Question

A number is selected at random from 1 to 50. What is the probability that it is not a perfect cube ?

Answer 1

Hint: We had to only find total number of numbers between 1 to 50 that are not the perfect cubes than we can easily apply the probability formula which states that probability of getting a favourable outcome = \[\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]

Complete step-by-step answer:
As we know that the number whose cube root is an integer is known as a perfect cube.
Like if x be any number and the cube root of x is y then for x to be a perfect cube y should be an integer or perfect number (i.e. y should not be in decimal or fraction).
So, the numbers between 1 and 50 that are perfect cubes are 1, 8, 27.
Now as we can see that from 1 to 50 there are only three numbers that are perfect cube.
So, the rest of the numbers must not be a perfect cube.
And from 1 to 50 there are a total of 50 numbers.
So, number of numbers from 1 to 50 that are not perfect cube will be 50 – 3 = 47
Now we can apply the probability formula which states that,
Probability of getting a favourable outcome = \[\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]
So, probability of not getting a perfect cube will be = \[\dfrac{{{\text{Number of numbers that are not perfect cube}}}}{{{\text{Total number of numbers}}}}\] = \[\dfrac{{47}}{{50}}\]
Hence, the probability of getting a number which is not a perfect cube between 1 to 50 will be \[\dfrac{{47}}{{50}}\].

Note: Whenever we come up with this type of question first, we should remember that perfect squares or cubes are those numbers whose square root or cube root is a perfect integer respectively. So, here we had to find the probability of getting a number which is not a perfect cube. So, to find that easily we can find a number of numbers that are perfect cube and then subtract that by 50 to get a number of numbers that are not perfect cube. After that we can apply a probability formula. This will be the easiest and efficient way to find the solution of the problem.