Question

Three distinct numbers a, b, c are chosen at random from the numbers 1, 2, ...., 100. The probability that

Answer 1

a, b, c can be chosen in ${}^{100}{C}_3$ ways.

A) a, b, c are in A.P. $\Rightarrow 2b=a+c$

$\Rightarrow$ a and c are both even or both odd.

Thus, number of favourable ways ${=}^{50}{C}_2{+}^{50}{C}_2=50\times 49$

$\therefore probability=\frac{50\times 49\times 3\times 2}{100\times 99\times 98}=\frac{1}{66}$

B) Taking $r=2,3,...,10$ we can show a, b, c can be in G.P. in 53 ways.

Thus, probability of required event $=\frac{53\times 3\times 2}{100\times 99\times 98}=\frac{53}{161700}$

C) $\frac{1}{a},\frac{1}{b},\frac{1}{c}$, are in G.P.

$\iff$ a, b, c are in G.P.

D) $a+b+c$ is divisible by $2$ when $a+b+c$ is even

$P(a+b+c$ is even)=1-P(a+b+c$isodd)$

$=1-\frac{{}^{50}{C}_3+{(}^{50}{C}_1\left){(}^{50}{C}_2\right)}{{}^{100}{C}_3}$

$=1-\frac{50\times 49\times 48\times 50\times 50\times 49\times 3}{100\times 99\times 98}=\frac{1}{2}$