Chapter 16: Probability Exercise 16.2

Solution:-

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

So, S = (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 ≠ φ … [because there is a common element in E and F]

Therefore, E and F are not mutually exclusive events.

Solution:-

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

So, S = (1, 2, 3, 4, 5, 6)

As per the conditions given in the question,

(i) A: a number less than 7

All the numbers in the die are less than 7.

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

(ii) B: a number greater than 7

There is no number greater than 7 on the die.

Then,

B= (φ)

(iii) C: a multiple of 3

There are only two numbers which are multiple of 3.

Then,

C= (3, 6)

(iv) D: a number less than 4

D= (1, 2, 3)

(v) E: an even number greater than 4

E = (6)

(vi) F: a number not less than 3

F= (3, 4, 5, 6)

Also, we have to find, A U B, A ∩ B, B U C, E ∩ F, D ∩ E, D – E, A – C, E ∩ F’, F’

So,

A ∩ B = (1, 2, 3, 4, 5, 6) ∩ (φ)

= (φ)

B U C = (φ) U (3, 6)

= (3, 6)

E ∩ F = (6) ∩ (3, 4, 5, 6)

= (6)

D ∩ E = (1, 2, 3) ∩ (6)

= (φ)

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)

= (φ)

Solution:-

Let us assume 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 die is thrown, so sample space will be

Now, we shall find whether pairs of these events are mutually exclusive or not.

(i) A∩ B = φ

There is no common element in A and B.

Therefore, A & B are mutually exclusive.

(ii) B ∩ C = φ

There is no common element between them.

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)} ≠ φ

A and C have common elements.

Therefore, A and C are mutually exclusive.

Solution:-

Either coin can turn up Head (H) or Tail (T), are the possible outcomes.

But, now three coins are tossed once, so the possible sample space contains

S = {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 first coin

D= (HHH, HHT, HTH, HTT)

(i) Mutually exclusive

A ∩ B = (HHH) ∩ (HHT, THH, HTH)

= φ

Therefore, A and C are mutually exclusive.

A ∩ C = (HHH) ∩ (TTT)

= φ

There, A and C are mutually exclusive.

A ∩ D = (HHH) ∩ (HHH, HHT, HTH, HTT)

= (HHH)

A ∩ D ≠ φ

So they are not mutually exclusive.

B ∩ C = (HHT, HTH, THH) ∩ (TTT)

= φ

There is no common element in B & C, so they are mutually exclusive.

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

= (HHT, HTH)

B ∩ D ≠ φ

There are common elements in B & D.

So, they are not mutually exclusive.

C ∩ D = (TTT) ∩ (HHH, HHT, HTH, HTT)

= φ

There is no common element in C & D.

So they are not mutually exclusive.

(ii) Simple event

If an event has only one sample point of a sample space, it is called a simple (or elementary) event.

A = (HHH)

C = (TTT)

Both A & C have only one element.

So they are simple events.

(iii) Compound events

If an event has more than one sample point, it is called a Compound event.

B= (HHT, HTH, THH)

D= (HHH, HHT, HTH, HTT)

Both B & D have more than one element.

So, they are compound events.

Solution:-

Either coin can turn up Head (H) or Tail (T), are the possible outcomes.

But, now three coins are tossed once, so the possible sample space contains

S= (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)

(i) Two events which are mutually exclusive.

Let us assume A be the event of getting only head.

A = (HHH)

Also, let us assume B be the event of getting only Tail.

B = (TTT)

So, A ∩ B = φ

There is no common element in A& B, so these two are mutually exclusive.

(ii) Three events which are mutually exclusive and exhaustive

Now,

Let us assume P be the event of getting exactly two tails.

P = (HTT, TTH, THT)

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

Q = (HHT, HTH, THH, HHH)

Let us assume R be the event of getting only one tail.

C= (TTT)

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

= φ

There is no common element in P and Q.

Therefore, they are mutually exclusive.

Q ∩ R = (HHT, HTH, THH, HHH) ∩ (TTT)

= φ

There is no common element in Q and R.

Hence, they are mutually exclusive.

P ∩ R = (HTT, TTH, THT) ∩ (TTT)

= φ

There is no common element in P and R.

So they are mutually exclusive.

Now, P and Q, Q and R, and P and R are mutually exclusive.

∴ P, Q, and R are mutually exclusive.

Also,

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

Hence P, Q and R are exhaustive events.

(iii) Two events which are not mutually exclusive

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

A = (HHH, HHT, THH, HTH)

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

B= (HHH)

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

= (HHH)

A ∩ B ≠ φ

There is a common element in A and B.

So they are not mutually exclusive.

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

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

P = (HHH)

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

Q = (TTT)

P ∩ Q = (HHH) ∩ (TTT)

= φ

There is no common element in P and Q.

So these are mutually exclusive events.

But,

P ∪ Q = (HHH) ∪ (TTT)

= {HHH, TTT}

P ∪ Q ≠ S

Since P ∪ Q ≠ S, these are not exhaustive events.

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

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

X = (HHH)

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

Y = (TTT)

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

Z= (HHT, THH, HTH)

Now,

X ∩ Y = (HHH) ∩ (TTT)

= φ

X ∩ Z = (HHH) ∩ (HHT, THH, HTH)

= φ

Y ∩ Z = (TTT) ∩ (HHT, THH, HTH)

= φ

Therefore, they are mutually exclusive.

Also

X ∪ Y ∪ Z = (HHH TTT, HHT, THH, HTH)

X ∪ Y ∪ Z ≠ S

So, X, Y and Z are not exhaustive.

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

Solution:-

Let us assume 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 die is thrown, so sample space will be

Solution:-

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

In the question is given that pair of die is thrown, so sample space will be

By referring the question 6 above,

(i) A and B are mutually exclusive.

So, (A ∩ B) = φ

So, A & B are mutually exclusive.

Hence, the given statement is true.

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

⇒ A ∪ B = S

From statement (i), we have A and B are mutually exclusive.

∴ A and B are mutually exclusive and exhaustive.

Hence, the statement is true.

(iii) A = B

Therefore, the statement is true.

(iv) A and C are mutually exclusive.

We have,

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

A ∩ C ≠ φ

A and C are not mutually exclusive.

Hence, the given statement is false.

(v) A and B^I are mutually exclusive.

We have,

A ∩ B^I = A ∩ A = A

∴ A ∩ B^I ≠ φ

So, A and B^I are not mutually exclusive.

Therefore, the given statement is false.

They are not mutually exclusive.

B^I and C are not mutually exclusive.

Therefore A’, B’ and C are not mutually exclusive and exhaustive.

So, the given statement is false.