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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

1.  $0$
2. $\dfrac{1}{4}$
3. $\dfrac{1}{2}$
4.  $1$
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Answer -$D$

let probability of Event $E1 = x =$ prob of $E2$

prob$($$E1 union E2$$)$ = prob$($$E1$$)$ + prob$($$E2$$)$ - prob$($$E1 intersect E2$$)$

$1 = x + x$ -$x^{2}$ (prob$($$E1 intersect E2$$)$ $=$ prob$($$E1$$)$ * prob$($$E2$$)$ as events are independent)

$x = 1$

edited

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.

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.

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