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21 Sep 2020, 05:15
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B
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69% (02:05) correct
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On a shelf there are 60 novels and 20 poetry books. Person A chooses a book at random off the shelf and leaves with it. Shortly after, Person B chooses another book at random. If it is known that Person B chose a novel, what is the probability that the book selected by Person A was a poetry book?
A. 59/237
B. 20/79
C. 59/79
D. 3/4
E. 60/79
Are You Up For the Challenge: 700 Level Questions
A. 59/237
B. 20/79
C. 59/79
D. 3/4
E. 60/79
Are You Up For the Challenge: 700 Level Questions
21 Sep 2020, 07:54
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Probability = \(\frac{Favorable \space Outcomes}{Total \space Outcomes}\)
By Bayes Theorem, P(A/B) = \(\frac{P(A \space and \space B)}{P(B)}\)
To Find: The probability that A chose a Poetry Given that B chooses a Novel = \(\frac{P(A \space chooses \space Poetry \space and \space then \space B \space chooses \space a \space novel)}{P(B \space Chooses \space a \space Novel)}\)
P(A chooses a Poetry book) = \(\frac{20}{80} = \frac{1}{4}\)
P(B Chooses a Novel after A chooses a poetry book) = \(\frac{60}{79}\)
Therefore P(A chooses a Poetry book and then B chooses a Novel) = \(\frac{1}{4} * \frac{60}{79} = \frac{60}{316}\)
P(B Chooses a Novel) = P(A chooses poetry and B chooses Novel) + P(A chooses Novel and B chooses Novel)
= \((\frac{1}{4} * \frac{60}{79}) + (\frac{3}{4} * \frac{59}{79}) = \frac{60}{316} + \frac{157}{316} = \frac{237}{316}\)
Therefore the required probability = \(\frac{\frac{60}{316}}{\frac{237}{316}} = \frac{60}{237} = \frac{20}{79}\)
Option B
Arun Kumar
By Bayes Theorem, P(A/B) = \(\frac{P(A \space and \space B)}{P(B)}\)
To Find: The probability that A chose a Poetry Given that B chooses a Novel = \(\frac{P(A \space chooses \space Poetry \space and \space then \space B \space chooses \space a \space novel)}{P(B \space Chooses \space a \space Novel)}\)
P(A chooses a Poetry book) = \(\frac{20}{80} = \frac{1}{4}\)
P(B Chooses a Novel after A chooses a poetry book) = \(\frac{60}{79}\)
Therefore P(A chooses a Poetry book and then B chooses a Novel) = \(\frac{1}{4} * \frac{60}{79} = \frac{60}{316}\)
P(B Chooses a Novel) = P(A chooses poetry and B chooses Novel) + P(A chooses Novel and B chooses Novel)
= \((\frac{1}{4} * \frac{60}{79}) + (\frac{3}{4} * \frac{59}{79}) = \frac{60}{316} + \frac{157}{316} = \frac{237}{316}\)
Therefore the required probability = \(\frac{\frac{60}{316}}{\frac{237}{316}} = \frac{60}{237} = \frac{20}{79}\)
Option B
Arun Kumar
General Discussion
21 Sep 2020, 07:41
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On a shelf there are 60 novels and 20 poetry books. Person A chooses a book at random off the shelf and leaves with it. Shortly after, Person B chooses another book at random. If it is known that Person B chose a novel, what is the probability that the book selected by Person A was a poetry book?
A. 59/237
B. 20/79
C. 59/79
D. 3/4
E. 60/79
Explanation-
This question is a typical example of conditional probability(Revise the topic if you have forgetten)
let x=novel,y=poetry
Then Probability that A chooses poetry book given B chose Novel is,
P(y/x)= P(y& x)/P(x)
Now Probability that A chooses poetry book AND B choose Novel is,
P(y&x)=(20/80)*(60/79)
Probability that B chooses novel is,
P(x)=P(A chooses poetry & B chooses Novel)+P(A chooses novel & B chooses Novel)
P(x)=(20/80)*(60/79)+(60/80)*(59/79)
Thus from P(y/x)= P(y& x)/P(x)
P(y/x)=(20/80)*(60/79)/(20/80)*(60/79)+(60/80)*(59/79)
=1200/4740
=20/79
Hence B is the correct answer.
A. 59/237
B. 20/79
C. 59/79
D. 3/4
E. 60/79
Explanation-
This question is a typical example of conditional probability(Revise the topic if you have forgetten)
let x=novel,y=poetry
Then Probability that A chooses poetry book given B chose Novel is,
P(y/x)= P(y& x)/P(x)
Now Probability that A chooses poetry book AND B choose Novel is,
P(y&x)=(20/80)*(60/79)
Probability that B chooses novel is,
P(x)=P(A chooses poetry & B chooses Novel)+P(A chooses novel & B chooses Novel)
P(x)=(20/80)*(60/79)+(60/80)*(59/79)
Thus from P(y/x)= P(y& x)/P(x)
P(y/x)=(20/80)*(60/79)/(20/80)*(60/79)+(60/80)*(59/79)
=1200/4740
=20/79
Hence B is the correct answer.
