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Question: Cards with numbers 2 to 101 are placed in a box. A card is selected at random from the box. Find the...

Cards with numbers 2 to 101 are placed in a box. A card is selected at random from the box. Find the probability that the selected card has a number which is a perfect cube.
A.1100\dfrac{1}{{100}}
B.2100\dfrac{2}{{100}}
C.3100\dfrac{3}{{100}}
D.4100\dfrac{4}{{100}}

Explanation

Solution

Hint : In terms of quantitative values ranging from zero to one, probability theory allows us to estimate the possibility of certain outcomes occurring as a result of a random experiment. An occurrence that is unlikely or impossible to occur has a probability of zero, while an occurrence that is likely or guaranteed to occur has a chance of one. Here we will use a formula for probability to answer this question.
FORMULA USED:
Probability=No.  of  Favourable  OutcomesTotal  No.  of  OutcomesProbability = \dfrac{{No.\;of\;Favourable\;Outcomes}}{{Total\;No.\;of\;Outcomes}}

Complete step-by-step answer :
Cards numbered 2 to 101 are placed in a box and presented to us.
Hence we have to find the total number of cards.
Total cards are given by: 1012=100101 - 2 = 100cards.
As a result, the total number of cards is 100.
Now we must determine the likelihood that the card chosen has a number that is a perfect cube.
Let us see how many perfect cubes are there from 2 to 101.
They are:
\left\\{ {{2^3},{3^3},{4^3}} \right\\}
Or, \left\\{ {8,27,64} \right\\}
Hence there are 3 outcomes.
Number of perfect cube cards = 3
And the total number of cards is 100.
Probability is given by: Probability=No.  of  Favourable  OutcomesTotal  No.  of  OutcomesProbability = \dfrac{{No.\;of\;Favourable\;Outcomes}}{{Total\;No.\;of\;Outcomes}}
Here favorable outcomes are 3 and the total number of outcomes is 100. Hence the probability will be:
Probability=3100Probability = \dfrac{3}{{100}}
Hence the correct answer to this question is option C.
So, the correct answer is “Option C”.

Note : A very common mistake that is made in such types of questions is that even 13{1^3}is considered in calculating the outcomes. Answer would be wrong if done so. We cannot consider 13{1^3} because the cards are numbered from 2 to 101.