Question

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 1

Answer: (B)

$\frac{2}{3}$

P(A solving)$=\frac{1}{3}\to$ P(A not solving)$=1-\frac{1}{3}=\frac{2}{3}$
P(B solving)$=\frac{1}{4}\to$ P(B not solving)$=1-\frac{1}{4}=\frac{3}{4}$
P(C solving)$=\frac{1}{5}\to$ P(C not solving)$=1-\frac{1}{5}=\frac{4}{5}$
P(D solving)$=\frac{1}{6}\to$ 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)$=1-\frac{1}{3}=\frac{2}{3}$.