Probability NOT a perfect square

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Probability NOT a perfect square

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#1 h Apr 1, 2025, 5:37 PM

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Mike decides to play a game with himself. He begins with a score of 0 and proceeds to flip a fair coin. If he lands on heads, he adds 2 to his score. If he lands on tails, he subtracts 1 from his score. After 5 flips, what is the probability that Mike’s score is not a perfect square?

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#2 h Apr 1, 2025, 6:29 PM • 2 Y

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sol

This post has been edited 2 times. Last edited by martianrunner, Apr 2, 2025, 3:55 PM
Reason: hiding solution

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#3 h Apr 1, 2025, 10:04 PM

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@above you should put that in a Click to reveal hidden text

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#4 h Apr 2, 2025, 1:43 AM • 1 Y

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My Approach

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ilikemath247365

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#5 h Apr 3, 2025, 5:12 AM

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If he gets $x$ heads and $y$ tails, his score will be: $2x - y$ . We also know $x + y = 5$ . Let's list out possible values of $x$ and $y$ and see which pairs give $2x - y$ to be a perfect square(we will be using complementary counting). If $x = 0, y = 5$ , this is impossible. If $x = 1, y = 4$ , this is also impossible. If $x = 2, y = 3$ , this will give us that $2x - y = 1$ , which is a perfect square. If $x = 3, y = 2$ , this will give us that $2x - y = 4$ , which is a perfect square. If $x = 4, y = 1$ , this is impossible. If $x = 5, y = 0$ , this is impossible. So, Mike will only get a perfect square if he gets either $2$ heads and $3$ tails or $3$ heads and $2$ tails. If Mike gets $2$ heads and $3$ tails, there are $10$ possible ways to do this( $5$ factorial ways to arrange the $5$ flips, then divide by the $2$ factorial ways to arrange the heads and the $3$ factorial ways to arrange the tails). Similarly, there are $10$ possible ways to do the second case. So we can have a total of $10 + 10 = 20$ total possible occurrences for which Mike WILL get a perfect square. The probability is simply $\frac{20}{2^{5}} = \frac{20}{32} = \frac{5}{8}$ . Taking the complementary of this, we get our final probability of $\boxed{\frac{3}{8}}$ .

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