(B) Given,$P(X) = 0.3$ and $P(Y) = 0.2$.Since the events are independent,the probability that both $X$ and $Y$ fail is $P(X \cap Y) = P(X) \times P(Y)

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(B) Given,$P(X) = 0.3$ and $P(Y) = 0.2$.Since the events are independent,the probability that both $X$ and $Y$ fail is $P(X \cap Y) = P(X) \times P(Y) = 0.3 \times 0.2 = 0.06$.The probability that either $X$ or $Y$ fails is given by the addition theorem of probability:$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$Substituting the values:$P(X \cup Y) = 0.3 + 0.2 - 0.06 = 0.44$.

Class 11 Mathematics Probability Odds in favor and odds against, Addition theorem on probability

Medium

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

IIT 1989 All IIT

A
$0.5$
B
$0.44$
C
$0.6$
D
None of these

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Odds in favor and odds against, Addition theorem on probability

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$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
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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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