Skip to content
Exercise 14.1 · Q4

Q.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?
Rajasthan RbseTextbookSubjective· 2mImportance★★★★★
8% · 7/93 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The key idea is to classify events based on their sample space relationships. Mutually exclusive events cannot occur together; simple events contain exactly one outcome; compound events contain more than one outcome. For three coin tosses: (i) A and C are mutually exclusive, A and B are mutually exclusive, B and C are mutually exclusive, A and D are not mutually exclusive, B and D are not mutually exclusive, C and D are mutually exclusive;

(ii) A and C are simple events;

(iii) B and D are compound events.

When three coins are tossed, the sample space has 8 equally likely outcomes:

S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}.

Let’s list each event explicitly:

  • A: 'three heads show' → {HHH}\{HHH\}
  • B: 'two heads and one tail show' → {HHT,HTH,THH}\{HHT, HTH, THH\}
  • C: 'three tails show' → {TTT}\{TTT\}
  • D: 'a head shows on the first coin' → {HHH,HHT,HTH,HTT}\{HHH, HHT, HTH, HTT\}

Now we classify them.


(i) Mutually exclusive events

Two events are mutually exclusive if they have no outcome in common — they cannot happen at the same time.

Check each pair:

  1. A and C: A∩C={HHH}∩{TTT}=∅A \cap C = \{HHH\} \cap \{TTT\} = \emptyset → mutually exclusive.
  2. A and B: A∩B={HHH}∩{HHT,HTH,THH}=∅A \cap B = \{HHH\} \cap \{HHT, HTH, THH\} = \emptyset → mutually exclusive.
  3. A and D: A∩D={HHH}∩{HHH,HHT,HTH,HTT}={HHH}≠∅A \cap D = \{HHH\} \cap \{HHH, HHT, HTH, HTT\} = \{HHH\} \neq \emptyset → not mutually exclusive.
  4. B and C: B∩C={HHT,HTH,THH}∩{TTT}=∅B \cap C = \{HHT, HTH, THH\} \cap \{TTT\} = \emptyset → mutually exclusive.
  5. B and D: B∩D={HHT,HTH,THH}∩{HHH,HHT,HTH,HTT}={HHT,HTH}≠∅B \cap D = \{HHT, HTH, THH\} \cap \{HHH, HHT, HTH, HTT\} = \{HHT, HTH\} \neq \emptyset → not mutually exclusive.
  6. C and D: C∩D={TTT}∩{HHH,HHT,HTH,HTT}=∅C \cap D = \{TTT\} \cap \{HHH, HHT, HTH, HTT\} = \emptyset → mutually exclusive.
Watch out

A common mistake is to think that if events are different, they must be mutually exclusive. But B and D share outcomes (HHT and HTH), so they can occur together — they are not mutually exclusive.

So the mutually exclusive pairs are: (A, C), (A, B), (B, C), (C, D).


(ii) Simple events

A simple event contains exactly one outcome from the sample space. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.