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Q5(v):
Three coins are tossed. Describe (v) Three events which are mutually exclusive but not exhaustive.

Solution :

Step 1: Establishing the Sample Space ($S$)

When three coins are tossed, each coin has two possible outcomes: Head ($H$) or Tail ($T$). For $n=3$ coins, the total number of elementary events in the sample space $S$ is given by $2^n = 2^3 = 8$.

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

Step 2: Defining Mutually Exclusive and Exhaustive Conditions

  • Mutually Exclusive: Events $A, B,$ and $C$ are mutually exclusive if $A \cap B = \phi$, $A \cap C = \phi$, and $B \cap C = \phi$. That is, no two events can occur simultaneously.
  • Exhaustive: Events are exhaustive if their union is equal to the sample space: $A \cup B \cup C = S$.
  • Requirement: We require events that are mutually exclusive but not exhaustive, meaning $A \cup B \cup C \subset S$.

Step 3: Constructing the Events

Let us define three specific events based on the number of heads appearing in the toss:

  • Event $E_1$: Getting exactly 3 heads.
    $E_1 = \{HHH\}$
  • Event $E_2$: Getting exactly 3 tails.
    $E_2 = \{TTT\}$
  • Event $E_3$: Getting exactly 2 heads and 1 tail.
    $E_3 = \{HHT, HTH, THH\}$

Step 4: Verification of Mutually Exclusive Property

We check for intersections between the defined events:

IntersectionResultStatus
$E_1 \cap E_2$$\phi$Mutually Exclusive
$E_1 \cap E_3$$\phi$Mutually Exclusive
$E_2 \cap E_3$$\phi$Mutually Exclusive

[By Definition of Disjoint Sets: Sets are mutually exclusive if they share no common elements].

Step 5: Verification of Non-Exhaustive Property

Calculate the union of the three events:

$E_1 \cup E_2 \cup E_3 = \{HHH, TTT, HHT, HTH, THH\}$

Comparing this to the Sample Space $S$:

$\{HHH, TTT, HHT, HTH, THH\} \neq \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

The elements $\{HTT, THT, TTH\}$ are missing from the union. Since $E_1 \cup E_2 \cup E_3 \subset S$, the events are not exhaustive.

Visual Representation of Sample Space Distribution

E1 E2 E3 Sample Space S (Not fully covered)

Final Solution: The three events $E_1 = \{HHH\}$, $E_2 = \{TTT\}$, and $E_3 = \{HHT, HTH, THH\}$ are mutually exclusive because they share no common elements, and they are not exhaustive because their union does not contain the elements $\{HTT, THT, TTH\}$.


More Questions from Class 11 Mathematics Probability EXERCISE 14.1


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