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A shop sells cigars of brand A as well as brand B. Let X be the event

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A shop sells cigars of brand A as well as brand B. Let X be the event  [#permalink]

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New post 08 Jan 2019, 22:40
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  95% (hard)

Question Stats:

33% (02:12) correct 67% (01:54) wrong based on 60 sessions

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A shop sells cigars of brand A as well as brand B, among other brands. Let X be the event that a customer buys at least one cigar of brand A, let Y be the event that a customer buys at least one cigar of brand B, and let Z be the event that a customer buys cigar of neither brand A nor brand B. Probability of event X is denoted as P(X), probability of event Y is denoted as P(Y), probability of event Z is denoted as P(Z). What is the probability of X U Y, P(XUY) i.e., what is the probability of either event X or event Y.

(1) P(Z) = 0.35.

(2) P(X) = 0.40 and P(Y) = 0.38.
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A shop sells cigars of brand A as well as brand B. Let X be the event  [#permalink]

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New post Updated on: 10 Jan 2019, 08:48
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#1:

p(XuY) = p(X) + p(Y) - p(XnY)
or
p(XuY) = 1 - p(XnY)' = 1 - Probability of neither of the events occurring = 1 - p(Z)
P(Z) = 0.35 - Sufficient


#2:
P(x)=P(a)=0.4 and P (y)=P(b)=0.38
P(a)+P(b)= 0.4+.38 = 0.78; not sufficient

IMO A


amanvermagmat wrote:
A shop sells cigars of brand A as well as brand B, among other brands. Let X be the event that a customer buys at least one cigar of brand A, let Y be the event that a customer buys at least one cigar of brand B, and let Z be the event that a customer buys cigar of neither brand A nor brand B. Probability of event X is denoted as P(X), probability of event Y is denoted as P(Y), probability of event Z is denoted as P(Z). What is the probability of X U Y, P(XUY) i.e., what is the probability of either event X or event Y.

(1) P(Z) = 0.35.

(2) P(X) = 0.40 and P(Y) = 0.38.

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Originally posted by Archit3110 on 09 Jan 2019, 06:28.
Last edited by Archit3110 on 10 Jan 2019, 08:48, edited 1 time in total.
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Re: A shop sells cigars of brand A as well as brand B. Let X be the event  [#permalink]

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New post 09 Jan 2019, 11:31
amanvermagmat wrote:
A shop sells cigars of brand A as well as brand B, among other brands. Let X be the event that a customer buys at least one cigar of brand A, let Y be the event that a customer buys at least one cigar of brand B, and let Z be the event that a customer buys cigar of neither brand A nor brand B. Probability of event X is denoted as P(X), probability of event Y is denoted as P(Y), probability of event Z is denoted as P(Z). What is the probability of X U Y, P(XUY) i.e., what is the probability of either event X or event Y.

(1) P(Z) = 0.35.

(2) P(X) = 0.40 and P(Y) = 0.38.


p(XuY) = p(X) + p(Y) - p(XnY)
or
p(XuY) = 1 - p(XnY)' = 1 - Probability of neither of the events occurring = 1 - p(Z)

(1) P(Z) = 0.35 - Sufficient

(2) P(X) = 0.40 and P(Y) = 0.38 - We don't know p(XnY) - Not sufficient.

Hence I'll go with A.

Cheers!
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Re: A shop sells cigars of brand A as well as brand B. Let X be the event   [#permalink] 09 Jan 2019, 11:31
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