Question

$S=\{1,2,3,.....30\}$ if $3$ numbers are chosen at random from $S$, the probability for they are consecutive

• A. $\frac{1}{145}$
• B. $\frac{35}{33}$
• C. $\frac{33}{35}$
• D. $\frac{1}{38}$

Answer 1

The correct option is A $\frac{1}{145}$

$totalnumberofsetssuchas(1,2,3),(2,3,4),..........(27,28,29)and(28,29,30)=28setsnownumberofchossingthreenumbersoutof30={}^{30}{C}_3=\left(\frac{30!}{27!3!}\right)=\left(\frac{30\times 29\times 28}{6}\right)=4060\therefore Probability=\left(\frac{28}{4060}\right)=\left(\frac{1}{145}\right)$