Difference between revisions of "2009 AIME I Problems/Problem 3"

Revision as of 16:12, 20 March 2009

Problem

A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution

If we let the odds of a tails ( $1-p$ ) equal $t$ , then the probability of three heads and five tails is $28{p^3}{t^5}$ The probability of five heads and three tails is $28{p^5}{t^3}$

$25*28{p^3}{t^5} = 28{p^5}{t^3}$ $25{t^2} = {p^2}$ $5t = p$ $5(1 - p) = p$ $5 - 5p = p$ $5 = 6p$ $p = \frac {5} {6}$ $5 + 6 = \boxed{11}$

@@ Line 5: / Line 5: @@
 == Solution ==
-If we let the odds of a tails <math>(1-p)</math> equal <math>t</math>, then the probability of three heads and five tails is <math>28{p^3}{t^5}</math>
+If we let the odds of a tails (<math>1-p</math>) equal <math>t</math>, then the probability of three heads and five tails is <math>28{p^3}{t^5}</math>
 The probability of five heads and three tails is <math>28{p^5}{t^3}</math>

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Difference between revisions of "2009 AIME I Problems/Problem 3"

Revision as of 16:12, 20 March 2009

Problem

Solution

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