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Erica and Alan each flip a fair coin 5 times. Suppose that Erica has flipped more heads than Alan after each has flipped the coin 2 times. The probability that Erica has flipped more heads than Alan after each has flipped the coin 5 times is m n , where m and n are relatively prime positive integers. Find m + n.

 Jun 30, 2024
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Let's analyze the scenario step-by-step:

 

Initial Flips: We are given that after 2 flips, Erica has more heads than Alan. This can happen in 3 ways:

 

Erica gets HH and Alan gets TT (2 heads for Erica, 0 for Alan).

 

Erica gets HT and Alan gets TT (1 head for Erica, 0 for Alan).

 

Erica gets TH and Alan gets TT (1 head for Erica, 0 for Alan).

 

Remaining Flips: There are 3 remaining flips for each person. Since the coin is fair (each flip has a 50% chance of heads or tails), there are 23=8 possible outcomes for each person's remaining flips.

 

Favorable Outcomes: We only care about the outcomes where Erica maintains more heads than Alan after 5 flips. Consider the 3 initial scenarios:

 

HH vs TT: In this case, Erica needs any combination of heads and tails in her remaining 3 flips. There are 8 possibilities for her flips. For Alan, he cannot have any heads, so he needs all tails. There is only 1 possibility for his flips. So, there are 8⋅1=8 favorable outcomes.

 

HT vs TT: Here, Erica needs at least 2 heads in her remaining flips. There are (23​)+(33​)=3+1=4 ways for her to achieve this (2 heads and 1 tail, or 3 heads). Again, Alan needs all tails, so there is only 1 possibility for his flips. This gives us 4⋅1=4 favorable outcomes.

 

TH vs TT: Similar to HT vs TT, Erica needs at least 2 heads. There are 4 favorable outcomes. Alan still needs all tails, resulting in 1 possibility. So, there are 4⋅1=4 favorable outcomes.

 

Total Favorable Outcomes: Summing up the favorable outcomes from each initial scenario, we get a total of 8+4+4=16 successful outcomes for Erica.

 

Total Outcomes: There are 23=8 possibilities for Erica's remaining flips and 23=8 possibilities for Alan's remaining flips, leading to a total of 8⋅8=64 total outcomes.

 

Probability: The probability that Erica has more heads than Alan after 5 flips is the number of favorable outcomes divided by the total number of outcomes: $ \frac{16}{64} = \frac{1}{4}$.

 

Converting the fraction to relatively prime integers, we have m=1 and n=4. Therefore, m+n=5​.

 Jul 1, 2024

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