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Difference between revisions of "2008 Mock ARML 1 Problems/Problem 4"

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== Problem ==
 
== Problem ==
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There are <math>4</math> black balls and <math>1</math> white ball in a hat. A turn consists of picking a ball from the hat and replacing it with one of the opposite color. Compute the probability that, after a sequence of turns, there are <math>5</math> black balls in the hat before there are <math>5</math> white balls.
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There are <math>4</math> black balls and <math>1</math> white ball in a hat. A turn consists of picking a ball from the hat and replacing it with one of the opposite color. Compute the [[probability]] that, after a sequence of turns, there are <math>5</math> black balls in the hat before there are <math>5</math> white balls.
  
 
== Solution ==
 
== Solution ==
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{{solution}} The answer is <math>\boxed{\frac{5}{8}}</math>.
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Let <math>f(n)</math> denote the probability of reaching <math>5</math> black balls before <math>5</math> white balls from a position of <math>n</math> black balls and <math>5-n</math> white balls. The probability that we have <math>5</math> black balls after a turn is <math>\frac 15</math> (note that <math>f(5) = 1</math>), and that we have <math>3</math> black balls is <math>\frac 45</math>. Similarly, if we start with <math>3</math> black balls, after a turn there is a <math>\frac 25</math> probability of ending with <math>4</math> black balls and a <math>\frac 35</math> probability of ending with <math>2</math> balls. Thus, we have the recursions
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<cmath>\begin{align}
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f(4) &= \frac 15 \cdot [1] + \frac 45 \cdot f(3) \\
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f(3) &= \frac 25\cdot f(4) + \frac 35 \cdot f(2)
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\end{align}</cmath>
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By [[symmetry]], <math>f(2) = 1-f(3)</math>; substitution into <math>(2)</math> yields <math>8f(3) = 2f(4) + 3</math>. Now, <math>(1)</math> and <math>(2)</math> form a two equation linear system which can be solved to find that <math>f(4) = \boxed{\frac{5}{8}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 09:03, 1 June 2008

Problem

There are $4$ black balls and $1$ white ball in a hat. A turn consists of picking a ball from the hat and replacing it with one of the opposite color. Compute the probability that, after a sequence of turns, there are $5$ black balls in the hat before there are $5$ white balls.

Solution

Let $f(n)$ denote the probability of reaching $5$ black balls before $5$ white balls from a position of $n$ black balls and $5-n$ white balls. The probability that we have $5$ black balls after a turn is $\frac 15$ (note that $f(5) = 1$), and that we have $3$ black balls is $\frac 45$. Similarly, if we start with $3$ black balls, after a turn there is a $\frac 25$ probability of ending with $4$ black balls and a $\frac 35$ probability of ending with $2$ balls. Thus, we have the recursions \begin{align} f(4) &= \frac 15 \cdot [1] + \frac 45 \cdot f(3) \\ f(3) &= \frac 25\cdot f(4) + \frac 35 \cdot f(2) \end{align} By symmetry, $f(2) = 1-f(3)$; substitution into $(2)$ yields $8f(3) = 2f(4) + 3$. Now, $(1)$ and $(2)$ form a two equation linear system which can be solved to find that $f(4) = \boxed{\frac{5}{8}}$.

See also

2008 Mock ARML 1 (Problems, Source)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8
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