AMC 8 · 2007 · #21

Grade 7 probability

probability-basiccombinations-basiccomplementary-counting caseworkcomplementary-counting ↑ Prerequisites: probability-basiccombinations-basic

📏 Short solution 💡 2 insights

Problem

Two cards are dealt from a deck of four red cards labeled $A$ , $B$ , $C$ , $D$ and four green cards labeled $A$ , $B$ , $C$ , $D$ . A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?

Pick an answer.

(A)

$\frac{2}{7}$

(B)

$\frac{3}{8}$

(C)

$\frac{1}{2}$

(D)

$\frac{4}{7}$

(E)

$\frac{5}{8}$

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Toolkit + CCSS Solution

Understand

Restated: A deck has $8$ cards: red $A, B, C, D$ and green $A, B, C, D$. Two cards are dealt at random. A "winning pair" means the two cards share a color or share a letter. Find the probability of a winning pair.

Givens: Four red cards labeled $A, B, C, D$; Four green cards labeled $A, B, C, D$; Two cards are dealt from the deck of $8$; A winning pair is two of the same color OR two of the same letter; Answer choices: (A) $\tfrac{2}{7}$, (B) $\tfrac{3}{8}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{4}{7}$, (E) $\tfrac{5}{8}$

Unknowns: The probability that the two dealt cards form a winning pair

Understand

Plan

Primary tool: #13 Count Systematically

Secondary: #7 Split into Subproblems

Probability with a small deck is a Grade 7 counting question: (favorable outcomes) $\div$ (total outcomes). Tool #13 (Count Systematically) fits because we can anchor on the first card and count how many of the remaining $7$ cards win with it. That "anchor and count" works for every starting card because of symmetry. As a check, Tool #7 (Split into Subproblems) breaks the winning event into two disjoint cases — "same letter" and "same color" — and adds their probabilities.

Execute — Answer: D

#13 Count Systematically 7.SP.C.7 Step 1

Anchor on the first card.
Whatever card is dealt first, there are $7$ cards left in the deck, all equally likely to be the second card.
So the probability is (number of winning second cards) $\div\, 7$.

$$P(\text{winning pair}) = \dfrac{\text{winning second cards}}{7}$$

💡 Grade 7 probability of a uniform model: each of the $7$ remaining cards has the same chance, so we just count the good ones.

#13 Count Systematically 7.SP.C.8 Step 2

Count the second cards that share the color.
Say the first card is red $A$.
The other three red cards — red $B$, red $C$, red $D$ — share its color.
That gives $3$ same-color matches.

$$\#\{\text{same color}\} = 3$$

💡 Each color has $4$ cards, so after removing the first card, $3$ same-color cards remain.

#13 Count Systematically 7.SP.C.8 Step 3

Count the second cards that share the letter.
The only other card with letter $A$ is green $A$.
That gives $1$ same-letter match, and it does not overlap with the same-color matches.

$$\#\{\text{same letter}\} = 1$$

💡 Each letter appears on exactly $2$ cards (one red, one green), so after removing the first card, $1$ same-letter card remains.

#7 Split into Subproblems 7.SP.C.8 Step 4

Add the disjoint counts and divide by $7$.
Same-color and same-letter matches are different cards (the first card is the only one that is both), so they do not overlap.
Total winners $= 3 + 1 = 4$.

$$P(\text{winning pair}) = \dfrac{3 + 1}{7} = \dfrac{4}{7} \;\Rightarrow\; \textbf{(D)}$$

💡 Grade 7 "probability of a compound event" by listing favorable outcomes — here, the $4$ second cards that produce a win.

[1] #13 7.SP.C.7 Anchor on the first card. Whatever card is dealt first, there are $7$ cards left

[2] #13 7.SP.C.8 Count the second cards that share the color. Say the first card is red $A$. The

[3] #13 7.SP.C.8 Count the second cards that share the letter. The only other card with letter $A

[4] #7 7.SP.C.8 Add the disjoint counts and divide by $7$. Same-color and same-letter matches ar

Review

Reasonableness: Cross-check by counting unordered pairs. Total pairs from $8$ cards: $\binom{8}{2} = 28$. Same-color pairs: $\binom{4}{2} + \binom{4}{2} = 6 + 6 = 12$. Same-letter pairs: one pair per letter, $4$ letters, so $4$. These sets do not overlap (two distinct cards cannot share both color and letter), so winners $= 12 + 4 = 16$. Probability $= \dfrac{16}{28} = \dfrac{4}{7}$, matching (D). Also $\tfrac{4}{7} \approx 0.57$, a bit above one-half — reasonable since each card already has $4$ partners out of $7$.

Alternative: Tool #11 (Find an Invariant): notice that no matter which card is drawn first, the count of winning partners is always $3 + 1 = 4$. The win probability does not depend on the first card, so by symmetry it equals $\dfrac{4}{7}$ directly — no casework, just the invariance of the count.

CCSS standards used (min grade 7)

7.SP.C.7 Develop a probability model and use it to find probabilities of events (Modeling the second card as uniformly chosen from the $7$ remaining cards, so probability $=$ (favorable count) $\div\, 7$.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Counting the favorable second cards by listing same-color cards ($3$) and same-letter cards ($1$) and adding the disjoint counts.)

⭐ Anchor on the first card, then count the winners among the $7$ left: $3$ share the color, $1$ shares the letter, total $4$ out of $7$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

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