Question

The probability that at least one of the events A and B occurs is 0.6.If A and B occur simultaneously with probability 0.2.Then P(ˉA)+P(ˉB) is; (A)0.4(B)0.8(C)1.2(D)1.6

Answer 1

p (AUB)=0.6,P (A intersection B)=0.2
P (AUB)=P (A)+P (B)-P (A intersection B)
0.6=P (A)+P (B)-0.2
0.6+0.2=P (A)+P (B)
P (A)+P (B)=0.8
P (A bar )+P (Bbar)=(1-p (A))+(1-p (B))
=2-(P(A)+P (B))
=2-(0.8)
=1.2

Answer 2

Answer:

Option (C) 1.2 is the correct answer

Step-by-step explanation:

Explanation:

Given that , probability that at least one of the events A and B occur  is

P(A ∪ B) = 0.6

and probability that A and B occur simultaneously is

P(A ∩ B) = 0.2

Step 1:

As we know that ,

$P(A) = 1 - P(\bar{A})$

and $P(B) = 1- P(\bar {B})$

Now ,we also know that  if A and B are any two event then

P(A ∪ B)  = P(A) + P(B)- P(A ∩ B)

put the value in the above formula we get ,

0.6 = $1 - P(\bar{A})$ + $1- P(\bar {B})$ - 0.2

⇒0.6  = 2 - 0.2 - $P(\bar{A}) - P(\bar{B})$

⇒1.2 = $P(\bar{A}) + P(\bar{B})$

Final answer :

Hence , the value of $P(\bar{A}) + P(\bar{B})$ is 1.2 .

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