AMC 10 · 2019 · #16
Grade 7 probabilityPick an answer.
Tool #1 (Diagram): draw the row of pads and cross out 3 and 6 — the picture makes the tight spots obvious. Tool #3 (Eliminate): a hop covers at most 2 pads, so getting past a crossed-out pad forces a specific landing; this kills almost every choice in the middle of the trip. Tool #7 (Subproblems): those forced landings act as checkpoints that cut the trip into three independent legs, and independent legs multiply. Tool #2 (Systematic List): each leg is short enough to list its routes outright. Tool #5 (Pattern): the pad-by-pad recursion p_n = 1/2p_n-1 + 1/2p_n-2 gives a second, mechanical route to the same number for checking.
Draw the row and mark the traps
Draw the row and mark the traps.
Each hop is one fair coin flip, so a route's probability is just 1/2 multiplied once per hop.
7.SP.C.7Draw A DiagramPassing a trap forces the landing
Passing a trap forces the landing.
With a maximum stride of 2, the only way over a hole is to stand right beside it and jump.
7.SP.C.7Eliminate PossibilitiesThe middle stretch has no choices left
The middle stretch has no choices.
Between the two predators there is exactly one safe road, so every survivor walks it.
7.SP.C.7Eliminate PossibilitiesCut the trip into three legs
Cut the trip into three legs.
Compulsory checkpoints chop one long random trip into short pieces that simply multiply.
Compulsory checkpoints chop one long random trip into short pieces that simply multiply.
▸ Why?
Each leg's hops tell you nothing about the next leg's, so the chances combine by multiplying.
▸ Why?
Every survivor's route is one choice per leg, so the routes fill a grid of independent choices.
Leg (a): reaching pad 2
Compute the first leg.
Multiply along one route, then add across the separate routes.
5.NF.A.1Make A Systematic ListLeg (b): the forced corridor costs three hops
The forced corridor still costs hops.
A forced move still has to be rolled, so each one costs a factor of 1/2.
5.NF.B.4Identify SubproblemsLeg (c): reaching pad 10 from pad 7
Compute the last leg.
Landing exactly on a pad is its own hurdle — a long stride can sail right over the target.
7.SP.C.8Make A Systematic ListMultiply the three legs
Multiplying the legs gives fifteen over two hundred fifty-six.
Independent legs of a journey multiply, so the three fractions collapse into one.
5.NF.B.4Identify SubproblemsWhen you can only step 1 or 2, the sole way past a bad pad is to stand right next to it and jump — those forced landings turn one long risky trip into three short legs whose probabilities you just multiply.
- Draw the row and mark the traps
- Passing a trap forces the landing
- The middle stretch has no choices left
- Cut the trip into three legs
- Leg (a): reaching pad 2
- Leg (b): the forced corridor costs three hops
- Leg (c): reaching pad 10 from pad 7
- Multiply the three legs