$E_1$ and $E_2$ are events in a probability space satisfying the following constraints:
- $\operatorname{Pr}(E_1)=\operatorname{Pr}(E_2)$
- $\operatorname{Pr}(E_1\cup E_2)=1$
- $E_1$ and $E_2$ are independent.
The probability of $E_1$ is …(a) $0$, (b) $1/4$, (c) $1/2$, (d) $1$
I think the answer should be (d) 1
Reasoning.
- E1 and E2 are equally likely.
- Sum of their probability is 1. This is possible if and only if both of their probabilities are either 1(edit: if the events are independent) or 0.5( edit: if these are dependent and exhaustive events)
- E1, E2 are independent events. This implies that they both doesn't belong to same same space.
Hence E1 and E2 are certain events with probability 1.
Please correct me if my reasoning or answer is wrong.
Best Answer
Your reasoning for (2) is in error. Imagine a 3-sided die. The probability of $A=\{1,2\}$ occurring and $B=\{2,3\}$ occurring is equal and their union is a certain event.
You can get the answer quickly by elimination:
$$1=P(E_1 \cup E_2) < P(E_1)+P(E_2) = 2P(E_1) = 2P(E_2)$$
The strict less-than comes from being independent. So $P(E_1),P(E_2)$ are both larger than $1/2$
You can explicitly prove this, too
$$1=P(E_1 \cup E_2) = P(E_1)+P(E_2)-P(E_1)P(E_2) = 2P(E_1) - P(E_1)^2$$
$$\iff P(E_1)^2 - 2P(E_1)+1=(P(E_1)-1)^2=0$$