It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Step-by-Step Solution
Key Concept: Use the complement principle. The probability that at least two students share the same birthday = 1 – (probability that all three birthdays are distinct). For three independent birthdays (ignoring leap year) the distinct‑birthday probability is \(\frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}\).
1. Assumption: There are 365 equally likely days in a year and birthdays are independent.
2. Probability that all three birthdays are different:
$$P(\text{all distinct}) = \frac{365}{365}\times\frac{364}{365}\times\frac{363}{365}
= \frac{364\times 363}{365^{2}}\approx 0.992.$$
This matches the given value (0.992).
3. Complementary event: The event "at least two students have the same birthday" is the complement of "all three birthdays are different".
$$P(\text{at least two same}) = 1 - P(\text{all distinct}).$$
4. Compute the required probability:
$$P(\text{at least two same}) = 1 - 0.992 = 0.008.$$
Hence the probability that two students (among the three) share the same birthday is \(0.008\) (or \(8\times10^{-3}\)).
5. Answer in fraction (optional):
$$0.008 = \frac{8}{1000} = \frac{1}{125}.$$
So the required probability can also be expressed as \(\frac{1}{125}\).
Correct Answer: 0.008 (or \(\frac{1}{125}\))