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Q5:
A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Solution :

Step 1: Analyzing the Sample Space for a Single Event

Each individual toss of a coin is an independent experiment. A fair coin has two possible outcomes for each toss, defined as the set $S = \{H, T\}$, where $H$ denotes 'Heads' and $T$ denotes 'Tails'.

Thus, the number of possible outcomes for a single toss, $n_1 = 2$.

Step 2: Applying the Fundamental Counting Principle

The Fundamental Counting Principle (also known as the Multiplication Principle) states that if there are $n$ ways to perform one task and $m$ ways to perform a second task, the total number of ways to perform both is $n \times m$. [Per the Fundamental Counting Principle of Combinatorics].

For $k$ independent events, the total number of outcomes is given by the product of the number of outcomes for each event: $N = n_1 \times n_2 \times n_3 \times \dots \times n_k$.

Step 3: Calculating Total Outcomes for Three Tosses

Given the coin is tossed $k = 3$ times:

  • Number of outcomes for toss 1 ($n_1$) = 2
  • Number of outcomes for toss 2 ($n_2$) = 2
  • Number of outcomes for toss 3 ($n_3$) = 2

Total possible outcomes = $2 \times 2 \times 2 = 2^3$.

Total possible outcomes = $8$.

Visual Representation of the Sample Space

Start H (2^2 remaining) T (2^2 remaining) Total outcomes calculated by: 2 * 2 * 2 = 8

Step 4: Enumerating the Sample Space ($S$)

To verify the calculation, we list the Cartesian product of the three sets $\{H, T\} \times \{H, T\} \times \{H, T\}$:

Outcome Index Sample Point
1(H, H, H)
2(H, H, T)
3(H, T, H)
4(H, T, T)
5(T, H, H)
6(T, H, T)
7(T, T, H)
8(T, T, T)

Final Solution: The total number of possible outcomes when a coin is tossed 3 times is $2^3 = 8$.


More Questions from Class 11 Mathematics Permutations and Combinations EXERCISE 6.1


CBSE Solutions for Class 11 Mathematics Permutations and Combinations


Chapters in CBSE - Class 11 Mathematics


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