Question

Numbers from 1 to 5000 are written on 5000 separate slips (one number on one slip). These slips are kept in a bag and mixed well. If one slip is chosen from the bag without looking into it, then the probability that the number on the slip is a perfect square, as well as a perfect cube, is
(A) \[\dfrac{1}{500}\]
(B) \[\dfrac{1}{1000}\]
(C) \[\dfrac{1}{1250}\]
(D) \[\dfrac{3}{5000}\]

Answer 1

Hint: The general term of the numbers which is a perfect square and a perfect cube is \[{{n}^{6}}\] . Now, put the values of n and check the number which is a perfect square and a perfect cube, and also lying under 5000. We have only four numbers from 1 to 5000 that are perfect squares as well as a perfect cube. We know the formula for the number of ways to select n objects out of m objects, \[^{m}{{C}_{n}}\] . Use this formula and get the value of the number of ways to select 1 number out of 4 numbers and the number of ways to select one number from 1 to 5000. We know that Probability = \[\dfrac{The\text{ }number\text{ }of\text{ }ways\text{ }to\text{ }select\text{ }1\text{ }number\text{ }out\text{ }of\text{ }4\text{ }numbers}{The\text{ }number\text{ }of\text{ }ways\text{ }to\text{ }select\text{ }one\text{ }number\text{ }from\text{ }1\text{ }to\text{ }5000}\] . Now, solve it further and calculate the probability.

Complete step-by-step answer:
According to the question, it is given that numbers from 1 to 5000 are written on 5000 separate slips in such a way that one number is written on one slip.
Now, the slips are kept in a bag and mixed well.
We know that every perfect square number can be written as \[{{n}^{2}}\] ……………………………………..(1)
We also know that every perfect cube number can be written as \[{{n}^{3}}\] ……………………………………………(2)
Here, we have to find that number which is a perfect square as well as a perfect cube.
On taking the L.C.M of the exponents of terms in equation (1) and equation (2), we get
L.C.M = \[6\]
The general term of the numbers which is a perfect square and a perfect cube is \[{{n}^{6}}\] ……………………………….(3)
Now, we have to find those numbers from 1 to 5000 whose exponent 6 is lying under 5000.
Now, putting \[n=1\] in equation (3), we get
\[{{n}^{6}}={{1}^{6}}=1\] …………………………………….(4)
Now, putting \[n=2\] in equation (3), we get
\[{{n}^{6}}={{2}^{6}}=64\] …………………………………….(5)
Now, putting \[n=3\] in equation (3), we get
\[{{n}^{6}}={{3}^{6}}=729\] …………………………………….(6)
Now, putting \[n=4\] in equation (3), we get
\[{{n}^{6}}={{4}^{6}}=4096\] …………………………………….(7)
Now, putting \[n=5\] in equation (3), we get
\[{{n}^{6}}={{5}^{6}}=15625>5000\] …………………………………….(8)
But we have to choose the numbers that are lying under 5000 and 15625 is greater than 5000. So, we can’t take this number.
So, the numbers which are a perfect square, as well as a perfect cube, are 1, 64, 729, and 4096 …………………………………….(9)
We know the formula for the number of ways to select n objects out of m objects, \[^{m}{{C}_{n}}\] ……………………………(10)
From equation (9), we have four numbers which are a perfect square, as well as a perfect cube.
The number of ways to select 1 number out of 4 numbers = \[^{4}{{C}_{1}}\] ………………………………….(11)
Now, the number of ways to select one number from 1 to 5000 = \[^{5000}{{C}_{1}}\] ………………(12)
The probability that the number is a perfect square as well as a perfect cube,
Probability = \[\dfrac{The\text{ }number\text{ }of\text{ }ways\text{ }to\text{ }select\text{ }1\text{ }number\text{ }out\text{ }of\text{ }4\text{ }numbers}{The\text{ }number\text{ }of\text{ }ways\text{ }to\text{ }select\text{ }one\text{ }number\text{ }from\text{ }1\text{ }to\text{ }5000}\] ………………………….(13)
Now, from equation (11), equation (12), and equation (13), we get
Probability = \[\dfrac{^{4}{{C}_{1}}}{^{5000}{{C}_{1}}}=\dfrac{4}{5000}=\dfrac{1}{1250}\] .
Therefore the probability that the number is a perfect square, as well as a perfect cube, is \[\dfrac{1}{1250}\] .

So, the correct answer is “Option C”.

Note: In this question, one might think that we have to find the probability that the number on the slip is either a perfect square or a perfect cube. This is wrong because in the question it is mentioned that we have to find the probability of a number which is both a perfect square and a perfect cube. Therefore, we have to find the probability that the number on the slip is a perfect square as well as a perfect cube.