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# help

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Two distinct numbers are randomly selected from among the first 64 positive integral perfect squares.  Find the probability that these two numbers are both cubes of integers.

Dec 11, 2019

#1
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These are the first 64 "perfect squares":
(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 3600, 3721, 3844, 3969, 4096).

Of the above 64, there are 4 cubes:1, 64, 729, 4096.
The probability that the first number is a cube is:
4/64 =1 / 16
The probability that the 2nd is also a cube is:
3/63 =1 / 21
Then the probability that both are cubes is:
1/16  x  1/21 =1 / 336
P.S. Somebody should check this, please. Thanks.

Dec 11, 2019
#2
+109064
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Thie guest appears to be correct

Number  of possible  sets formed by choosing two cubes from four  = C(4, 2)   =  6

Number  of possible sets of choosing  two squares from sixty-four  =  C(64,2)  =  2016

Probability  =     6  / 2016    =    1   / 336

Dec 11, 2019