Question

The probability of getting a number chosen from 1,2,…, 100 as a cube is
[a] $\dfrac{1}{25}$
[b] $\dfrac{2}{25}$
[c] $\dfrac{3}{25}$
[d] $\dfrac{4}{25}$

Answer 1

Hint: Probability of event E = $\dfrac{n(E)}{n(S)}=\dfrac{\text{Favourable cases}}{\text{Total number of cases}}$ where S is called the sample space of the random experiment. Assume that E be the event that the chosen number is a perfect cube. Find the number of cubes between 1 and 100 and hence determine n(E). Hence using the above formula determines the probability of event E.

Complete step-by-step answer:
Let E be the event: The chosen number is a perfect cube.
Finding the number of perfect cubes between 1-100:
The perfect cubes between 1- 100 are 1, 8,27 and 64.
Hence there are 4 perfect cubes between 1-100 (Both inclusive)
Since there are 4 perfect cubes, the total number of cases favourable to E = 4
Hence, we have n (E) = 4
The total number of ways in which we can choose a number = 100
Hence, we have n (S) = 100
Hence, P (E) = $\dfrac{4}{100}=\dfrac{1}{25}$
Hence the probability that the chosen number is a perfect cube is $\dfrac{1}{25}$
Hence option [a] is correct.

Note: [1] It is important to note that choosing uniformly at random is important for the application of the above formula. If the choice is not random, then there is a bias factor in choosing, and the above formula is not applicable. In those cases, we use the conditional probability of an event.
[2] The probability of an event always lies between 0 and 1.
[3] The number of perfect cubes between 1 to a is $\left[ \sqrt[3]{a} \right]$, where [x] denotes the greatest integer less or equal to x.
Hence, we have
$n\left( E \right)=\left[ \sqrt[3]{100} \right]=4$, which is the same as obtained above.
Following a similar procedure as above, we have $P\left( E \right)=\dfrac{1}{25}$.
Hence option [a] is correct.