To find a book, a student decided to visit three libraries.…
To find a book, a student decided to visit three libraries. For each library it is equally likely that the book is or is not in its collection; if the library has the book, it is equally likely that the book is or is not checked out by other readers. The libraries acquire books independently of each other. What is more likely — that the student will find the book or will not?
Explanation of Method
In one library the student gets the book iff the library has it and it is not checked out: $\frac12\cdot\frac12=\frac14$. The student fails to find the book only if all three independent libraries fail, with probability $\left(\frac34\right)^3$. Comparing this with $\frac12$ answers the question.
Step 1: One library
$H$ = "the library has the book", $F$ = "the book is available (not checked out)". Given: $P(H)=\frac12$, $P(F\mid H)=\frac12$.
\[ P(\text{book obtained in one library}) = P(H)\,P(F\mid H) = \frac12\cdot\frac12=\frac14. \]
Step 2: Three independent libraries
The student does not get the book only if it is unavailable in all three:
\[ P(\text{not found}) = \left(1-\frac14\right)^3 = \frac{27}{64}\approx0.422. \]
Step 3: The answer
\[ \boxed{P(\text{found}) = 1-\frac{27}{64}=\frac{37}{64}\approx0.578\gt\frac12.} \]
It is more likely that the student will find the book.
Step 4: How many libraries are needed?
| Libraries | $P(\text{found})=1-(3/4)^k$ |
|---|---|
| $1$ | $0.250$ |
| $2$ | $0.438$ |
| $3$ | $0.578$ |
| $4$ | $0.684$ |
| $5$ | $0.763$ |
With two libraries the chances are still below $\frac12$ ($0.4375$); with three they exceed $\frac12$.
Result
| Answer | |
|---|---|
| One library | $1/4$ |
| $P(\text{found})$ | $1-(3/4)^3=37/64\approx0.578$ |
| Conclusion | finding the book is more likely |