AMC 10 · 2005 · #21
Grade 7 probabilityForty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let p be the probability that all four slips bear the same number. Let q be the probability that two of the slips bear a number a and the other two bear a number b=a. What is the value of q/p?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A hat holds forty slips: the numbers $1$ through $10$, each printed on exactly four slips. Draw four slips at random without replacement. Let $p$ be the chance all four show the same number, and $q$ the chance two show one number $a$ and the other two show a different number $b$. Find $q/p$.
Givens: Forty slips total, numbers $1$ to $10$, each number on exactly four slips; Four slips are drawn at once, without replacement; $p$ = probability all four drawn slips share one number; $q$ = probability the four split as two of number $a$ and two of number $b\neq a$; Answer choices: (A) $162$, (B) $180$, (C) $324$, (D) $360$, (E) $720$
Unknowns: The ratio $q/p$ of the two probabilities
Understand
Restated: A hat holds forty slips: the numbers $1$ through $10$, each printed on exactly four slips. Draw four slips at random without replacement. Let $p$ be the chance all four show the same number, and $q$ the chance two show one number $a$ and the other two show a different number $b$. Find $q/p$.
Givens: Forty slips total, numbers $1$ to $10$, each number on exactly four slips; Four slips are drawn at once, without replacement; $p$ = probability all four drawn slips share one number; $q$ = probability the four split as two of number $a$ and two of number $b\neq a$; Answer choices: (A) $162$, (B) $180$, (C) $324$, (D) $360$, (E) $720$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus / Count the Complement, #2 Make a Systematic List
Both $p$ and $q$ are (favorable draws) divided by the same total number of four-slip draws, $\binom{40}{4}$. Tool #16 (Change Focus) turns the whole problem on this fact: in the ratio $q/p$ that shared total cancels, so $q/p$ equals just the count of two-pair draws divided by the count of four-of-a-kind draws — the giant $\binom{40}{4}$ never has to be computed. That leaves two clean counting subproblems (Tool #7), each handled by Tool #2 (Make a Systematic List): count the four-of-a-kind draws, then count the two-and-two draws, and divide.
Execute — Answer: A
6.RP.A.1 Step 1 Both chances share one total
- Every choice of four slips out of forty is equally likely, and there are $\binom{40}{4}$ such choices in all.
- So $p$ is (number of four-of-a-kind draws) divided by $\binom{40}{4}$, and $q$ is (number of two-and-two draws) divided by the very same $\binom{40}{4}$.
- When you form the ratio $q/p$, that common bottom cancels.
- This means $q/p$ is just one count divided by another count — you never need the actual value of $\binom{40}{4}$.
💡 Two fractions with the same denominator compare by their tops alone, so the huge total drops out of the ratio.
7.SP.C.8 Step 2 Count four-of-a-kind draws
- For all four slips to match, they must be the entire group of four slips carrying one number.
- First pick which number it is: there are $10$ choices, one for each number $1$ through $10$.
- Once the number is chosen, the four slips are forced — you must take all four of them, and there is exactly one way to do that.
- So the number of four-of-a-kind draws is $N_p = 10$.
💡 A perfect match uses up a number's whole set of four slips, so only the choice of number is free.
7.SP.C.8 Step 3 Count two-and-two draws
- Now build a draw with two of number $a$ and two of number $b$.
- First choose the pair of numbers $\{a,b\}$: since order does not matter, that is $\binom{10}{2}=45$ ways.
- From number $a$'s four slips, pick which two you take: $\binom{4}{2}=6$ ways.
- From number $b$'s four slips, likewise $\binom{4}{2}=6$ ways.
- These independent choices multiply, so $N_q = 45 \times 6 \times 6 = 1620$.
💡 Pick the two numbers, then pick two slips from each number's four — the choices stack by multiplication.
6.RP.A.1 Step 4 Divide the two counts
- From Step 1, $q/p = N_q / N_p$.
- Substitute the two counts: $1620$ divided by $10$ is $162$.
- So $q/p = 162$, which is choice (A).
- Notice the answer is a clean whole number, exactly as expected when a shared total cancels away.
💡 With the totals gone, the ratio is just the bigger favorable count over the smaller one.
6.RP.A.1 Every choice of four slips out of forty is equally likely, and there are $\binom 7.SP.C.8 For all four slips to match, they must be the entire group of four slips carryin 7.SP.C.8 Now build a draw with two of number $a$ and two of number $b$. First choose the 6.RP.A.1 From Step 1, $q/p = N_q / N_p$. Substitute the two counts: $1620$ divided by $10 Review
Reasonableness: The value $162$ is a whole number, which fits: because $p$ and $q$ share the denominator $\binom{40}{4}$, the ratio had to reduce to $N_q/N_p$ with $N_p=10$ dividing evenly into $N_q=1620$. It is also reasonable that $q$ is far larger than $p$ — a two-and-two split can happen in many more ways ($1620$) than a rare perfect match ($10$), so $q/p \gg 1$. The trap answers come from miscounting: forgetting that $\{a,b\}$ is unordered doubles $1620$ to $3240$ and gives $324$ (C); using $2\binom{10}{2}$ style slips-order overcounts toward $360$ (D) or $720$ (E). Only (A) matches the careful count.
Alternative: Skip counting and use probabilities directly. Draw slips one at a time. For $p$, after any first slip, the next three must match it: $\frac{3}{39}\cdot\frac{2}{38}\cdot\frac{1}{37}$. For $q$, the first slip sets $a$; one of the next three positions must be $a$'s partner and the two others form the $b$ pair — carefully, $q = \binom{3}{1}\cdot\frac{3}{39}\cdot\frac{36}{38}\cdot\frac{3}{37}$. Dividing, all the $/39,/38,/37$ factors cancel and $q/p = \frac{3\cdot 3\cdot 36 \cdot 3}{3\cdot 2\cdot 1} = 162$, the same (A).
CCSS standards used (min grade 7)
6.RP.A.1Understand the concept of a ratio and use ratio language (Recognizing that $q/p$ equals the ratio of favorable counts $N_q/N_p$ once the shared total $\binom{40}{4}$ cancels, then evaluating $1620/10=162$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Counting the $10$ four-of-a-kind draws and the $\binom{10}{2}\binom{4}{2}\binom{4}{2}=1620$ two-and-two draws as compound events.)
⭐ Since $p$ and $q$ are both counts over the same total, $q/p$ is just one count over the other: $\frac{1620}{10}=162$.
⭐ Since $p$ and $q$ are both counts over the same total, $q/p$ is just one count over the other: $\frac{1620}{10}=162$.
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