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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.  $1$
asked in Probability by Veteran (59.5k points)
edited by | 1.1k views

2 Answers

+22 votes
Best answer

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)

$x = 1$

answered by Loyal (9k points)
edited by
+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.

answered by Loyal (8.3k points)


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