Question

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?

Answer 1

Hint : We will first make the sample space for tossing of three coins, then for the events. Using these sample spaces, we can find the events that are independent of each other, these are known as mutually exclusive events and their intersection is a null subset, the events that are simple and compound. Simple are the ones with one and compound are the events with multiple sample points

** Complete step-by-step answer** :
The possible outcomes of a coin are 2 i.e. heads and tails but when three coins are tossed once, the possible outcomes and sample space is:
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
The sample space for respective events can be given as:
A: three heads show
A = {HHH}
B: Two heads and one tail show
B = {HHT, HTH, THH}
C: three tails show
C = {TTT}
D: A head shows on the first coin
D = {HHH, HHT, HTH, THH, HTT}
Answering the given questions:
(i) Mutually exclusive events: The events where no component of sample space are same. Mathematically:
$\left( {A \cap B} \right) = \phi$ where A and B are two events, $' \cap '$ shows their intersection and $\phi$ denotes null set.
a.{\text{ }}\left( {A \cap B} \right) = \left\\{ {HHH} \right\\} \cap \left\\{ {HHT,{\text{ }}HTH,{\text{ }}THH} \right\\} \\\ \Rightarrow \left( {A \cap B} \right) = \phi \;
As there is no common element between A and B, these are mutually exclusive events.
b.{\text{ }}\left( {A \cap C} \right) = \left\\{ {HHH} \right\\} \cap \left\\{ {TTT} \right\\} \\\ \Rightarrow \left( {A \cap C} \right) = \phi \;
As there is no common element between A and C, these are mutually exclusive events.
c.{\text{ }}\left( {A \cap D} \right) = \left\\{ {HHH} \right\\} \cap \left\\{ {HHH,{\text{ }}HHT,{\text{ }}HTH,{\text{ }}THH,{\text{ }}HTT} \right\\} \\\ \Rightarrow \left( {A \cap D} \right) = \left\\{ {HHH} \right\\} \\\ \Rightarrow \left( {A \cap D} \right) \ne \phi \;
As there is one common element between A and D, these are not mutually exclusive events.
d.{\text{ }}\left( {B \cap C} \right) = \left\\{ {HHT,{\text{ }}HTH,{\text{ }}THH} \right\\} \cap \left\\{ {TTT} \right\\} \\\ \Rightarrow \left( {B \cap C} \right) = \phi \;
As there is no common element between B and C, these are mutually exclusive events.
e.{\text{ }}\left( {B \cap D} \right) = \left\\{ {HHT,{\text{ }}HTH,{\text{ }}THH} \right\\} \cap \left\\{ {HHH,{\text{ }}HHT,{\text{ }}HTH,{\text{ }}THH,{\text{ }}HTT} \right\\} \\\ \Rightarrow \left( {B \cap D} \right) = \left\\{ {HHT,{\text{ }}HTH} \right\\} \\\ \Rightarrow \left( {B \cap D} \right) \ne \phi \;
As there is one common element between B and D, these are not mutually exclusive events.
f.{\text{ }}\left( {C \cap D} \right) = \left\\{ {TTT} \right\\} \cap \left\\{ {HHH,{\text{ }}HHT,{\text{ }}HTH,{\text{ }}THH,{\text{ }}HTT} \right\\} \\\ \Rightarrow \left( {C \cap D} \right) \ne \phi \;
As there is no common element between C and D, these are mutually exclusive events.
Therefore, the mutually exclusive pair of events are:
A and B, A and C, B and C, C and D
So, the correct answer is “ A and B, A and C, B and C, C and D”.

(ii) Simple events: The events having only one component called sample point in sample space are known as simple events.
The only events having one sample point are A and C because:
A = {HHH}
C = {TTT}
Therefore, A and C events are simple.
So, the correct answer is “ A and C”.

(ii) Compound events: The events having more than one component called sample point in sample space are known as compound events.
The only events having more than one sample points are B and D because:
B = {HHT, HTH, THH}
D = {HHH, HHT, HTH, THH, HTT}
Therefore, B and D events are compound.
So, the correct answer is “ B and D”.

Note : In general, if a coin is tossed n number of times, then the possible outcomes are given by 2 raised to the power n. here, the coin was tossed 3 times, then n is 3 . So, possible outcomes or sample points in sample space will be:
${2^3} = 8$ ,We always represent sample space with the curly brackets ‘{}’. It is the subset of all the possible outcomes of the experiment known as the events.