(b) \(\frac{2}{3}\)
P (A solving ) = \(\frac{1}{3}\) \(\Rightarrow\) P (A not solving) =1- \(\frac{1}{3}\) = \(\frac{2}{3}\)
P (B solving ) = \(\frac{1}{4}\) \(\Rightarrow\) P (B not solving) =1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)
P (C solving) = \(\frac{1}{5}\) \(\Rightarrow\) P (C not solving) = 1- \(\frac{1}{5}\) = \(\frac{4}{5}\)
P (D solving) = \(\frac{1}{6}\) \(\Rightarrow\) P(D not solving) = 1- \(\frac{1}{6}\) = \(\frac{5}{6}\)
\(\therefore\) Probability (problem not solved)
= \(\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\) = \(\frac{1}{3}\)
\(\Rightarrow\) Probability (problem solved) =1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)