Question

$AprobleminStatisticsisgiventofourstudentsA,B,C,andD.$
$Theirchancesofsolvingitare\frac{1}{3},\frac{1}{4},\frac{1}{5},and\frac{1}{6}respectively.$
$Whatistheprobabilitythattheproblemwillbesolved?$

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{4}{5}$
• D. None of these

Answer 1

The correct option is B $\frac{2}{3}$

Probability that A fails to solve the problem is
$1-\frac{1}{3}=\frac{2}{3}$

Probability that B fails to solve the problem is
$1-\frac{1}{4}=\frac{3}{4}$

Probability that C fails solve the problem is
$1-\frac{1}{5}=\frac{4}{5}$

Probability that D fails to solve the problem is
$1-\frac{1}{6}=\frac{5}{6}$

Since the events are independent, the probability that all the four students to solve the problem is

$\frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times \frac{5}{6}=\frac{1}{3}$

$\therefore$ The probability that the problem will be solved
$=1-\frac{1}{3}=\frac{2}{3}$