclusive and exhaustive:

A’ = {(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)}

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

(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

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

A’ ∩ B’ = ∅

Hence, there is no common element between A’ and B’

So A’ and B’ are mutually exclusive.

B’ ∩ C ={{(2, 1), (2, 2), (2, 3), (4, 1)}

As, B’ ∩ C ≠ ∅, B and C are not mutually exclusive.

Thus, it is proved that A’, B’ and C are not mutually exclusive and exhaustive.

Hence, the given statement is false.

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