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Difference between revisions of "2022 SSMO Speed Round Problems/Problem 1"

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== solution 1 ==
  
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Consider the probability <math>P(</math> win <math>)</math> as the sum of the probabilities of all sequences where Bobby wins:
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<math>P(</math> win <math>)=P(2</math> heads and then 1 tails <math>)+P(4</math> heads and then 1 tails <math>)+</math> <math>P(6</math> heads and then 1 tails <math>)+\ldots</math>
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For any sequence with <math>2 k</math> heads followed by a tail, the probability is:
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<cmath>
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\left(\frac{1}{2}\right)^{2 k} \times\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{2 k+1}
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</cmath>
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We sum this for <math>k=1,2,3, \ldots</math> :
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<cmath>
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P(\text { win })=\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^{2 k+1}
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</cmath>
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Factor out the constant term <math>\frac{1}{2}</math> :
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<cmath>
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P(\text { win })=\frac{1}{2} \sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^k
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</cmath>
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This is a geometric series with the first term <math>a=\left(\frac{1}{4}\right)</math> and common ratio
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<cmath>
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r=\left(\frac{1}{4}\right)
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</cmath>
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<cmath>
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\sum_{k=0}^{\infty} r^k=\frac{a}{1-r}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}
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</cmath>
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Thus:
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<cmath>
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P(\text { win })=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}
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</cmath>
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The probability <math>P(</math> win) can be expressed as:
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<cmath>
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\frac{1}{6}
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</cmath>
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In this case, <math>m=1</math> and <math>n=6</math>. Therefore, <math>m+n=1+6=7</math>.
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Thus, the value of <math>m+n</math> is:
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<cmath>
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\boxed{\text{7}}
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</cmath>

Latest revision as of 15:23, 24 May 2024

solution 1

Consider the probability $P($ win $)$ as the sum of the probabilities of all sequences where Bobby wins: $P($ win $)=P(2$ heads and then 1 tails $)+P(4$ heads and then 1 tails $)+$ $P(6$ heads and then 1 tails $)+\ldots$

For any sequence with $2 k$ heads followed by a tail, the probability is: \[\left(\frac{1}{2}\right)^{2 k} \times\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{2 k+1}\]

We sum this for $k=1,2,3, \ldots$ : \[P(\text { win })=\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^{2 k+1}\]

Factor out the constant term $\frac{1}{2}$ : \[P(\text { win })=\frac{1}{2} \sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^k\]

This is a geometric series with the first term $a=\left(\frac{1}{4}\right)$ and common ratio \[r=\left(\frac{1}{4}\right)\] \[\sum_{k=0}^{\infty} r^k=\frac{a}{1-r}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\]

Thus: \[P(\text { win })=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}\]

The probability $P($ win) can be expressed as: \[\frac{1}{6}\]

In this case, $m=1$ and $n=6$. Therefore, $m+n=1+6=7$. Thus, the value of $m+n$ is: \[\boxed{\text{7}}\]

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