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?
Let E be the event that 2 students have the same birthday, and not E be the event that 2 students do not have the same birthday.
Given: P(not E) = 0.992
Since the events 'having the same birthday' and 'not having the same birthday' are complementary events, we know that:
P(E) + P(not E) = 1
Therefore, P(E) = 1 - P(not E)
Calculation: P(E) = 1 - 0.992 P(E) = 0.008
Hence, the probability that the 2 students have the same birthday is 0.008.
Explanation
This question tests the understanding of complementary events in probability. The context from Example 5 and Example 6 clearly demonstrates that for complementary events (like winning/losing or same birthday/different birthday), P(E) + P(not E) = 1. The student needs to recognize that 'having the same birthday' and 'not having the same birthday' are complementary events, and apply the formula P(E) = 1 - P(not E).
Solution Steps
Step 1: Identify that 'same birthday' and 'not same birthday' are complementary events
Step 2: Apply the formula P(E) = 1 - P(not E)
Step 3: Calculate: P(E) = 1 - 0.992 = 0.008