A number is selected at random from $1$ to $50$. What is the probability that it is not a perfect cube?
Answer
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Hint: Probability is the term mathematically with the events that occur, which is the number of favorable events that is divided by the total number of the outcomes.
If we divide the probability and then multiplied with the hundred then we will determine its percentage value.
$\dfrac{1}{6}$ which means the favorable event is $1$ and the total outcome is $6$
Formula used:
$P = \dfrac{F}{T}$ where P is the overall probability, F is the possible favorable events and T is the total outcomes from the given.
Complete step-by-step solution:
Since from the given that A number is selected at random from $1$ to $50$. Which means there are total $50$ numbers and hence we have the total event outcome as $50$
Now we need to find the favorable event, that is not a perfect cube.
Perfect cube means the number can be represented in both the cube and cube root, like ${2^3} = 8,\sqrt[3]{8} = 2$
Hence the perfect cubes from the given numbers from $1$ to $50$ are $1,8,27$ because which can be expressed as $({1^3} = 1,\sqrt[3]{1} = 1),({2^3} = 8,\sqrt[3]{8} = 2),({3^3} = 27,\sqrt[3]{{27}} = 3)$ and after we get the number four as ${4^3} = 64$ which is not from $1$ to $50$
Hence there are three perfect cubes $1$ to $50$
Therefore, the non-perfect cubes from $1$ to $50$ are $50 - 3 = 47$ (overall subtracts the perfect cube to get the non-perfect cube) which is the favorable event.
Hence, we get $P = \dfrac{F}{T} \Rightarrow \dfrac{{47}}{{50}}$
Thus, the probability that it is not a perfect cube is $\dfrac{{47}}{{50}}$.
Note: Since we just need to know such things about the square root numbers and perfect square numbers, A perfect square is the numbers that obtain by multiplying any whole numbers (zero to infinity) twice, or the square of the given numbers yields a whole number like $\sqrt {25} = 5$or $25 = {5^2}$
Similarly, the perfect cube can be expressed as $\sqrt[3]{3} = 27,27 = {3^3}$
If we divide the probability and then multiplied with the hundred then we will determine its percentage value.
$\dfrac{1}{6}$ which means the favorable event is $1$ and the total outcome is $6$
Formula used:
$P = \dfrac{F}{T}$ where P is the overall probability, F is the possible favorable events and T is the total outcomes from the given.
Complete step-by-step solution:
Since from the given that A number is selected at random from $1$ to $50$. Which means there are total $50$ numbers and hence we have the total event outcome as $50$
Now we need to find the favorable event, that is not a perfect cube.
Perfect cube means the number can be represented in both the cube and cube root, like ${2^3} = 8,\sqrt[3]{8} = 2$
Hence the perfect cubes from the given numbers from $1$ to $50$ are $1,8,27$ because which can be expressed as $({1^3} = 1,\sqrt[3]{1} = 1),({2^3} = 8,\sqrt[3]{8} = 2),({3^3} = 27,\sqrt[3]{{27}} = 3)$ and after we get the number four as ${4^3} = 64$ which is not from $1$ to $50$
Hence there are three perfect cubes $1$ to $50$
Therefore, the non-perfect cubes from $1$ to $50$ are $50 - 3 = 47$ (overall subtracts the perfect cube to get the non-perfect cube) which is the favorable event.
Hence, we get $P = \dfrac{F}{T} \Rightarrow \dfrac{{47}}{{50}}$
Thus, the probability that it is not a perfect cube is $\dfrac{{47}}{{50}}$.
Note: Since we just need to know such things about the square root numbers and perfect square numbers, A perfect square is the numbers that obtain by multiplying any whole numbers (zero to infinity) twice, or the square of the given numbers yields a whole number like $\sqrt {25} = 5$or $25 = {5^2}$
Similarly, the perfect cube can be expressed as $\sqrt[3]{3} = 27,27 = {3^3}$
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