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...") |
|||
(4 intermediate revisions by 2 users not shown) | |||
Line 1: | Line 1: | ||
− | + | ==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 | ||
+ | ==See Also== | ||
+ | {{AHSME box|year=1978|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:24, 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.