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

### 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: B

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

_{2 }= {(1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}(iii) 2 may come in both the die simultaneously: B

_{3 }= {(2, 2)}Hence, B = {B

_{1 }∪ B_{2 }∪ B_{3}} = {(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: ‘three heads’

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 eventIf 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 eventsIf 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 exhaustiveNow,

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 exclusiveLet 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 exhaustiveLet 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 exhaustiveLet 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.