GSEB Solutions Class 11 Maths Chapter 16 Probability Ex 16.2

   

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 16 Probability Ex 16.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 16 Probability Ex 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:
When we throw a die, it can result in any one of the six numbers, 1, 2, 3, 4, 5, 6 and
S = {1, 2, 3, 4, 5, 6}.
E (die shows 4) = {4}.
F (die shows an even number) = {2, 4, 6).
∴ E ∩ F = {4} ⇒ E ∩ F ≠ ϕ.
⇒ E and F are not mutually exclusive.

GSEB Solutions Class 11 Maths Chapter 16 Probability Ex 16.2

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, E ∪ F, D ∩ E, A – C, D – E, F’ and E ∩ F’.
Solution:
When we throw a die, it can result in any one of the six numbers 1, 2, 3, 4, 5, 6 and
S = {1, 2, 3, 4, 5, 6}.
(i) A : a number less than 7 = {1, 2, 3, 4, 5, 6}
(ii) B : a number greater than 7 = { } = ϕ
(iii) C : a multiple of 3 = (3, 6}
(iv) D : a number less than 4 = (1, 2, 3}
(v) E : an even number greater than 4 = {6}
(vi) F : a number not less than 3 = (3, 4, 5, 6}
∴ A ∪ B = {1, 2, 3, 4, 5, 6} ∪ ϕ = {1, 2, 3, 4, 5, 6}
A ∩ B = (1, 2, 3, 4, 5, 6} ∩ ϕ = ϕ
E ∪ F = {6} ∪ (3, 4, 5, 6} = (3, 4, 5, 6}
D ∩ E = {1, 2, 3} ∩ {6} = ϕ
A – C = {1, 2, 3, 4, 5, 6} – (3, 6} = (1, 2, 4, 5}
D – E = (1, 2, 3} – {6} = {1, 2, 3}
F’ = {1, 2, 3, 4, 5, 6} – {3, 4, 5, 6) = {1, 2}
E ∩ F’ = {6} ∩ {1, 2} = ϕ.

GSEB Solutions Class 11 Maths Chapter 16 Probability Ex 16.2

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.
Also, find A ∩ B, B ∩ C and A ∩ C.
Are

  1. A and B mutually exclusive?
  2. B and C mutually exclusive?
  3. A and C mutually exclusive?

Solution:
When two dice are thrown, there are 6 × 6 = 62 possible outcomes and
S = {(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 = {(3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}.
B : 2 occurs on either die = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6)}.
C : the sum is at least 7 and a multiple of 3
= {(3, 6), (6, 3), (5, 4), (4, 5), (6, 6)}.
A ∩ B = ϕ, B ∩ C = ϕ
and A ∩ C = {(3, 6), (6, 3), (5, 4), (4, 5), (6, 6)}.

  1. Since A ∩ B = ϕ, so A and B are mutually exclusive.
  2. Since B ∩ C = ϕ so B and C are mutually exclusive.
  3. Since A ∩ C ≠ ϕ, so A and C are not mutually exclusive.

GSEB Solutions Class 11 Maths Chapter 16 Probability Ex 16.2

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

  1. mutually exclusive?
  2. simple?
  3. compound?

Solution:
When three coins are tossed, then the simple space S is
= {HHH, HHT, HTH, HIT, THH, THT, TTH, TTT}
∴ A : Three heads show = {HHH}
B : Two heads and one tail show = {HHT, HTH, THH}
C : A three tails show = {TTT}
D : A head show on the first coin = {HHH, HHT, HTH, HTT}

  1. Since A ∩ B = ϕ, A ∩ C = ϕ, B ∩ C = ϕ, C ∩ D = ϕ, A ∩ B ∩ C = ϕ.
    ⇒ A and B; A and C; B and C; C and D and A, B and C are mutually exclusive.
  2. A and C are simple events.
  3. B and D are compound events.

