Question

Numbers $1,2,3,...,100$ are written down on each of the cards $A, B$ and $C$. One number is selected at random from each of the cards. The probability that the numbers so selected can be the measures (in cm) of three sides of a right angled triangle, is

• A. $\frac{4}{100^{3}}$
• B. $\frac{3}{50^{3}}$
• C. $\frac{3 !}{100^{3}}$
• D. None of these

Answer 1

Answer: (D)

Number of ways of selecting three numbers one on each card
$=100 \times 100 \times 100=100^{3}$
We know that $(2 n+1),\left(2 n^{2}+2 n\right)$
and $\left(2 n^{2}+2 n+1\right)$ are Pythagorean triplets.
$\therefore $ For $n=1,2,3,4,5,6$, we get the lengths of three sides of a right angled triangle such that its hypotenuse is less than or equal to $100\, cm$.
Now, one Pythagorean triplet (e.g. $3, 4, 5; 5, 12,13$; etc.) can be chosen in $3!$ ways.
$\therefore $ Number of ways of selecting $6$ Pythagorean triplets $=6 \times 3!$
Hence, required probability $=\frac{6 \times 3 !}{100^{3}}$