A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once? [Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment] 3 m 2 m PROBABILITY 217
Step-by-Step Solution
Key Concept: Use the concept of equally likely outcomes for independent trials. The total number of outcomes when a die is thrown twice is $6 \times 6 = 36$. For part (i) count the outcomes where 5 does not appear in either throw (5 choices for each throw). For part (ii) use the complement rule: \(P(\text{at least one 5}) = 1 - P(\text{no 5})\).
1. Sample space: When a die is thrown twice, each throw can give any of the numbers $1,2,3,4,5,6$. Hence the total number of equally likely outcomes is
$$\Omega = 6 \times 6 = 36.$$
2. Part (i) – 5 does not appear either time:
- For the first throw, the favourable outcomes are $\{1,2,3,4,6\}$ – 5 choices.
- For the second throw, the favourable outcomes are also $\{1,2,3,4,6\}$ – 5 choices.
- Number of favourable ordered pairs = $5 \times 5 = 25$.
- Therefore,
$$P(\text{no 5 in both throws}) = \frac{25}{36}.$$
3. Part (ii) – 5 appears at least once:
- This is the complement of the event in part (i). Hence,
$$P(\text{at least one 5}) = 1 - P(\text{no 5}) = 1 - \frac{25}{36} = \frac{11}{36}.$$
- (Alternatively, count directly):
- First throw is 5 and second is not 5: $1 \times 5 = 5$ outcomes.
- First throw is not 5 and second is 5: $5 \times 1 = 5$ outcomes.
- Both throws are 5: $1$ outcome.
- Total favourable outcomes = $5 + 5 + 1 = 11$.
- Hence $P = \frac{11}{36}$.
4. Answer:
- (i) $\displaystyle \frac{25}{36}$
- (ii) $\displaystyle \frac{11}{36}$
Correct Answer: (i) \(\frac{25}{36}\) ; (ii) \(\frac{11}{36}\)