Maths

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, THT, TTH, TTT}∪ {HHT, HTH, THH}∪ {HHH} = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

= S

Therefore, A, B and C are three events which are mutually exclusive and exhaustive.

(iii) Let A and B be two such that

A: head occurs

So, A = {HHH, HHT, HTH, HTT, THH, THT, TTH}

B: tail occurs

So, B = {HHT, HTH, HTT, THH, THT, TTH, TTT}

Now, A ∩B = {HHH, HHT, HTH, HTT, THH, THT, TTH}∩ {HHT, HTH, HTT, THH, THT, TTH, TTT}

= {HHT, HTH, HTT, THH, THT, TTH} =∅

Hence A and B are not mutually exclusive.

(iv) Let A and B be two events such that

A: only head occurs

So, A = {HHH}

B = only tail occurs

So, B = {TTT}

Hence, A∩ B = {HHH}∩ {TTT} =∅

So, A ∩B =∅

But A∪ B = {HHH}∪ {TTT} = {HHH, TTT}≠ S

Therefore A and B are two events which are mutually exclusive but not exhaustive.

(v) Let A, B and C be three events such that

A: only head occurs

So, A = {HHH}

B: only tail occurs

So, B = {TTT}

C: getting one tail

So, C = {THH, HTH, HHT}

Hence, A∩ B = {HHH} ∩ {TTT} =∅

A∩ C = {HHH} ∩ {THH, HTH, HHT} =∅

B∩C = {TTT}∩ {THH, HTH, HHT} =∅

And A ∪B∪C = {HHH} ∪ {TTT}∪ {THH, HTH, HHT} = {HHH, TTT, THH, HTH, HHT} ≠ S

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

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 compound event. So, B and D are compound event.

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∩ B = { (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)}

A ∩B =∅

B∩ C = { (1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)}∩ { (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}

A∩C = { (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}

= { (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}

So, (A and B) and (B and C) are mutually exclusive.

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, 5, 6} – {3, 4, 5, 6} = {1, 2}