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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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2 Answers

35 votes
 
Best answer
Answer -$D$

Let probability of Event $E_1 = x =$ prob of $E_2$

prob$(E_1 \cup E_2) =$ prob$(E_1) +$ prob$(E_2) -$ prob$(E_1 \cap E_2)$

$1 = x + x -x^{2}$ (prob$(E_1$ intersect $E_2)$ $=$ prob$(E_1) \times $ prob$(E_2)$ as events are independent)

$\implies x = 1$
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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.

Answer:

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