Question

if p(aub)=0.8,p(anb)=0.3 then p(a')+p(b')=​

Answer 1

$\textbf{Given:}$

$P(A{\cup}B)=0.8$

$P(A{\cap}B)=0.3$

$\textbf{To find:}$

$P(A')+P(B')$

$\textbf{Solution:}$

$\text{We know that,}$

$\textbf{Addition theorem of probability:}$

$\text{If A and B are any two events, then}$

$P(A{\cup}B)=P(A)+P(B)-P(A{\cap}B)$

$\implies\,0.8=P(A)+P(B)-0.3$

$\implies\,0.8+0.3=P(A)+P(B)$

$\implies\,P(A)+P(B)=1.1$

$\text{Now,}$

$P(A')+P(B')$

$=[1-P(A)]+[1-P(B)]$

$=2-[P(A)+P(B)]$

$=2-1.1$

$=0.9$

$\textbf{Answer:}$

$\boxed{\bf\,P(A')+P(B')=0.9}$

Find more:

Let A and B be two events such that P(A)=0.6, P(B) = 0.2 and P(A/B)=0.5. Then P (A'/B')

equals

(a) 1/10

(b) 3/10

(C) 3/8

(d) 6/7​

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A and b are events such that p(a u b)=3/4 p(a n b)=1/4 p(a bar)=2/3 then p(abar n b)is

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