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Q2:
A coin is tossed twice, what is the probability that atleast one tail occurs?

Solution :

Step 1: Definition of the Sample Space ($S$)

When a fair coin is tossed, there are two possible outcomes: Heads ($H$) or Tails ($T$). For a compound experiment involving two independent tosses, the sample space $S$ is the Cartesian product of the individual outcomes: $\{H, T\} \times \{H, T\}$.

The total number of outcomes $n(S)$ is given by $2^n$, where $n=2$ tosses: $2^2 = 4$.

$S = \{(H, H), (H, T), (T, H), (T, T)\}$

[By the Fundamental Counting Principle, $2 \times 2 = 4$ total outcomes.]

Step 2: Defining the Favorable Event ($E$)

Let $E$ be the event of getting "at least one tail." This condition is satisfied if the outcome contains one tail or two tails.

  • $(H, T)$: Contains one tail.
  • $(T, H)$: Contains one tail.
  • $(T, T)$: Contains two tails.

Thus, $E = \{(H, T), (T, H), (T, T)\}$.

The number of favorable outcomes $n(E) = 3$.

Step 3: Calculating Probability ($P(E)$)

The probability of an event $E$ is defined by the ratio of favorable outcomes to total possible outcomes in the sample space [Per the Classical Definition of Probability]:

$P(E) = \frac{n(E)}{n(S)}$

$P(E) = \frac{3}{4}$

Visual Representation of Sample Space

S = { (H,H), (H,T), (T,H), (T,T) } Favorable outcomes (At least 1 Tail): { (H,T), (T,H), (T,T) }

Alternative Method: Complementary Probability

Using the Complement Rule: $P(E) = 1 - P(E^c)$, where $E^c$ is the event "no tails occur" (i.e., only heads).

$E^c = \{(H, H)\}$, so $n(E^c) = 1$.

$P(E^c) = \frac{1}{4}$

$P(E) = 1 - \frac{1}{4} = \frac{3}{4}$

Final Solution: The probability of getting at least one tail is $\frac{3}{4}$ or $0.75$.


More Questions from Class 11 Mathematics Probability EXERCISE 14.2


CBSE Solutions for Class 11 Mathematics Probability


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