Dylan Chandler

2023-03-06

If $A$ and $B$ are two independent events such that $P\left(B\right)=\frac{2}{7}$, $P\left(A\cup B\text{'}\right)=0.8$, then $P\left(A\right)$ is equal to...

phonelookup1l6x

$P\left(A\cup B\text{'}\right)=P\left(A\right)+P\left(B\text{'}\right)–P\left(A\cap B\text{'}\right)$ …..(1)
Also, $P\left(A\cap B\text{'}\right)=P\left(A\right)-P\left(A\cap B\right)$
$P\left(A\cup B\text{'}\right)=P\left(A\right)+P\left(B\text{'}\right)–\left(P\left(A\right)-P\left(A\cap B\right)\right)$
$⇒$$P\left(A\cup B\text{'}\right)=P\left(A\right)+P\left(B\text{'}\right)–P\left(A\right)+P\left(A\cap B\right)$
$⇒$$P\left(A\cup B\text{'}\right)=P\left(B\text{'}\right)+P\left(A\cap B\right)$
$⇒$$P\left(A\cup B\text{'}\right)=\left(1-P\left(B\right)\right)+P\left(A\right).P\left(B\right)$ ….(2) $\left(\because P\left(B\text{'}\right)=1-P\left(B\right),P\left(A\cap B\right)=P\left(A\right).P\left(B\right)\right)$
$0.8=\left(1-\frac{2}{7}\right)+P\left(A\right).\left(\frac{2}{7}\right)$
$⇒$ $0.8=\frac{5}{7}+\frac{2}{7}×P\left(A\right)$
$⇒$$5+2P\left(A\right)=5.6$
$⇒$ $2P\left(A\right)=0.6$
$\therefore P\left(A\right)=0.3$

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