Competition · AMC preparation · step 4 of 4

AMC 10 · 2020B · #11

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype Two-Set Inclusion-Exclusion: Solve for Missing Region

combinations-basicprobability-basicfraction-arithmetic easier-related-problemidentify-subproblemscomplementary-counting ↑ Prerequisites: combinations-basicprobability-basic

📏 Medium solution 💡 2 insights

Problem

There is a reading list of 10 books. Harold picks an unordered set of 5 books at random; independently, Betty picks an unordered set of 5 books at random. What is the probability that Harold's set and Betty's set share exactly 2 books?

Pick an answer.

(A)

$\frac{1}{8}$

(B)

$\frac{5}{36}$

(C)

$\frac{14}{45}$

(D)

$\frac{25}{63}$

(E)

$\frac{1}{2}$

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem): the symmetry lets us freeze Harold's 5 books in place and only worry about Betty's choice — the answer doesn't depend on which 5 books Harold actually picked. Tool #7 (Subproblems): split Betty's pick into two independent sub-counts — '2 books from Harold's 5' and '3 books from the other 5' — and multiply. Tool #3 (Eliminate): match the simplified fraction to the five answer choices.

1STEP 1

Fix Harold's five books

By symmetry, fix Harold's 5 books as group H and the other 5 as group N — the answer ignores which books he picked.

|H| = 5, |N| = 5

2STEP 2

Count all of Betty's picks

Betty picks 5 of the 10 books; the denominator counts all such sets: 252.

C(10, 5) = 252

3STEP 3

Count the favorable picks

Favorable: Betty takes 2 books from H and 3 from N; the sub-choices are independent, so multiply to 100.

C(5, 2) · C(5, 3) = 10 · 10 = 100

4STEP 4

Form and simplify the probability

Form $\frac{100}{252}$ and reduce by 4 to get the probability $\frac{25}{63}$ .

100/252 = 25/63

5STEP 5

Match against the choices

Only choice (D) equals $\frac{25}{63}$ ; the other denominators can't come from 252.

25/63 → (D)

Answer

25/63

\frac{25}{63}

≈ 0.397, which is the largest single share count out of 0,1,2,3,4,5 shared books — and that makes sense: with 5 picks from 10, a Betty draw shares about

\frac{(5 · 5)}{10}

= 2.5 books on average, so 'exactly 2' should be the most common outcome. The answer being close to

\frac{2}{5}

matches intuition.

💡Key takeaway

This AMC 10 problem only needs Grade 7 probability you already know! Lock in Harold's 5 books and ask: how often does Betty's 5-pick share exactly 2 with Harold's 5? Betty needs 2 from Harold's 5 (C(5, 2) = 10 ways) and 3 from the other 5 (C(5, 3) = 10 ways), so 10 · 10 = 100 good picks out of C(10, 5) = 252 total. Simplify $\frac{100}{252} = \frac{25}{63}$ , answer (D).

Fix Harold's five books
Count all of Betty's picks
Count the favorable picks
Form and simplify the probability
Match against the choices

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Fix Harold's five books

Count all of Betty's picks

Count the favorable picks

Form and simplify the probability

Match against the choices

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