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. | + | ==Problem== |
| + | A positive integer <math>n</math> not exceeding <math>100</math> is chosen in such a way that if <math>n\le 50</math>, then the probability of choosing <math>n</math> is <math>p</math>, and if <math>n > 50</math>, then the probability of choosing <math>n</math> is <math>3p</math>. The probability that a perfect square is chosen is | ||
| + | |||
| + | <math>\textbf{(A) }.05\qquad \textbf{(B) }.065\qquad \textbf{(C) }.08\qquad \textbf{(D) }.09\qquad \textbf{(E) }.1</math> | ||
| + | |||
| + | ==Solution== | ||
| + | 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 | ||
| + | ==See Also== | ||
| + | {{AHSME box|year=1978|num-b=18|num-a=20}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 15:24, 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
See Also
| 1978 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
