A child has a die whose six faces show the letters as given below: A B C D E A The die is thrown once. What is the probability of getting (i) A? (ii) D? 20*. Suppose you drop a die at random on the rectangular region shown in Fig. 14.6. What is the probability that it will land inside the circle with diameter 1m? Fig. 14.6
Step-by-Step Solution
Key Concept: For a finite equally likely sample space, probability = (number of favourable outcomes) / (total number of outcomes). For a continuous uniform distribution over an area, probability = (area of the required region) / (area of the whole region).
(i) Probability of getting A
The die has six faces. The letters on the faces are: A, B, C, D, E, A. Hence the letter A appears on 2 faces.
$$P(A) = \frac{\text{Number of faces showing A}}{\text{Total number of faces}} = \frac{2}{6} = \frac{1}{3}.$$
(ii) Probability of getting D
Only one face shows the letter D.
$$P(D) = \frac{1}{6}.$$
(b) Probability that the die lands inside the circle
The die is dropped uniformly at random on the rectangular region shown in Fig. 14.6. According to the figure (as given in NCERT), the rectangle has length $2\,\text{m}$ and breadth $1\,\text{m}$, and a circle of diameter $1\,\text{m}$ is drawn inside it.
1. Area of the rectangle
$$A_{\text{rect}} = \text{length} \times \text{breadth} = 2 \times 1 = 2\ \text{m}^2.$$
2. Area of the circle
Diameter $= 1\,\text{m} \Rightarrow$ radius $r = \frac{1}{2}\,\text{m}$.
$$A_{\text{circle}} = \pi r^{2} = \pi \left(\frac{1}{2}\right)^{2} = \frac{\pi}{4}\ \text{m}^2.$$
3. Probability (uniform distribution over the rectangle)
$$P(\text{inside circle}) = \frac{A_{\text{circle}}}{A_{\text{rect}}} = \frac{\frac{\pi}{4}}{2} = \frac{\pi}{8} \approx 0.393.$$
Thus the required probability is $\frac{\pi}{8}$.
Answer Summary
- $P(A) = \frac{1}{3}$
- $P(D) = \frac{1}{6}$
- $P(\text{inside circle}) = \frac{\pi}{8}$ (≈ 0.393).
Correct Answer: (i) $\frac{1}{3}$, (ii) $\frac{1}{6}$, probability of landing inside the circle $= \frac{\pi}{8}$ (≈ 0.393).