There are 6 peas in a glass, 4 floating on the top and 2 sitting on the bottom. At each five second interval, a random number of peas from 0 to 2 sink from the top to the bottom and a random number from 0 to 2 rise from the bottom to the top. (If there is only 1 pea left, it moves or stays put with equal probability.) The probability that all six peas are on the top before all six are on the bottom can be expressed in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Question

Noel Lara

Mathematics

30 March, 14:07

There are 6 peas in a glass, 4 floating on the top and 2 sitting on the bottom. At each five second interval, a random number of peas from 0 to 2 sink from the top to the bottom and a random number from 0 to 2 rise from the bottom to the top. (If there is only 1 pea left, it moves or stays put with equal probability.) The probability that all six peas are on the top before all six are on the bottom can be expressed in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

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Austin Willis · Answer 1

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Step-by-step explanation:

probability that all six peas are on top = $m/n$

probability that they are at the bottom = 1 - ($m/n$)

sum of the two probabilities = 1