Ask & Answer: India's Largest Education Community

 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, 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)}

= { (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

Hence, A B = S and A∩ B =∅

A and B are mutually exclusive and mutually exhaustive.

So, the give statement is true.

(iii) 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 = S – 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)} – { (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)}

B' = A

Hence, the given relation is true.

(iv) A ∩C = { (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), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, (1)}

= { (2, 1), (2, 2), (2, 3), (4, 1)}≠∅

So, A∩ C≠∅

i.e., A and C are not mutually exclusive.

Hence, the given statement is false.

(v) A∩ B' = A∩ A = A≠∅ { B' = A}

Hence A∩B'≠∅ i.e., A and B are not mutually exclusive.

So, the given statement is false.

(vi) As A' = B

B = A

So, A'∩B = B∩A =∅

A'∩ C = B∩C = { (1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}=∅

B'∩C = A∩C = { (2, 1), (2, 2), (2, 3), (4, 1)}=∅

Hence, A', B' and C are not mutually exclusive.

The given statement is false.

After completing your 10th, you can get admission to the Diploma in Computer Engineering at M. H. Saboo Siddik Polytechnic, Mumbai, through the Centralized Admission Process (CAP) conducted by the Maharashtra Directorate of Technical Education (DTE). The selection is based on your SSC marks, with importance given to your overall percentage and performance in Maths and Science. Once the CAP registration opens, usually in June, you need to fill out the online form, list M. H. Saboo Siddik Polytechnic as one of your preferences, and submit your details. DTE will release a merit list, and if your rank matches the cutoff for Computer Engineering, you will be allotted a seat. You then confirm your admission by visiting the college with original documents and paying the required fees.

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), (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) A? = S – A = Sample space – getting even number on 1st dice = getting odd number on 1st dice = B

(ii) B' = S – B = Sample space – getting odd number on 1st dice = getting even number on 1st dice = A

(iii) A or B = A∪B = (getting even on 1st die)∪ (getting odd o 1st dice) = Sample space = S

(iv) A and B = A∩ B = (getting even on 1st dice)∩ (getting odd on 1st dice) =∅

(v) A but not C = { (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), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

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), (3, 1), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}∪ { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

= { (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), (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)}∩ { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)} = { (1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2), (4, 1)}

(viii) A∩B ∩C = A ∩A ∩C (as B' = A)

= A ∩C' (A ∩A = A)

So, C' = S – C = { (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)} – { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

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)}

So, A ∩B ∩ C' = (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, 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)}

= { (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)}

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}

If your AIIMS B.Sc. Nursing application form was rejected because you entered the wrong passing month, the only way to fix it is through the AIIMS correction process — but AIIMS is very strict, and corrections are only allowed during the official application correction window.

Here's what you should do:

1. Check the AIIMS exam portal immediately – Log in to your candidate account and see if the correction window is currently open.
2. Use the correction facility – If the window is active, update the correct passing month and save the changes.
3. If the correction window is closed – Contact the AIIMS exam helpdesk right away through:
• Email: usually mentioned in the prospectus (e.g., aiims.bsc [year]@aiimsexams.ac.in)
• Helpline numbers listed on the AIIMS exam website.
4.  
5. Explain clearly – Mention your registration number, full name, application ID, and the exact mistake, and request them to allow a correction.
6. Keep proof – Save screenshots or email copies of all your communication with AIIMS.

If the correction window is already closed and AIIMS refuses changes, you may have to reapply next year — but in many cases, if the error is minor and you inform them on time, they allow it to be fixed before document verification.