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If the probability of $X$ to fail in the examination is 0.3 and that for $Y$ is 0.2 , then the probability that either $X$ or $Y$ fail in the examination is
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The correct answer is:
$0.44$
Here $P(X)=0.3 ; \quad P(Y)=0.2$
Now $P(X \cup Y)=P(X)+P(Y)-P(X \cap Y)$
Since these are independent events, so
$P(X \cap Y)=P(X) \cdot P(Y)$
Thus required probability $=0.3+0.2-0.06=0.44$
Now $P(X \cup Y)=P(X)+P(Y)-P(X \cap Y)$
Since these are independent events, so
$P(X \cap Y)=P(X) \cdot P(Y)$
Thus required probability $=0.3+0.2-0.06=0.44$
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