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Let $X,Y$ be two firms. An MBA applies for the job in $X$ and $Y$. The probability he being selected in firm $X$ is $0.7$ and being rejected at $Y$ is $0.5$. Probability of at least one of his applications being rejected is $0.6$. What is the probability that he will be selected in one of the firms?

i.e.,

$P(X)=0.7$, $P(\bar{Y})=0.5$, $1-P(X\cap Y)=0.6$

So, my doubt is,

$P$(he will be selected in one of the firms)

=$P(X\cap \bar{Y})+P(\bar{X}\cap Y)$

or

=$P(X\cup Y)$ {but this means atleast one firm, am i correct}

So, what does the question actually mean.

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2 Answers 2

up vote 0 down vote accepted

I would interpret the question "what is the probability that he will be selected in one of the firms?" as "what is the probability that he will be selected in $X$ or $Y$?" with an inclusive 'or' ($X$ or $Y$ or both). Then the probability you want is $P(X \cup Y)$.

If the other option was the case, I feel that it would be more natural to ask "what is the probability that he will be selected in only one of the firms?".

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Your first suggestion is the formula for "accepted at exactly one firm" (but not at both), the second for "accepted in at least one firm" (and even possibly at both). I vote for the latter.

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but the question says one of the firms. In such case how can you consider the possibility at both? –  amul28 Mar 27 '11 at 14:02
1  
@amul28: If you are accepted at both firms then you have been accepted to one firm. –  PEV Mar 27 '11 at 14:05
    
oh ok, i got it. –  amul28 Mar 27 '11 at 14:15

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