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

(Created page with "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> pr...")
(No difference)

Revision as of 19:41, 21 May 2021

Let's say that we will have $3$ slips for every number not exceeding $100$ but bigger than $50.$ This is to account for the $3p$ probability part. Let's now say that we will only have one slip for each number below or equal to $50.$ The probability(or $p$) will then be $\frac{1}{200}.$ Now let's have all the squares under $50,$ which are $1,4,9,16,25,36,49.$ The probability for these are $\frac{7}{200}.$ The numbers above $50$ that are squares are $64,81,100.$ We then need to multiply the probability by $3$ so the probability of these are $\frac{9}{200}.$ The answer is $\frac{7}{200}+\frac{9}{200}=0.008\implies\boxed{\textbf{(C).}}$

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