GSEB Solutions Class 11 Maths Chapter 16 Probability Ex 16.2

Question 5.
Three coins are tossed. Describe:

  1. two events which are mutually exclusive.
  2. three events which are mutually exclusive and exhaustive.
  3. two events which are not mutually exclusive.
  4. two events which are mutually exclusive but not exhaustive.
  5. three events which are mutually exclusive but not exhaustive.

Solution:
When three coins are tossed, then the samole space S is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

1. two events A and B which are mutually exclusive are
A : “getting at least two heads” and
B : “getting at least two tails”.

2. Three events A, B and C which are mutually exclusive and exhaustive are
A : “getting at most one head”
B : “getting exactly two heads” and
C : “getting exactly three heads”

Alternatively
Getting no head,
Getting exactly one head Getting at least two heads

3. Two events A and B which are not mutually exclusive are
A : “getting at most two tails” and
B : “getting exactly two heads” or “getting exactly two tails”

4. Two events A and B which are mutually exclusive but not exhaustive are
A : “getting exactly one head” and
B : “getting exactly two heads”.

5. Three events A, B and C which are mutually exclusive but not exhaustive are
A : “getting exactly one tail”
B : “getting exactly two tails” and
C : “getting exactly three tails”.

GSEB Solutions Class 11 Maths Chapter 16 Probability Ex 16.2

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 of 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:
When two dice are thrown, there are 6 × 6 = 36 possible outcomes and
S = {(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 : getting an even number on the first die
= {(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
= {(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 numbers on the dice ≤ 5
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
(i) A’ = getting an odd number on the first die = B.

(ii) not B : getting an even number on the first die = A.

(iii) A = getting an even number on the first die
B : getting an odd number on the first die.
A or B = A ∪ B = S = {1, 2, 3, 4, 5, 6} appear on the first die as well as on the second die.
A ∪ 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), (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)} = S.

(iv) A and B = A ∩ B = ϕ

(v) A but not C
A : getting are even number on the first die.
not C : getting the sum of numbers on two dice > 5.
= {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}
∴ A but not C = A – C : {(2, 4), (4, 2), (2, 5), (4, 3), (6, 1), (2, 6), (4, 4), (6, 2), (4, 5), (6, 3), (4, 6), (6, 4), (6, 5), (6, 6)}.

(vi) B : getting an odd number of first die.
C : {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
B or C = B ∪ C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (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 : getting an even number on the first die.
B’ = getting an even number on the first die.
C’ = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}
A ∩ B’ ∩ 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)}

GSEB Solutions Class 11 Maths Chapter 16 Probability Ex 16.2

Question 7.
Refer to question 6 above, state true or false:
(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 B’are mutually exclusive.
(vi) A’, B’, C are mutually exclusive and exhaustive.
Solution:
(i) True.
Since A : getting an even number on the first die.
B = getting an odd number on the first die.
There is no common element in A and B.
⇒ A ∩ B = ϕ.
∴ A and B are mutually exclusive.

(ii) True. A and B are mutually exclusive.
A ∪ B = {1, 2, 3, 4, 5, 6} × (1, 2, 3, 4, 5, 6} = {(1, 1), (1, 2), …, (1, 6), (2, 1), (2, 2), …, (2, 6) … (6, 1), (6, 2), …, (6, 6)}
∴ A ∪ B is exhaustive.

(iii) True. B = getting an odd number on the first die.
B’ = getting an even number on first die.
= A
∴ A = B’.

(iv) False. Since A ∩ C = {(2, 1), (2, 2), (2, 3), (4,1)} ≠ ϕ.

(v) False. Since B’ = A, So, A ∩ B’ = A ∩ A = A ≠ ϕ.

(vi) False. Since A’and B’are mutually exclusive.
A’ ∩ B’ = ϕ ⇒ A’ ∩ B’ ∩ C = ϕ.
But A’ ∩ C ≠ ϕ, B’ ∩ C ≠ ϕ ⇒ A’, B’ and C are not mutually exclusive.

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