AMC 10 · 2008 · #16

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype Parity Partition: Complement via Odd-Even Split

probability-basicparitycasework casework ↑ Prerequisites: probability-basic

📏 Long solution 💡 3 insights

Problem

Two fair coins are tossed. For every coin that lands heads, one fair die is rolled. Add up the numbers on the dice that were rolled (if no die is rolled, the total is 0). Find the probability that this total is odd.

Pick an answer.

(A)

$\ {{{\frac{3} {8}}}}$

(B)

$\ {{{\frac{1} {2}}}}$

(C)

$\ {{{\frac{43} {72}}}}$

(D)

$\ {{{\frac{5} {8}}}}$

(E)

$\ {{{\frac{2} {3}}}}$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

How many dice get rolled is itself random, so tool #7 (Identify Subproblems) leads: split the problem by the number of heads — 0, 1, or 2 — because each case has a different, much simpler sub-question about parity. Tool #2 (Make a Systematic List) pins the chance of each case from the coin toss: 0 heads and 2 heads each happen 1/4 of the time, 1 head happens 1/2 of the time. Tool #5 (Look for a Pattern) handles the two-dice case, where a clean parity fact — the sum of two dice is odd exactly when the two dice disagree in parity — gives 1/2 without listing all 36 outcomes. Weighting each case's odd-probability by how often the case occurs and adding gives the answer.

1STEP 1

Split by number of heads

The dice count equals the head count. Two coins give 0 heads with chance $\frac{1}{4}$ , 1 head $\frac{1}{2}$ , 2 heads $\frac{1}{4}$ .

P(0)=1/4, P(1)=1/2, P(2)=1/4

2STEP 2

Zero heads: sum is 0

Both tails means no die at all, so the total is 0 — even. This case can never produce an odd sum.

P(odd ∣ 0 heads)=0

3STEP 3

One head: one die

One head means one die. Three of its six faces are odd, so the total is odd with chance $\frac{1}{2}$ .

P(odd ∣ 1 head)=3/6=1/2

4STEP 4

Two heads: two dice

Two dice sum to an odd number exactly when their parities differ, and the second die differs from the first with chance $\frac{1}{2}$ .

P(odd ∣ 2 heads)=1/2

5STEP 5

Weight each case

Multiply each case's chance by its odd-sum chance: $\frac{1}{4}\times 0=0$ , $\frac{1}{2}\times\frac{1}{2}=$ $\frac{1}{4}$ , $\frac{1}{4}\times\frac{1}{2}=$ $\frac{1}{8}$ .

0, 1/2·1/2=1/4, 1/4·1/2=1/8

6STEP 6

Add the cases

Add the three weighted pieces over a common denominator of 8: $0+\frac{2}{8}+\frac{1}{8}=$ $\frac{3}{8}$ , choice (A).

0+1/4+1/8=2/8+1/8=3/8

Answer

{{{3/8}}}

The answer 3/8 is below 1/2, which makes sense: the 0-heads case (probability 1/4) can never give an odd sum, dragging the total down from the 1/2 you would get if a die were always rolled. It is a legal probability between 0 and 1, and it matches choice (A). As a sanity check, the chance of an even sum is 1-3/8=5/8, exactly choice (D); the extra weight on 'even' comes entirely from the no-die case, confirming the arithmetic.

💡Key takeaway

When the number of dice is itself random, split by how many dice get rolled, find the odd-sum chance in each case, and weight each by how often that case happens.

Split by number of heads
Zero heads: sum is 0
One head: one die
Two heads: two dice
Weight each case
Add the cases