Question

Show that the probability that exactly one of the events E or F occurs is equal to P(E)+P(F)−2P(EF)

Answer 1

To show that the probability that exactly one of the events E or F occurs is equal to P(E) + P(F) - 2P(EF), we can use the principle of inclusion-exclusion.

Step-by-step explanation:

To show that the probability that exactly one of the events E or F occurs is equal to P(E) + P(F) - 2P(EF), we can use the principle of inclusion-exclusion.

First, let's define the events. E and F are two events. We want to find the probability that exactly one of them occurs.

To do this, we need to subtract the probability that both events occur (EF) from the combined probabilities of each event occurring (E and F separately). The formula is:

P(E or F) = P(E) + P(F) - 2P(EF)

Answer 2

To show that the probability that exactly one of the events E or F occurs is equal to P(E) + P(F) - 2P(EF), we need to consider the events and their probabilities.

Step-by-step explanation:

To show that the probability that exactly one of the events E or F occurs is equal to P(E) + P(F) - 2P(EF), we need to consider the events and their probabilities.

Let's assume E and F are two events. The probability of event E occurring is represented as P(E), the probability of event F occurring is represented as P(F), and the probability of both events occurring together is represented as P(EF).

The probability that exactly one of the events E or F occurs can be represented as P(E or F) - P(EF), which means we consider the probability of either E or F occurring and subtract the probability of both E and F occurring together.

Therefore, the probability that exactly one of the events E or F occurs is equal to P(E) + P(F) - 2P(EF).