AMC 10 · 2015 · #15

Grade 7 probability
probability-basicindependent-eventsmean-median-mode-range systematic-enumerationwork-backwards ↑ Prerequisites: probability-basicindependent-events
📏 Medium solution 💡 3 insights
Problem
Two of four grades are locked and the other two are random and independent. Find the chance of clearing an average.

Pick an answer.

(A)
$\frac{11}{72}$
(B)
$\frac{1}{6}$
(C)
$\frac{3}{16}$
(D)
$\frac{11}{24}$
(E)
$\frac{1}{2}$
How to solve
Strategy Make a Systematic List

Work backwards from the GPA target first: two of the four grades are already settled, so the condition on the GPA collapses into a single condition on the English and History points. Each of those classes has exactly three possible grades, so the whole sample space is only nine pairs — small enough to write out completely, which is the safest way to be certain about the boundary. The boundary is the entire problem. An A paired with a C scores exactly 6 and passes, while a B paired with a C scores 5 and fails, so the intuitive reading 'she needs to do well in both' is wrong. Before any of that, the unstated C probabilities have to be recovered as the leftovers from 1, which is a subproblem in its own right. Independence turns each cell into a product, and computing the three failing cells as well as the six passing ones gives a check that costs almost nothing.

1STEP 1

Turn the GPA target into a point target

The average becomes a plain point target.

(P_M + P_S + P_E + P_H)/4 ≥ 3.5 ⇔ P_M + P_S + P_E + P_H ≥ 14 ⇔ P_E + P_H ≥ 6
2STEP 2

Recover the missing C probabilities

Each missing probability is recovered.

P_E(C) = 1 - 1/6 - 1/4 = 7/12, P_H(C) = 1 - 1/4 - 1/3 = 5/12
3STEP 3

List all nine grade pairs

There are only nine grade pairs.

(4,4)=8, (4,3)=7, (3,4)=7, (4,2)=6, (3,3)=6, (2,4)=6 ≥ 6; (3,2)=5, (2,3)=5, (2,2)=4 < 6
4STEP 4

Give each passing pair its probability

Independence prices each pair directly.

1/6 · 1/4 = 1/24, 1/6 · 1/3 = 1/18, 1/6 · 5/12 = 5/72, 1/4 · 1/4 = 1/16, 1/4 · 1/3 = 1/12, 7/12 · 1/4 = 7/48
5STEP 5

Add the passing cells

Adding the passing cells gives 11/24.

(6 + 8 + 10 + 9 + 12 + 21)/144 = 66/144 = 11/24
6STEP 6

Cross-check with the failing cells

The failing cells confirm it, choice (E).

(15 + 28 + 35)/144 = 78/144, 66/144 + 78/144 = 1, 1 - 78/144 = 11/24
Answer
11/24
The value 11/24 sits just under 1/2, which is exactly where it belongs. Any A in either class already clears the bar, since 4 + 2 = 6, and the probability of at least one A is 1 - 5/6 · 3/4 = 3/8, so the answer must be strictly greater than 3/8 = 9/24. That alone eliminates (A) 11/72, (B) 1/6 and (C) 3/16, all of which are below 3/8. And the failing pairs total 78/144, strictly more than half, so the answer must be strictly below 1/2, which rules out (E). Only 11/24 survives. The nine cell probabilities also add to exactly 1, confirming the probability model itself is consistent.
💡Key takeaway

Turn the GPA target into a point target first, then list every grade pair — the problem hides right on the boundary, where an A next to a C squeaks through but a B next to a C does not.

  • Turn the GPA target into a point target
  • Recover the missing C probabilities
  • List all nine grade pairs
  • Give each passing pair its probability
  • Add the passing cells
  • Cross-check with the failing cells