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Brice arranges an infinite number of mice in a line and numbers them $1,2,3, \dots$. Brice distributes three indistinguishable bowls of rice among the mice such that the $n$th mouse has a $\frac 1 {2^n}$ chance of receiving each bowl of rice. The probability that no mouse receives more than one bowl of rice can be expressed as a simplified fraction $\frac p{q}.$ Compute $p+q$.

My work:

The probability that one mouse got all 3 bowls is $(\frac 1 {2})^{3n}$ where n is the position of the mouse. I summed this and got a probability of $1/7.$ My next case had one mouse get 2 bowls and another mouse get the other bowl. I found this to be $(\frac 1 2)^{2n}(1-(\frac 1/2)^n).$ I summed this expression to be $\frac 1{3} - \frac 1{7}.$ So i got an answer of 2/3, but this answe is wrong. Am i undercounting?

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    $\begingroup$ So I tried using complimentary counting. I found the probability that one mouse got three bowls and one mouse got two bowls. But I got the wrong answer so I am little confused if i am overcounting. $\endgroup$ – MathIsLit Jul 29 '17 at 17:33
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    $\begingroup$ It will be impossible for us to tell if you are over counting without you showing your work $\endgroup$ – TomGrubb Jul 29 '17 at 18:18
  • $\begingroup$ Yes, you undercounted $\frac{1}{2}^{2n}$ is the probabilty that mouse $n$ gets bowl 1 and 2 and $(1 - \frac 12^n)$ is the probability that mouse $n$ does not get bowl 3. $\endgroup$ – fleablood Jul 29 '17 at 20:23
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Let $b_n$ be the number of bowls received by mouse $n$. \begin{align*} P(b_n = 3) &= \frac{1}{2^{3n}} = \frac{1}{8^n} \\ P(b_n = 2) &= \binom{3}{2}\frac{1}{2^{2n}}\left(1 - \frac{1}{2^n}\right)= \frac{3}{4^n} - \frac{3}{8^n} \end{align*} The factor of $\binom{3}{2}$ comes in because the mouse can receive any two of the three bowls. Note that the events $\{b_n = 3\}, \{b_n = 2\}$ ranging over all $n$ are disjoint. \begin{align*} P(\exists n, b_n = 2 \text{ or } 3) &= \sum_{n=1}^\infty \frac{1}{8^n} + \frac{3}{4^n} - \frac{3}{8^n} \\ &= \frac{3/4}{1 - 1/4} - \frac{2/8}{1 - 1/8} = 5/7 \\ P(\forall n, b_n = 1 \text{ or }0) &= 1 - P(\exists n, b_n = 2 \text{ or } 3) = \boxed{\frac{2}{7}} \end{align*}

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  • $\begingroup$ ohh i see my error. thank you. $\endgroup$ – MathIsLit Jul 29 '17 at 18:53

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