Let A and B be two independent events such that P(A 0.26 and P(B 0.54 What is P(A or B)))

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Misc 7 A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (ii) P(A′ ∩ B′) P(A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) = 1 − 0.88 = 0.12 (By Demorgan law) Demorgan’s law "(A’"∩"B’) = (A "∪" B)’" "or (A’ "∪" B’) = (A "∩" B)’"

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

Given $P(A)=P(B)=x$,

so $x+x-(x\times x)=0.5\Rightarrow 2x-x^2=\frac{1}{2}\Rightarrow 4x-2x^2-1=0$.

We have to solve, $2x^2-4x+1=0$. Here, $a=2,b=-4,c=1$.

Now,$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\pm\sqrt{16-4\times2\times(-1)}}{2\times2}=\frac{4\pm\sqrt{8}}{4}=1\pm\frac{1}{\sqrt{2}}$$

We have $x=1.707 \& 0.292$.

$\because, x $ is probability $x {\not>} 2$.

$\therefore \boxed{x= P(A)= 0.292}$