Question

Three students A, B, C can solve the problem independently, the probabilities of solving the problem of each of them are $\frac{1}{2},\frac{1}{3},\frac{1}{5}$ respectively, then the probability that at least one of them solves the problem is

• A. $\frac{1}{30}$
• B. $\frac{4}{15}$
• C. $\frac{29}{30}$
• D. $\frac{11}{15}$

Answer 1

The correct option is D $\frac{11}{15}$

Portability of atleast one solve problem $=$

$=1-$ (no one solve problem)

$P\left(\overline{A}\right)=1-\frac{1}{2}=\frac{1}{2}$ $P\left(\overline{B}\right)=1-\frac{1}{3}=\frac{2}{3},$ $P\left(\overline{C}\right)=1-\frac{1}{5}=\frac{4}{5}$

no one solve problem: $P=\frac{1}{2}\times \frac{2}{3}\times \frac{4}{5}=\frac{4}{15}$ [independent terms]

$\therefore P$ (atleast one) $=-1-\frac{4}{15}=\frac{11}{15}$