Question

23. Consider the following experiment. Pick a random integer from 1 to 10¹².

    1. (a) What is the probability that it is either a perfect square (1, 4, 9, 16, ...) or a perfect cube (1, 8, 27, 64,...)?

    2. (b) What is the probability that it is either a perfect fourth power (1, 16, 81, 256, ...) or a perfectsixth power (1, 64, 729, 4096,...)?

Answer 1

(a)

We see that :- (10⁶)² = 10¹² ;  (10⁶+1)² > 10¹²

Thus from 1 to 10¹² , there are 10⁶ numbers which are perfect square.

Thus probability that the number picked is perfect square = (10⁶/10¹²) = 0.000001

We see that :- (10⁴)³ = 10¹² ;  (10⁴+1)³ > 10¹²

Thus from 1 to 10¹² , there are 10⁴ numbers which are perfect cube.

Thus probability that the number picked is perfect cube = (10⁴/10¹²) = 0.00000001

We see that :- (1²)³ = (1³)² ;  (2²)³ = (2³)² ; (3²)³ = (3³)² ; ........... ;  (100²)³ = (100³)² ;  (101²)³ = (101³)² > 10¹²

Thus from 1 to 10¹² , there are 100 numbers which are perfect square as well as perfect cube

Thus probability that the number picked is perfect square as well as perfect cube = (100/10¹²) = 0.0000000001

Thus probability that the number picked is perfect square or perfect cube

= 0.000001+0.00000001-0.0000000001 = 0.0000010099

.

(b)

We see that :- (10³)⁴ = 10¹² ;  (10³+1)⁴ > 10¹²

Thus from 1 to 10¹² , there are 10³ numbers which are perfect fourth power.

Thus probability that the number picked is perfect fourth power = (10³/10¹²) = 0.000000001

We see that :- (10²)⁶ = 10¹² ;  (10²+1)⁶ > 10¹²

Thus from 1 to 10¹² , there are 10² numbers which are perfect sixth power.

Thus probability that the number picked is perfect sixth power = (10²/10¹²) = 0.0000000001

We see that :- (1⁴)⁶ = (1⁶)⁴ ;  (2⁴)⁶ = (2⁶)⁴ ; (3⁴)⁶ = (3⁶)⁴ ;  (4⁴)⁶ = (4⁶)⁴ > 10¹²

Thus from 1 to 10¹² , there are 3 numbers which are perfect fourth power as well as perfect sixth power.

Thus probability that the number picked is perfect square as well as perfect cube = (3/10¹²) = 0.000000000003

Thus probability that the number picked is perfect square or perfect cube

= 0.000000001+0.0000000001-0.000000000003 = 0.000000001097