On a shelf there are 60 novels and 20 poetry books. Person A chooses

Question

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

CrackverbalGMAT · Accepted Answer

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

Akp880 · Answer

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

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.

TestPrepUnlimited · Answer

This is a typical conditional probability problem, the formula we are interested in is:

P(A Poetry, given B novel) = \frac{P(A Poetry & B Novel) }{ P(B Novel)} .

The probability that A took Poetry and B took Novel is  \frac{20}{80} * \frac{60}{79} = \frac{15}{79} .

The probability that B took a novel however depends on if A took a novel or not, so we need to separate into two cases:

\frac{60}{80}*\frac{59}{79} + \frac{20}{80} * \frac{60}{79} = \frac{3}{4}*\frac{59}{79} + \frac{15}{79} = \frac{3*59 + 60}{4*79} .

Finally our answer is  \frac{15}{79} *\frac{4*79}{3*59+60} = \frac{60}{3*59+60} = \frac{20}{59+20} = \frac{20}{79} .

Ans: B

HWPO · Answer

Theoretically, shouldn't the probability be 20/80 - 1/4 ? 
Person B chooses a book only AFTER person A does. So if person A is to first choose a poetry and there are currently 80 books, wouldn't the probability simply be 1/4?

Dragon805 · Answer

Option B is the correct answer.

I solved it by reasoning it in the following way: if we know that person B chose a novel (lets say specifically : "The Republic by Plato") then we know for sure person A did not choose this book. Thus person A in effect had 59 novels and 20 poetry books to choose from. Therefore, the probability of person A choosing a poetry book would be 20/(59+20) = 20/79

Pr4n · Answer

But I think even if you reached the right answer, it is almost similar to the fact if A came after B i.e., B already made his pick and as these events are dependent on each other hence probabilty of A is 20/79. Even though Baye's theorem also gives this answer, but I don't think this kind of pattern will always hold. And if it does, then no need to do baye's theorem complex calculations

Dragon805 · Answer

You would not be able to solve it the way I did and will have to use Bayes theorem in questions where the probability is directly given like lets say P(B) = 0.45 but the number of favourable outcomes and total number of outcomes is unknown (it could have been 45, 100 or 90, 200 etc.. we don't know) then in cases like this the only way to solve for conditional probability say probability of A given B is using Bayes theorem. But the way I solved it will always work for questions like the one we are concerned with where we also have information about the number of favourable outcomes as well as the number of total outcomes. Solving it this way will also save you a lot of time on your test day!

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