A problem in statistics is given to four students A, B, C and D. Their

1.	A problem in statistics is given to four students A, B, C and D. Their chances of solving it are \(\frac{1}{3}\),\(\frac{1}{4}\),\(\frac{1}{5}\) and \(\frac{1}{6}\) respectively. What is the probability that the problem will be solved?(a) \(\frac{1}{3}\)(b) \(\frac{2}{3}\)(c) \(\frac{4}{5}\)(d) None of these
Answer» (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 - \(\frac{1}{3}\) = \(\frac{2}{3}\)

A problem in statistics is given to four students A, B, C and D. Their chances of solving it are \(\frac{1}{3}\),\(\frac{1}{4}\),\(\frac{1}{5}\) and \(\frac{1}{6}\) respectively. What is the probability that the problem will be solved?(a) \(\frac{1}{3}\)(b) \(\frac{2}{3}\)(c) \(\frac{4}{5}\)(d) None of these

Answer»

(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 - \(\frac{1}{3}\) = \(\frac{2}{3}\)