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New answer posted
4 months agoContributor-Level 10
25. When a coin is tossed twice we have the sample space
S = {TT, TH, HT, HH}
So, n (S) = 4
Let A be the event of getting at least one tail.
Then, A = {TH, HT, TT}
So, n (A) = 3
Therefore, P (A)= Numver of outcome favorable to A/Total possible outcomes = .
New answer posted
4 months agoContributor-Level 10
24. (a) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)
P (S) = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6
P (S) = 1
As the probability of sample space is 'one' the given assignment of probabilities is valid.
(b) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)
P (S) =
.
P (S) = 1
Hence, the given assignment of probability is valid.
(c) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)
= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.5 + 0.6 + 0.7
= 2.8
i.e., P (S) > 1
As probability of the sample space S should always be '1'. The given assignment is invalid.
(d) Here P (W1) = –0.1 is negative.
As p
New answer posted
4 months agoContributor-Level 10
23. (i) A ∩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)}∩ { (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)}
A ∩B =∅
Hence, A and B are mutually exclusive.
The given statement is true.
(ii) A ∪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)}∩ { (1, 1), (1, 2), (1,
New answer posted
4 months agoContributor-Level 10
22. The sample space of the experiment is
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
New answer posted
4 months agoContributor-Level 10
21. The sample space of the experiment is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(i) Let A and B be two events such that
A: only head occurs
A = {HHH}
B: only tail occurs
B = {TTT}
And A ∩B = {HHH}∩ {TTT} =∅
(ii) Let A, B and C be two events such that
A: at most one head (i.e., the event in which we get maximum one head; one head or no head at all) occurs
So, A = {HTT, THT, TTH, TTT}
B: exactly two head occurs
So, B = {HHT, HTH, THH}
C: all are heads
So, C = {HHH}
Hence, A ∩ B = {HTT, THT, TTH, TTT} ∩ {HHT, HTH, THH} =∅
A ∩C = {HTT, THT, TTH, TTT} ∩ {HHH} =∅
B∩ C = {HHT, HTH, THH}∩ {HHH} =∅
And A∪ B ∪C = {HTT, T
New answer posted
4 months agoContributor-Level 10
20. The sample space of the experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Now, A = {HHH}
B = {HHT, HTH, THH}
C = {TTT}
D = {HHH, HHT, HTH, HTT}
(i) A∩ B = {HHH}∩ {HHT, HTH, THH} =∅
A ∩ C = {HHH}∩ {TTT} =∅
A∩ D = {HHH}∩ {HHH, HTH, HTT} = {HHH}
B ∩ C = {HHT, HTH, THH}∩ {TTT} =∅
B∩ D = {HHT, HTH, THH}∩ {HHH, HHT, HTH, HTT} = {HHT, HTH}
C∩ D = {TTT}∩ {HHH, HHT, HTH, HTT} =∅
Hence, (A, B), (A, C), (B, C) and (C, D) are pairs of mutually exclusive events.
(ii) Simple event are those which has only one sample point. So, the simple event are A and C.
(iii) Event having more than one sample point are called com
New answer posted
4 months agoContributor-Level 10
19. The sample space of the experiment is
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)}
And 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)}
Now, A∩
New answer posted
4 months agoContributor-Level 10
18. The sample space of the experiment is
S = {1, 2, 3, 4, 5, 6}
(i) A = {1, 2, 3, 4, 5, 6}
(ii) B =∅
(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}∩ {∅}= ∅h
= {1, 2, 3, 4, 5, 6}
A ∩B = {1, 2, 3, 4, 5, 6}∩ {∅}
B ∪C = ∅∪ {3, 6) = {3, 6}
E∩ F = {6}∩ {3, 4, 5, 6} = {6}
D∩ E = {1, 2, 3}∩ {6} =∅
A – C = {1, 2, 3, 4, 5, 6} – {3, 6} = {1, 2, 4, 5}
D – E = {1, 2, 3} – {6} = {1, 2, 3}
E ∩F' = E∩ (S – F) = {6}∩ [ {1, 2, 3, 4, 5, 6} – {3, 4, 5, 6}]
= {6} ∩ {1, 2}
=∅
F' = S – F = {1, 2, 3, 4,
New answer posted
4 months agoContributor-Level 10
17. The sample space of the experiment is
S = {1, 2, 3, 4, 5, 6} and E = {4}
F = {2, 4, 6}
So, E∩ F = {4} ∩ {2, 4, 6} = {4} ≠
Therefore E and F are not mutually exclusive.
New answer posted
4 months agoContributor-Level 10
16. When a die is drawn we can have 1, 2, 3, 4, 5 and 6. So, sample space of throwing dice until a six comes up is
S = { (6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 6), (1, 2, 6), (1, 3, 6), (1, 4, 6), (1, 5, 6), (1, 6)……., (1, 5, 6), (1, 6)……… (1, 5, 6)………}
Hence, the sample space is indefinite.
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