Problem 4 · AMC 8 Stretch
Stretch
Counting & Probability
accounting-for-all-possibilitieslogical-reasoning
You have 3 cards. Card A is red on both sides. Card B is blue on both sides. Card C is red on one side, blue on the other. You shuffle them in a bag, pull one out, and lay it down. The side facing up is RED. What is the probability that the hidden side (the bottom) is also red?
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Answer: \(\frac{2}{3}\)
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Hint 1 of 4
The tempting wrong answer is \(\frac{1}{2}\) ('it's either the red-red card or the red-blue card, 50-50'). But that ignores something. Count SIDES, not cards.
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Hint 2 of 4
Make a list of every individual face that could be the red one facing up. The red-red card has TWO red faces; the red-blue card has only ONE. So red faces aren't all on the same kind of card.
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Hint 3 of 4
List the equally likely red faces that could be up, and write what's underneath each: (A-side1 / A-side2 red), (A-side2 / A-side1 red), (C-red / C-blue).
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Approach: Accounting for all possibilities — count faces, not cards
- The trap answer is \(\frac{1}{2}\): 'the card is either A (red-red) or C (red-blue), so 50-50.' That's wrong because it counts cards, but the right thing to count is faces.
- There are three red faces in the whole set: Card A side 1 (other side red), Card A side 2 (other side red), and Card C red side (other side blue).
- Seeing red means one of these three equally likely red faces is up. In 2 of those 3 cases (both faces of card A), the hidden side is also red; in only 1 case (card C) is it blue.
- So \(P(\text{other side is red})=\frac{2}{3}\). The key move: condition on what you actually saw (a red face) and count faces, not cards.
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