There are 20 books in a bookcase. If one book is selected at random

Question

There are 20 books in a bookcase. If one book is selected at random, what is the probability that the book is either a hardback or a novel?

(1) 8 books in the bookcase are novel books and 10 books are hardbacks.
(2) 3 books in the bookcase are hardback novels

DavidTutorexamPAL · Answer

We'll use numbers to help us understand the logic behind the question.
This is an Alternative approach.

(1) Say all 8 of the novels are hardbacks. Then 10 of the books are hardbacks or novels and the probability is 10/20. Say none of the novels are hardbacks. Then 18 books are novels or hardbacks and the answer is 18/20.
Insufficient.

(2) We have no information on the number of non-hardback novels and can pick easy numbers such as 0 or to 1. In the first case the probability is 3/20 and in the second 4/20.
Insufficient.

Combined:
So 3 books are hardback novels meaning that 10 - 3 = 7 are hardbacks that aren't novels and 8 - 3 = 5 are novels that aren't hardbacks.
So there are 3 + 7 + 5 = 15 books that are novels or hardbacks and the probability must be 15/20.
Sufficient.

(C) is our answer.

EgmatQuantExpert · Answer

Solution

Given:  
• Number of books in the bookcase = 20

To find:  
• If one book is selected at random, the probability of it being either hardback or novel

Analysing Statement 1  
• As per the information provided in Statement 1, 8 books are novel books and 10 books are hardbacks
 o But from this statement, we cannot find out whether there are any books which are novel and have hardcover or vice versa.

Hence, statement 1 is not sufficient to answer

Analysing Statement 2  
• As per the information provided in Statement 2, 3 novels are hardback
 o But we don’t know whether there are books which are only novel without hardcover or books with hardcover which are not novel.

Hence, statement 2 is not sufficient to answer

Combining Both Statements  
We already know 8 novel books and 10 hardbacks books are there, out of which 3 novels are hardbacks. 
• Therefore the probability =  \frac{(8 + 10 – 3)}{20} = \frac{3}{4}

Hence, the correct answer is option C.

Answer: C

jackspire · Answer

When a question asks the probability of an either-or event, then we have a simple formula -

P(A or B) = P(A) + P(B) - P(A and B)

1. From statement 1 we can derive P(A) and P(B), but we can't derive P(A & B) as we don't know the number of novel books with a hardback. So it's insufficient.

2. From statement 2 we can't derive P(A) and P(B), but we know the P(A & B). So it's insufficient.

When we combine both statements, we can have P(A or B).
So, C is the answer.

JeffTargetTestPrep · Answer

We need to answer the question:

P(hardback or novel) = ?

Probability = Favorable/Total = #(hardback or novel)/20 = ?

Statement One Alone:

=> 8 books in the bookcase are novel books and 10 books are hardbacks.

#(hardback or novel) = #(hardback) + #(novel) - #(hardback and novel) = 10 + 8 - #(hardback and novel)

We don’t know the exact value of #(hardback and novel). Statement one is not sufficient. Eliminate answer choices A and D.

Statement Two Alone:

=> 3 books in the bookcase are hardback novels.

#(hardback or novel) = #(hardback) + #(novel) – 3

We don’t know the exact value of #(hardback) + #(novel). Statement two is not sufficient. Eliminate answer choice B.

Statements One And Two Together:

#(hardback or novel) = 10 + 8 – 3 = 15

Probability = 15/20 = 3/4

The two statements together are sufficient.

Answer: C

pappal · Answer

sir value of neither hardcover nor novel is not known,
so IMO E is the answer. please correct me?

DUMDUM21 · Answer

The question stem never said that it has only two types of books in the bookcase. Shouldn't the answer be E?

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There are 20 books in a bookcase. If one book is selected at random

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