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

The correct option is B

$\frac{2}{3}$

Explanation for the correct option.

The chances of each student solving a problem are:

$P\left(A\right)=\frac{1}{3},P\left(B\right)=\frac{1}{4},P\left(C\right)=\frac{1}{5},P\left(D\right)=\frac{1}{6}$

The probability of each student not solving the problem is:

P(A not solving) $\begin{array}{rcl}& =& 1-P\left(A\right)\end{array}$

$\begin{array}{rcl}& \Rightarrow & 1-\frac{1}{3}=\frac{2}{3}\end{array}$

Similarly, P(B not solving)$\begin{array}{rcl}& =& 1-P\left(B\right)\end{array}$

$\Rightarrow 1-\frac{1}{4}=\frac{3}{4}$,

P(C not solving)$\begin{array}{rcl}& =& 1-P\left(C\right)\end{array}$

$\Rightarrow 1-\frac{1}{5}=\frac{4}{5}$,

P(D not solving)$\begin{array}{rcl}& =& 1-P\left(D\right)\end{array}$

$\Rightarrow 1-\frac{1}{6}=\frac{5}{6}$

The total probability of not solving the problem is:

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

The total probability of solving the problem is:

$1-\frac{1}{3}=\frac{2}{3}$

Therefore, option (B) $\frac{2}{3}$ is the correct option.