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There are \(20\) books in a bookcase. If one book is selected at random, what is the probability that the book is either

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by AAPL » Mon Sep 07, 2026 5:53 pm
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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

OA C
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Source: — Data Sufficiency |

What we see: a Data Sufficiency probability question. There are \(20\) books. What is the chance a random book is a hardback or a novel?

Step 1. Write the rule for "or."
What we see: the question asks for hardback OR novel.
What it means: books that are hardback or novel \(=\) hardbacks \(+\) novels \(-\) books that are both. We subtract "both" because those books were counted once as hardbacks and again as novels.
What we do and why: we need all three numbers: hardbacks, novels, and hardback novels. Then we divide by \(20\).

Step 2. Statement 1 alone.
What we see: \(8\) novels and \(10\) hardbacks.
What it means: we do not know how many books are both. If none are, \(18\) books fit, a chance of \(\dfrac{18}{20}\). If all \(8\) novels are hardbacks, only \(10\) fit, a chance of \(\dfrac{10}{20}\).
What we do and why: two different answers. Not sufficient.

Step 3. Statement 2 alone.
What we see: \(3\) books are hardback novels.
What it means: we know the overlap but not how many hardbacks or novels there are in total.
What we do and why: not sufficient.

Step 4. Both statements together.
What we see: hardbacks \(10\), novels \(8\), both \(3\).
What it means: books that are hardback or novel \(= 10 + 8 - 3 = 15\).
What we do and why: the chance is \(\dfrac{15}{20} = \dfrac{3}{4}\). One answer, so together they are sufficient.

The trap: adding \(10 + 8 = 18\) from Statement 1 and calling it done. That counts every hardback novel twice, and Statement 1 never says how many there are.
Takeaway: "A or B" \(=\) A \(+\) B \(-\) both. Until you know the overlap, you cannot count "or."

Answer: C
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