Question

Number 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 right-angled triangles no two of which are similar, is -

• A. $\frac { 4 } { 100 ^ { 3 } }$
• B. $\frac { 3 } { 50 ^ { 3 } }$
• C. $\frac { 36 } { 100 ^ { 3 } }$
• D. None of these

Answer 1

Answer: (C)

$\frac { 36 } { 100 ^ { 3 } }$

Answer 2

n(S) = 100 × 100 × 100. We know that (2n + 1)² +

(2n² + 2n)² = (2n² + 2n + 1)² for all n Ī N. \ For n = 1, 2, 3, 4, 5, 6, we get lengths of the three sides of a right-angle triangle whose longest side £ 100.

For e.g. When n = 1, sides are 3, 4, 5; when n = 2, sides are 5, 12, 13 and so on.

The number of selections of 3, 4, 5 from the three cards by taking one from each is 3!.

\ n (5) = 6(3!).

Hence, P(5) = $\frac { 6 ( 3 ! ) } { 100 \times 100 \times 100 }$=$\frac { 1 } { 100 }$ $\left( \frac { 3 } { 50 } \right) ^ { 2 }$

Hence (3) is the correct answer