Probability

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New answer posted

4 months ago

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A
alok kumar singh

Contributor-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 = =n (A)n (S) .

=34

New answer posted

4 months ago

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A
alok kumar singh

Contributor-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) = 17+17+17+17+17+17+17

=1+1+1+1+1+1+17=77=1 .

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

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New answer posted

4 months ago

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A
alok kumar singh

Contributor-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,

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New answer posted

4 months ago

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A
alok kumar singh

Contributor-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

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New answer posted

4 months ago

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A
alok kumar singh

Contributor-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

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New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-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

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New answer posted

4 months ago

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A
alok kumar singh

Contributor-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∩

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New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-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,

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New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-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 ago

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A
alok kumar singh

Contributor-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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