AMC 10 · 2015 · #17

Grade 11 probability

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

combinations-basicprobability-basicindependent-events convert-to-algebracasework ↑ Prerequisites: combinations-basicprobability-basic

📏 Long solution 💡 3 insights

Problem

A biased coin gives the same chance of two successes as of three. Find the number of tosses.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Convert to Algebra

The sentence "exactly two heads is as likely as exactly three heads" is an equation waiting to be written, so write each probability in the count-times-weight form and set the two equal. Both sides are built from the same powers of 1/4 and 3/4, so almost everything cancels, and clearing the common factors turns a probability equation into a plain whole-number statement about C(n, 2) and C(n, 3). One cancellation is not free, though. Dividing by C(n, 2) is legal only when C(n, 2) is not 0, and it is 0 exactly when n is 0 or 1 — which is precisely the case the condition n ＞ 1 removes, because a single toss makes two heads and three heads both impossible and therefore trivially equally likely. So test that boundary first, then compare C(n, 3) with C(n, 2) by a short double count instead of grinding factorials, solve the linear equation that falls out, and finish by confirming the value works and that no other n can.

1STEP 1

Write both probabilities

Both probabilities share almost every factor.

P(2)=C(n, 2)(1/4)²(3/4)ⁿ⁻², P(3)=C(n, 3)(1/4)³(3/4)ⁿ⁻³, P(2)=P(3)

2STEP 2

Strip the shared weight away

Stripping them leaves a ratio of counts.

C(n, 2) 3^ n-2=C(n, 3) 3^ n-3 ⟺ 3C(n, 2)=C(n, 3)

3STEP 3

Test the boundary the condition blocks

The stated bound blocks the degenerate case.

n=1: C(1, 2)=C(1, 3)=0 → P(2)=P(3)=0; n ≥ 2: C(n, 2)=(n(n-1))/2 ≥ 1

4STEP 4

Compare the counts by double counting

Comparing counts is a plain identity.

C(n, 2)(n-2)=3C(n, 3) ⟹ (C(n, 3))/(C(n, 2))=(n-2)/3

5STEP 5

Solve for n and pin down uniqueness

Solving gives 11 tosses.

3C(n, 2)=C(n, 3)=C(n, 2)·(n-2)/3 ⟹ (n-2)/3=3 ⟹ n-2=9 ⟹ n=11

6STEP 6

Confirm it works and rule out the rest

The ratio rules out every other choice, (D).

P(3)/P(2)=(n-2)/9: 3/9, 6/9, 8/9, 9/9, 11/9 for n=5,8,10,11,13

Answer

Compute both probabilities outright at n = 11. P(2) = 55·3⁹/4¹¹ = (55 · 19683)/4194304 = 1082565/4194304 and P(3) = 165·3⁸/4¹¹ = (165 · 6561)/4194304 = 1082565/4194304. They match exactly, both about 0.258. The scale is sensible too: with heads coming up 1/4 of the time, the average number of heads in 11 tosses is 11/4 = 2.75, sitting between 2 and 3, exactly where the two probabilities should be neck and neck. Had the answer been 5 or 8, the average number of heads would be 1.25 or 2, too low for three heads to catch two; at 13 the average is 3.25, already past the tie.

💡Key takeaway

When two probabilities are said to be equal, divide one by the other instead of subtracting: the shared weights vanish and one small equation for n is left — but first make sure the thing you cancel is not zero.

Write both probabilities
Strip the shared weight away
Test the boundary the condition blocks
Compare the counts by double counting
Solve for n and pin down uniqueness
Confirm it works and rule out the rest