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E1 and E2 are events in a probability space satisfying the following constraints:

  • Pr(E1) = Pr(E2)
  • Pr(E1 ∪ E2) = 1
  • E1 and E2 are independent

The value of Pr(E1), the probability of the event E1, is

  1.  $0$
  2. $\dfrac{1}{4}$
  3. $\dfrac{1}{2}$
  4.  $1$
asked in Probability by Veteran (68.8k points)
edited by | 810 views

2 Answers

+16 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 -x2 (prob(E1 intersect E2) = prob(E1) * prob(E2) as events are independent)

x = 1

answered by Boss (9.3k points)
selected by
+4 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 Boss (7.5k points)


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