AMC 10 · 2020 · #11

Grade 7 probability

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

📏 Medium solution 💡 2 insights

Problem

Ms. Carr asks her students to read any $5$ of the $10$ books on a reading list. Harold randomly selects $5$ books from this list, and Betty does the same. What is the probability that there are exactly $2$ books that they both select?

Pick an answer.

(A)

$\frac{1}{8}$

(B)

$\frac{5}{36}$

(C)

$\frac{14}{45}$

(D)

$\frac{25}{63}$

(E)

$\frac{1}{2}$

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Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: Reading list has $10$ books; Harold picks a $5$-book subset uniformly at random; Betty picks a $5$-book subset uniformly at random, independently of Harold; Answer choices: $\tfrac{1}{8},\; \tfrac{5}{36},\; \tfrac{14}{45},\; \tfrac{25}{63},\; \tfrac{1}{2}$

Unknowns: $P(\text{exactly } 2 \text{ shared books})$

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities

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.

Execute — Answer: D

#9 Solve an Easier Related Problem 7.SP.C.7 Step 1

Freeze Harold's choice.
Because every $5$-subset is equally likely for both choosers, the probability that Betty shares exactly $2$ books with Harold does not depend on which $5$ Harold picked.
So treat Harold's $5$ books as a fixed labeled group H (the 'on-list' books) and the other $5$ as group N (the 'off-list' books).

$$|H| = 5,\;\; |N| = 5$$

💡 Grade 7 probability: by symmetry, only Betty's draw matters; Harold's identities cancel.

#7 Identify Subproblems 7.SP.C.8 Step 2

Count the total number of $5$-subsets Betty can pick from the $10$ books.
This is the denominator of the probability.

$$\binom{10}{5} = 252$$

💡 Grade 7 compound events: count all equally likely outcomes for Betty.

#7 Identify Subproblems 7.SP.C.8 Step 3

Count the favorable picks.
Betty must take exactly $2$ books from H and exactly $3$ books from N.
The two sub-choices are independent, so multiply.

$$\binom{5}{2} \cdot \binom{5}{3} = 10 \cdot 10 = 100$$

💡 Grade 7 compound events: independent sub-choices multiply.

#7 Identify Subproblems 4.NF.A.1 Step 4

Form the probability and simplify.
Divide $100$ by $252$ — both share a factor of $4$.

$$\dfrac{100}{252} = \dfrac{25}{63}$$

💡 Grade 4 equivalent fractions: divide top and bottom by $4$.

#3 Eliminate Possibilities 4.NF.A.2 Step 5

Match $\tfrac{25}{63}$ to the choices: option (D).
The other choices have wrong denominators ($8, 36, 45, 2$) that would never arise from $\binom{10}{5} = 252$.

$$\dfrac{25}{63} \;\Rightarrow\; \textbf{(D)}$$

💡 Grade 4 fraction comparison: only one option equals $\tfrac{25}{63}$.

[1] #9 7.SP.C.7 Freeze Harold's choice. Because every $5$-subset is equally likely for both choo

[2] #7 7.SP.C.8 Count the total number of $5$-subsets Betty can pick from the $10$ books. This i

[3] #7 7.SP.C.8 Count the favorable picks. Betty must take exactly $2$ books from H and exactly

[4] #7 4.NF.A.1 Form the probability and simplify. Divide $100$ by $252$ — both share a factor o

[5] #3 4.NF.A.2 Match $\tfrac{25}{63}$ to the choices: option (D). The other choices have wrong

Review

Reasonableness: $\tfrac{25}{63} \approx 0.397$, 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 $\tfrac{5 \cdot 5}{10} = 2.5$ books on average, so 'exactly $2$' should be the most common outcome. The answer being close to $\tfrac{2}{5}$ matches intuition.

Alternative: Tool #6 (Guess and Check) via the full distribution. List $P(k \text{ shared}) = \binom{5}{k}\binom{5}{5-k} / \binom{10}{5}$ for $k = 0,1,2,3,4,5$: numerators $1, 25, 100, 100, 25, 1$ summing to $252$. This not only confirms $\tfrac{100}{252} = \tfrac{25}{63}$ but also verifies the count is right because the six cases sum to $1$.

CCSS standards used (min grade 7)

4.NF.A.1 Explain why a fraction is equivalent to another fraction (Reducing $\tfrac{100}{252}$ to $\tfrac{25}{63}$ by dividing top and bottom by $4$.)
4.NF.A.2 Compare two fractions with different numerators and different denominators (Matching the simplified fraction $\tfrac{25}{63}$ to the five answer choices.)
7.SP.C.7 Develop probability models and use them to find probabilities of events (Using the symmetry of uniformly random $5$-subsets to freeze Harold's choice.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Counting Betty's favorable picks via $\binom{5}{2}\binom{5}{3}$ and dividing by $\binom{10}{5}$.)

⭐ 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 $5$ ($\binom{5}{2} = 10$ ways) and $3$ from the other $5$ ($\binom{5}{3} = 10$ ways), so $10 \cdot 10 = 100$ good picks out of $\binom{10}{5} = 252$ total. Simplify $\tfrac{100}{252} = \mathbf{\tfrac{25}{63}}$, answer (D).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

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