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2020 AMC 10B Problem 11

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2020 AMC 10B Problem 11AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 3 of 5 · about 3 min · position P11-15

Ms. Carr asks her students to read any 55 of the 1010 books on a reading list. Harold randomly selects 55 books from this list, and Betty does the same. What is the probability that there are exactly 22 books that they both select?

  1. A)

    18\frac{1}{8}

  2. B)

    536\frac{5}{36}

  3. C)

    1445\frac{14}{45}

  4. D)

    2563\frac{25}{63}

  5. E)

    12\frac{1}{2}

Answer

D

Key insight

Fix Harold's five books; Betty must choose 2 of those 5 and 3 of the other 5 out of C(10,5) equally likely selections.

Solution

Whatever Harold picks, the list splits into his 55 books and the other 55. Betty's selection is a uniformly random 55-subset, (105)=252\binom{10}{5} = 252 possibilities in all.

For exactly 22 books in common, Betty must take 22 of Harold's 55 and 33 of the remaining 55:
(52)(53)=10⋅10=100.\binom{5}{2}\binom{5}{3} = 10 \cdot 10 = 100.

The probability is 100252=2563\dfrac{100}{252} = \dfrac{25}{63}.

The answer is (D) 2563\boxed{\textbf{(D)}\ \frac{25}{63}}.

Why this works

When two people choose independently and uniformly, condition on the first choice: by symmetry the answer does not depend on which set Harold picked, so treat his set as fixed and count only Betty's options. The count is then a standard "choose some from group A and the rest from group B" product, a hypergeometric setup.

The trap

Trying to count both selections simultaneously, or forgetting the C(5,3) factor for the books Betty picks outside Harold's set.

Common mistakes

  • Trying to count both selections simultaneously, or forgetting the C(5,3) factor for the books Betty picks outside Harold's set.
  • Using (102)\binom{10}{2} for the shared books, or not reducing 100252\frac{100}{252} and failing to match a choice.

Techniques

Set up the equation/formula and compute; no special trick needed