A die is thrown once. Find the probability of getting (i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.
Step-by-Step Solution
Key Concept: For an experiment with equally likely outcomes, the probability of an event = (Number of favourable outcomes) รท (Total number of outcomes). Here the sample space for a single throw of a fair die is \(S = \{1,2,3,4,5,6\}\) with \(|S| = 6\).
1. Identify the sample space
\[ S = \{1,2,3,4,5,6\} \]
Total outcomes, \(n(S) = 6\).
2. (i) Prime number
Prime numbers on a die: \(\{2,3,5\}\).
Number of favourable outcomes, \(n(A) = 3\).
\[ P(\text{prime}) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2} \]
3. (ii) Number lying between 2 and 6 (strictly between)
Numbers greater than 2 and less than 6: \(\{3,4,5\}\).
Favourable outcomes, \(n(B) = 3\).
\[ P(2 < X < 6) = \frac{3}{6} = \frac{1}{2} \]
4. (iii) Odd number
Odd numbers on a die: \(\{1,3,5\}\).
Favourable outcomes, \(n(C) = 3\).
\[ P(\text{odd}) = \frac{3}{6} = \frac{1}{2} \]
5. Conclusion
All three required probabilities are \(\frac{1}{2}\).
Correct Answer: (i) \(\frac{1}{2}\) ; (ii) \(\frac{1}{2}\) ; (iii) \(\frac{1}{2}\)