Difference between revisions of "2008 Mock ARML 1 Problems/Problem 4"
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== Problem == | == Problem == | ||
− | 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. | + | 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 == | ||
− | {{ | + | 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 |
+ | <cmath>\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}</cmath> | ||
+ | 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 08:03, 1 June 2008
Problem
There are black balls and 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 black balls in the hat before there are white balls.
Solution
Let denote the probability of reaching black balls before white balls from a position of black balls and white balls. The probability that we have black balls after a turn is (note that ), and that we have black balls is . Similarly, if we start with black balls, after a turn there is a probability of ending with black balls and a probability of ending with balls. Thus, we have the recursions By symmetry, ; substitution into yields . Now, and form a two equation linear system which can be solved to find that .
See also
2008 Mock ARML 1 (Problems, Source) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 |