Consider the following experiment. Pick a random integer from 1 to 1012 . (a) What is the probability that it is either a perfect square (1, 4, 9, 16, …) or a perfect cube (1, 8, 27, 64,…)? (b) What is the probability that it is either a perfect fourth power (1, 16, 81, 256, …) or a perfect sixth power (1, 64, 729, 4096,…)?

Respuesta :

Answer:

a.) 0.0402

b.) 0.00789

Step-by-step explanation:

In between the numbers 1 and 1012,

The number of perfect squares we have is 31, that is all numbers between 1 and 31 inclusive have their squares between the numbers 1 and 1012

The number of perfect cubes we have is 10, that is all numbers between 1 and 10 Inclusive have their perfect cube between the numbers 1 and 1012.

The number of perfect Fourth we have is 5, that is all numbers between 1 and 5 inclusive have their perfect fourth between the numbers 1 and 1012.

The number of perfect Sixth we have is 3, that is all numbers between 1 and 3 inclusive have their perfect sixth between the numbers 1 and 1012.

Hence,

a.) probability of choosing a perfect square = 31/1012

Probability of choosing a perfect cube = 10/1012

Probability of choosing a perfect square or a perfect cube = (31/1012) + (10/1012) - [(31/1012)*(10/1012)]

=41/1012 - 310/1024144

=0.0405 - 0.0003

=0.0402.

b.) Probability of choosing a perfect fourth or a perfect Sixth = (5/1012) + (3/1012) - [(5/1012)*(3/1012)]

= 8/1012 - 15/1024144

= 0.00789.