SATHEE: Probability Question 195

Probability Question 195

Question: If $ P(A\cup B)=0.8 $ and $ P(A\cap B)=0.3, $ then $ P(\bar{A})+P(\bar{B})= $

[EAMCET 2003]

Options:

A) 0.3

B) 0.5

C) 0.7

D) 0.9

Show Answer

Answer:

Correct Answer: D

Solution:

$P(A∪B)=P(A)+P(B)−P(A∩B)$

$⇒P(A)+P(B)=0.8+0.3=1.1$

And P $ (\bar A)$ + P $(\bar B)$= $1−P(A)+1−P(B)=2−1.1=0.9$

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Probability Question 195

Question: If $ P(A\cup B)=0.8 $ and $ P(A\cap B)=0.3, $ then $ P(\bar{A})+P(\bar{B})= $

Options:

Answer:

Solution:

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