Question

Number 1, 2, 3 . . . . . . . . 100 are written down on each of the cards A, B and C. One of the numbers is selected at random from each of the cards. The probability that the number so selected can be measure (in cm) of those sides of a right angled triangle is
\[\begin{align}
  & A.\dfrac{4}{{{100}^{3}}} \\
 & B.\dfrac{3}{{{50}^{3}}} \\
 & C.\dfrac{3!}{{{100}^{3}}} \\
 & D.\text{ None of these} \\
\end{align}\]

Answer 1

Hint: In this question, we have to find probability, therefore, we will use formula:
\[\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}}\]
For this we will first count total outcomes. After that, we will calculate favorable outcomes. Since, the selected number should be the side of the right angled triangle. Therefore, we have to find Pythagorean triplets and find total ways Pythagorean triplet can be selected from three cards.

Complete step-by-step answer:
Here, we are given three cards with number 1, 2, 3 . . . . . . . 100 written on them. Since, any number can be selected from the first card, similarly for the second card and third day. Therefore, total ways of selecting three numbers $\Rightarrow 100\times 100\times 100={{100}^{3}}$
Now, we have to find favorable outcomes, which means we have to find ways in which three numbers are selected in a way such that they are sides of the right angled triangle which means we have to find the number of Pythagorean triplet and total ways in which they can be selected. As we know, for integers, Pythagorean triplet are given by
\[\left( 2n+1 \right),\left( 2{{n}^{2}}+2n \right)\text{ and }\left( 2{{n}^{2}}+2n+1 \right)\]
Now, let us find how many such Pythagorean triplets exist such that all three numbers are between 1 to 100.
For n = 1, we have triplets 3, 4 and 5
For n = 2, we have triplets 5, 12 and 13
For n = 3, we have triplets 7, 24 and 25
For n = 4, we have triplets 9, 40 and 41
For n = 5, we have triplets 11, 60 and 61
For n = 6, we have triplets 13, 84 and 85
For n = 7, we have triplets 15, 112 and 113
As we can see, after n = 7, the number becomes greater than 100, hence, we can have only six Pythagorean triplets.
Now, for every triplet there are 3 slots. Hence, there will be 3! ways to every triplet.
For six triplets there will be $6\times 3!$ triplets. Hence, favorable outcome $\Rightarrow 6\times 3!$
\[\begin{align}
  & \text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{Number of total outcomes}} \\
 & \text{Probability}=\dfrac{\text{6}\times \text{3!}}{\text{10}{{\text{0}}^{\text{3}}}}=\dfrac{2\times 2\times 2\times 3!}{100\times 100\times 100}=\dfrac{3!}{50\times 50\times 50}=\dfrac{\text{3!}}{\text{5}{{\text{0}}^{\text{3}}}} \\
\end{align}\]

So, the correct answer is “Option B”.

Note: Students should learn formula for finding Pythagoras triples. After finding the number of triplets, don't forget to consider ways of selecting their numbers for a particular triplet. Students can also check Pythagorean triplet by applying Pythagoras theorem on them. For example, ${{3}^{2}}+{{4}^{2}}={{5}^{2}}\Rightarrow 9+16=25\Rightarrow 25=25$
Hence, it is a Pythagorean triplet and thus forms the side of a triangle.