Referring to the AP Inter 1st Year Maths Study Material Chapter 14 Probability Exercise 14a Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Probability Solutions Exercise 14a
I.
Question 1.
A dice is rolled. Let E be the event “dice shows 4” and F be the event “dice shows even number”. Are E and F mutually exclusive ?
Solution:
No. When a dice is rolled, the sample space is given by S = {1, 2, 3, 4, 5, 6} Accordingly, E = {4} and F = {2, 4, 6}. It is observed that E ∩ F = {4} ≠ Φ
Therefore, E and F are not mutually exclusive events.
Question 2.
In the experiment of throwing a dice, consider the following events.
A = {1, 3, 5} B = {2, 4, 6} C = {1, 2, 3}
Are these events equally likely ?
Solution:
Yes, chances of occurring the events A, B and C are equal. Hence, clearly A, B, C are equally likely.
Question 3.
In the experiment of throwing a dice, consider the following events.
A = {1, 3, 5} B = {2, 4} C = {6}
Are these events mutually exclusive ?
Solution:
Yes, because of one of the given events A, B and C prevents the happening of other two. Hence A, B, C are mutually exclusive.
Question 4.
In the experiment of throwing a dice, consider the following events.
A = {2, 4, 6} B = {3, 6} C = {1, 5, 6}
Are these events exhaustive ?
Solution:
Yes. Let ‘S’ be the sample space for the random experiment of throwing a dice. Then S = {1, 2, 3, 4, 5, 6}
Given A = {2, 4, 6}, B = {3, 6}, C = {1, 5, 6}
Clearly, events are exhaustive. If their union covers the entire sample space.
∴ A ∪ B ∪ C = S
Hence clearly A, B, C are exhaustive.
II.
Question 1.
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 dice
C : the sum is at least 7 and a multiple of 3.
Which pairs of these events are mutually exclusive ?
Solution:
When a pair of dice is rolled, the sample space is given by
S = {(x, y) : X, y = 1, 2, 3, 4, 5, 6} = {(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)}
Now, A = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}
B = {(1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)}
C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
Here, A ∩ B = Φ, B ∩ C = Φ and A ∩ C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6) ≠ Φ
Hence, events A and B as well as events B and C are mutually exclusive.
Question 2.
Three coins are tossed once. Let A denote the event ‘three heads show’, B denote the event “two heads and one tail show”, C denote 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:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Now, A = {HHH}, B = {HHT, HTH, THH}, C = {TTT}, D = {HHH, HHT, HTH, HTT} We now observe that, A ∩ B = Φ, A ∩ C = Φ, A ∩ D = {HHH} ≠ Φ
B ∩ C = Φ, B ∩ D = {HHT}, ≠ Φ, C ∩ D = Φ
i) Events A and B, events A and C, events B and C, and events C and D are all mutually exclusive.
ii) If an event has only one sample point of a sample space, it is called a sample event. Thus, A and C are simple events.
iii) If an event has more than one sample point of a sample space, it is called a compound event. Hence, B and D are compound events.
Question 3.
Two dice are thrown. The events A, B and C are as follows.
A : getting an even number on the first dice,
B : getting an odd number on the first dice and
C : getting the sum of numbers on the dice < 5.
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 B’ are mutually exclusive.
vi) A’, B’, C are mutually exclusive and exhaustive.
Solution:
When a pair of dice is rolled the sample space is given by
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)}
Now,
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)}
C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
i) It is observed that A ∩ B = Φ
A and B are mutually exclusive. Thus, the given statement is true.
ii) It is observed that A ∩ B = Φ and A ∪ B = S
A and B are mutually exclusive and exhaustive. Thus, the given statement is true.
iii) It is observed that
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 Thus, the given statement is true.
iv) It is observed that A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)}
A and C are not mutually exclusive.
Thus, the given statement is false.
v) A ∩ B’ = A ∩ A = A ⇒ A ∩ B’ ≠ Φ
∴ A and B’ are not mutually exclusive. Thus the given statement is false.
vi) It can be observed that : A’ ∪ B’ ∪ C = S However B’ ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠ Φ
Therefore, events A’, B’ and C are not mutually exclusive and exhaustive. Thus, the given statement is false,
III.
Question 1.
A dice 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 ∩ F’, F’.
Solution:
When a dice is thrown, the sample space is given by S = {1, 2, 3, 4, 5, 6} Accordingly,
i) A = {1, 2, 3, 4, 5, 6}
ii) B = Φ
iii) C = {3, 6}
iv) D = {1, 2, 3}
v) E = {6}
vi) F = {3, 4, 5, 6}
A ∪ B = {1, 2, 3, 4, 5, 6}, A ∩ B = Φ, B ∪ C = {3, 6}, E ∩ F = {6}, D ∩ E = Φ,
A – C = {1, 2, 4, 5}, D – E = {1, 2,3}, E ∩ F’ = Φ, F’ = S – F = {1, 2}.
Question 2.
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 three coins are tossed, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
i) Two events that are mutually exclusive can be
A : getting no heads and B : getting no tails
Therefore, A = {TTT} and B = {HHH} are disjoint because AnB = f
∴ A, B are mutually exclusive.
ii) Three events that are mutually exclusive and exhaustive can be
A : getting no heads, C : getting at least two heads,
B : getting exactly one head
Therefore, A = {TTT}, B = {HTT, THT, TTH}, C = {HHH, HHT, HTH, THH}
This is because A ∩ B = B ∩ C = C ∩ A = Φ and A ∪ B ∪ C = S.
∴ A, B, C are mutually exclusive and exhaustive.
iii) Two events that are not mutually exclusive can be
A : getting three heads,
B : getting at least 2 heads,
Therefore, A = {HHH}, B : {HHH, HHT, HTH, THH}
This is because A ∩ B = {HHH} ≠ Φ.
∴ A, B are not mutually exclusive.
iv) Two events which are mutually exclusive but not exhaustive can be
A : getting exactly one head
B : getting exactly one tail
Therefore, A = {HTT, THT, TTH}, B = {HHT, HTH, THH}
It is because, A ∩ B = Φ, but A ∪ B ∪ S.
A, B are mutually exclusive but not exhaustive.
v) Three events that are mutually exclusive but not exhaustive can be
A : getting exactly three heads,
B : getting one head and two tails,
C : getting one tail and two heads.
Therefore, A = {HHH}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH}
This is because A ∩ B = B ∩ C = C ∩ A = Φ, but A ∪ B ∪ C ≠ S.
∴ A, B, C are mutually exclusive but not exhaustive.
Question 3.
Two dice are thrown. The events A, B and C are as follows.
A : getting an even number on the first dice.
B : getting an odd number on the first dice.
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:
When a pair of dice is rolled, the sample space is given by
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)}.
Now,
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)}.
C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}.
Therefore,
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) not 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.
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)} = S.
iv) A and B = A ∩ B = Φ.
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), (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) 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’ = {(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)}.