# Difference between revisions of "1978 AHSME Problems/Problem 19"

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− | Let's say that we will have <math>3</math> slips for every number not exceeding <math>100</math> but bigger than <math>50.</math> This is to account for the <math>3p</math> probability part. Let's now say that we will only have one slip for each number below or equal to <math>50.</math> The probability(or <math>p</math>) will then be <math>\frac{1}{200}.</math> Now let's have all the squares under <math>50,</math> which are <math>1,4,9,16,25,36,49.</math> The probability for these are <math>\frac{7}{200}.</math> The numbers above <math>50</math> that are squares are <math>64,81,100.</math> We then need to multiply the probability by <math>3</math> so the probability of these are <math>\frac{9}{200}.</math> The answer is <math>\frac{7}{200}+\frac{9}{200}=0. | + | Let's say that we will have <math>3</math> slips for every number not exceeding <math>100</math> but bigger than <math>50.</math> This is to account for the <math>3p</math> probability part. Let's now say that we will only have one slip for each number below or equal to <math>50.</math> The probability(or <math>p</math>) will then be <math>\frac{1}{200}.</math> Now let's have all the squares under <math>50,</math> which are <math>1,4,9,16,25,36,49.</math> The probability for these are <math>\frac{7}{200}.</math> The numbers above <math>50</math> that are squares are <math>64,81,100.</math> We then need to multiply the probability by <math>3</math> so the probability of these are <math>\frac{9}{200}.</math> The answer is <math>\frac{7}{200}+\frac{9}{200}=0.08\implies\boxed{\textbf{(C).}}</math> |

~volkie thangy | ~volkie thangy |

## Revision as of 15:19, 18 June 2021

## Problem

A positive integer not exceeding is chosen in such a way that if , then the probability of choosing is , and if , then the probability of choosing is . The probability that a perfect square is chosen is

## Solution

Let's say that we will have slips for every number not exceeding but bigger than This is to account for the probability part. Let's now say that we will only have one slip for each number below or equal to The probability(or ) will then be Now let's have all the squares under which are The probability for these are The numbers above that are squares are We then need to multiply the probability by so the probability of these are The answer is

~volkie thangy