Let E1,E2 be two mutually exclusive events of an experiment with P(not E2)=0.6=P(E1∪ E2). Then P(E1) is equal to

Question

Let E1,E2 be two mutually exclusive events of an experiment with P(not E2)=0.6=P(E1∪ E2). Then P(E1) is equal to (A)  0.1  (B)  0.3  (C)  0.4  (D)  0.2

Tardigrade Admin · Accepted Answer

Given, P( barE2)=0.6=P(E1∪ E2) Since, E1, E2 are mutually exclusive events, then P(E1∩ E2)=0 Now, P(E1∪ E2)=P(E1)+P(E2)-P(E1∩ E2) ⇒ 0.6 = P(E1)+1-P( barE2)-0 ⇒ 0.6=P(E1)+0.4 ⇒ P(E1)=0.2

Q. Let $E_{1}, E_{2}$ be two mutually exclusive events of an experiment with $P (n o t E_{2}) = 0.6 = P (E_{1} \cup E_{2}) .$ Then $P (E_{1})$ is equal to

$0.1$

$0.3$

$0.4$

$0.2$

Q. Let E1​,E2​ be two mutually exclusive events of an experiment with P(notE2​)=0.6=P(E1​∪E2​). Then P(E1​) is equal to

0.1

0.3

0.4

0.2

Given, P(Eˉ2​)=0.6=P(E1​∪E2​) Since, E1​,E2​ are mutually exclusive events, then P(E1​∩E2​)=0 Now, P(E1​∪E2​)=P(E1​)+P(E2​)−P(E1​∩E2​) ⇒ 0.6=P(E1​)+1−P(Eˉ2​)−0 ⇒ 0.6=P(E1​)+0.4 ⇒ P(E1​)=0.2

Q. Let $E_{1}, E_{2}$ be two mutually exclusive events of an experiment with $P (n o t E_{2}) = 0.6 = P (E_{1} \cup E_{2}) .$ Then $P (E_{1})$ is equal to

$0.1$

$0.3$

$0.4$

$0.2$