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$E_{1}$ and $E_{2}$ are events in a probability space satisfying the following constraints:

- $Pr$$\left ( E_{1} \right )$ = $Pr$$\left ( E_{2} \right )$
- $Pr$$\left ( E_{1}\cup E_{2} \right )$ = $1$
- $E_{1}$ and $E_{2}$ are independent

The value of $Pr$$\left ( E_{1} \right )$, the probability of the event $E_{1}$, is

- $0$
- $\dfrac{1}{4}$
- $\dfrac{1}{2}$
- $1$

5 votes

Given Constraints:

1. Pr(E1) = Pr(E2)

2. Pr( E1 U E2) = 1

3. E1 and E2 are independent

As we know: Pr(E1 U E2) = Pr(E1) + Pr(E2) – Pr(E1 ∩ E2) As E1 and E2 are independent events. (cond.3)

So Pr(E1 ∩ E2) = Pr(E1) Pr(E2) Pr(E1) = Pr(E2) (cond.2)

let probability of Event E1 = x = prob of E2 So, Pr(E1 U E2) = Pr(E1) + Pr(E2) – Pr(E1) Pr(E2) 1 = x + x -x* x (cond. 1) 1=2x-x^2 x^2-2x+1 = 0 (x-1)^2 = 0 x = 1 So, Pr(E1) = Pr(E2) = 1 Thus, option (D) is the answer.

Reference : https://people.richland.edu/james/lecture/m170/ch05-rul.html

**Another Solution :** E1 and E2 are independent events.

Pr(E1 U E2) = Pr(E1) + Pr(E2) – Pr(E1) Pr(E2)

Pr(E1) = Pr(E2) (given)

So,

2 * Pr(E1) – Pr(E1)^{2} = Pr( E1 U E2)

2 * Pr(E1) – Pr(E1)^{2} = 1

So, Pr(E1) = Pr(E2) = 1

Thus, option (D) is the answer.