AMC 10 · 2004 · #20

Grade 7 probability

Archetype Geometric Probability by Area └ Sub-archetype Uniform Draws to Favorable Region

geometric-probabilityprobability-basicarea-triangles caseworkcomplementary-counting ↑ Prerequisites: probability-basic

📏 Long solution 💡 2 insights

Problem

Two numbers a and b are picked independently and uniformly from the interval from 0 to 1, and c=a+b. Each of a, b, c is rounded to the nearest integer, giving A, B, C. Find the probability that rounding survives the addition, that is, that A+B=C.

Pick an answer.

(A)

$\frac14$

(B)

$\frac13$

(C)

$\frac12$

(D)

$\frac23$

(E)

$\frac34$

How to solve

Strategy Draw a Diagram

There is no finite list of outcomes to count here — a and b vary continuously — so the counting has to become measuring. Tool #1 (Draw a Diagram) does exactly that: plot the outcome as the point (a,b) in the unit square, and probability becomes area. Rounding is what makes this work, because every condition in the problem (a ＜ 1/2, b ≥ 1/2, a+b ＜ 3/2) is a straight-line boundary, so the favourable set is a union of polygons whose areas are easy. Tool #2 (Make a Systematic List) organises the work: the two cut lines a=1/2 and b=1/2 split the square into four quarters, one for each possible pair (A,B), and each quarter is handled separately so nothing is double counted and nothing is skipped. Tool #16 (Change Focus) gives the independent check at the end by measuring the failures instead.

1STEP 1

Turn rounding into inequalities

Each rounding becomes an inequality against its halfway points; the sum has two of them.

A= 0,& a ＜ 1/2 ; 1,& a ≥ 1/2 C= 0,& c ＜ 1/2 ; 1,& 1/2 ≤ c ＜ 3/2 ; 2,& c ≥ 3/2

2STEP 2

Make the unit square the sample space

A trial is a point in the unit square, so probability equals area.

P(region R)=area(R)/area([0,1]²)=area(R)

3STEP 3

Cut the square into four quarters

The two halfway lines cut the square into four quarters, one per rounded pair.

[0,1]²=[0,1/2)²_A+B=0 ∪ [1/2,1]×[0,1/2)_A+B=1 ∪ [0,1/2)×[1/2,1]_A+B=1 ∪ [1/2,1]²_A+B=2

4STEP 4

Lower-left quarter: both round down

In the low corner only a triangle of area 1/8 succeeds.

area{a,b ≥ 0, a+b ＜ 1/2}=1/2·1/2·1/2=1/8

5STEP 5

The two mixed quarters: one rounds up, one rounds down

Both mixed quarters succeed entirely, contributing 1/2.

a ＜ 1/2 ≤ b → 1/2 ≤ a+b ＜ 3/2 → C=1=A+B; 1/4+1/4=1/2

6STEP 6

Upper-right quarter: both round up

The high corner mirrors the low one, another 1/8.

area{a,b ≤ 1, a+b ≥ 3/2}=1/2·1/2·1/2=1/8

7STEP 7

Add the four pieces and check against the failures

Adding gives 3/4, and counting the failures confirms it, choice (E).

1/8+1/4+1/4+1/8=3/4; 1-(1/8+1/8)=3/4 → (E)

Answer

3/4

The probability has to be well above 1/2: half the square is the two mixed quarters, which never fail, so the answer is at least 1/2 before the corner triangles are even counted. That alone rules out (A) 1/4, (B) 1/3, and (C) 1/2, and (C) is precisely the trap of stopping after the mixed quarters. It also has to be below 1, since a=b=0.4 gives A+B=0 but c=0.8 rounds to C=1, a genuine failure. A quick numerical sanity check agrees: the failure region is where both numbers land on the same side of 1/2 and their sum crosses a halfway point, which happens for about a quarter of random pairs. The value 3/4 fits, and choice (D) 2/3 would require the two failure triangles to have combined area 1/3, more than the 1/4 that the corner quarters could even hold between them once the mixed quarters are known to be entirely successful.

💡Key takeaway

Draw the pair (a,b) as a point in a unit square: probability turns into area, and every rounding rule turns into a straight line you can measure against.

Turn rounding into inequalities
Make the unit square the sample space
Cut the square into four quarters
Lower-left quarter: both round down
The two mixed quarters: one rounds up, one rounds down
Upper-right quarter: both round up
Add the four pieces and check against the failures