If A and B are two independent events such that $$P(A$$∩$$B)=2/15$$ and $$P(A$$∩$$B)=1/6$$, then $$P(B)$$ is

Question

If A and B are two independent events such that $$P(A$$∩$$B)=2/15$$ and $$P(A$$∩$$B)=1/6$$, then $$P(B)$$ is

Infinity Learn · Accepted Answer

Let $$P(A)=x$$ and $$P(B)=y$$. Since A and B are  independent events, $$P(A$$∩$$B)=2/15$$⇒    $$P(A)P(B)=2/15$$⇒    (1$$−$$P(A))P(B)=2/15$$⇒    $$(1$$−$$x)y=2/15----1$$ and $$P(A$$∩$$B)=16$$⇒$$P(A)P(B)=16$$⇒ $$x(1$$−$$y)=16$$⇒ x−$$xy=16----2Subtracting$$ $$(1) from (2), we getx−$$y=130$$⇒$$x=130+yPutting$$ this value of x in (1), we get    y−$$y130+y=215$$⇒$$30y$$−y−$$30y2=2/5$$⇒     $$30y2$$−$$29y+4=0$$⇒$$(6y$$−$$1)(5y$$−$$4)=0$$⇒     $$y=1/6$$ or $$y=4/5$$⇒$$P(B)=1/16$$ or $$P(B)=4/5$$

$If A and B are two independent events such that P (A \cap B) = 2 / 15 and P (A \cap B) = 1 / 6, then P (B) is$

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If A and B are two independent events such that P(A∩B)=2/15 and P(A∩B)=1/6, then P(B) is

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$If A and B are two independent events such that P (A \cap B) = 2 / 15 and P (A \cap B) = 1 / 6, then P (B) is$