Question

A problem is given to four students A, B, C and D. Their chances of solving it are
$\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6}$ respectively. What is the probability that the problem is solved?
(a) $\dfrac{1}{3}$
(b) $\dfrac{2}{3}$
(c) $\dfrac{4}{5}$
(d) none of the above

Answer 1

First we need to find the chances of not solving the problem by each student. Then we multiply all the probability of not solving a problem by each student to get probability of not solving the problem then we get probability of solving the problem. We multiply all the probability of not solving the problem because there will be only one choice of not solving by all students. But solving a problem will include solving at least by one student. So we should not multiply the probability of solving the problem.

Complete step-by-step answer:
Now, let us consider that the probability of problem solving by ‘A’ as
$P\left( A \right)=\dfrac{1}{3}$
The probability of not solving the problem by ‘A’ is calculated as

& \Rightarrow P\left( {\bar{A}} \right)=1-P\left( A \right) \\\ & \Rightarrow P\left( {\bar{A}} \right)=1-\dfrac{1}{3} \\\ & \Rightarrow P\left( {\bar{A}} \right)=\dfrac{2}{3} \\\ \end{aligned}$$ Now, let us consider that the probability of problem solving by ‘B’ as $$P\left( B \right)=\dfrac{1}{4}$$ The probability of not solving the problem by ‘B’ is calculated as $$\begin{aligned} & \Rightarrow P\left( {\bar{B}} \right)=1-P\left( B \right) \\\ & \Rightarrow P\left( {\bar{B}} \right)=1-\dfrac{1}{4} \\\ & \Rightarrow P\left( {\bar{B}} \right)=\dfrac{3}{4} \\\ \end{aligned}$$ Now, let us consider that the probability of problem solving by ‘C’ as $$P\left( C \right)=\dfrac{1}{5}$$ The probability of not solving the problem by ‘C’ is calculated as $$\begin{aligned} & \Rightarrow P\left( {\bar{C}} \right)=1-P\left( A \right) \\\ & \Rightarrow P\left( {\bar{C}} \right)=1-\dfrac{1}{5} \\\ & \Rightarrow P\left( {\bar{C}} \right)=\dfrac{4}{5} \\\ \end{aligned}$$ Now, let us consider that the probability of problem solving by ‘D’ as $$P\left( D \right)=\dfrac{1}{6}$$ The probability of not solving the problem by ‘D’ is calculated as $$\begin{aligned} & \Rightarrow P\left( {\bar{D}} \right)=1-P\left( A \right) \\\ & \Rightarrow P\left( {\bar{D}} \right)=1-\dfrac{1}{6} \\\ & \Rightarrow P\left( {\bar{D}} \right)=\dfrac{5}{6} \\\ \end{aligned}$$ Now, let us consider that ‘E’ is the event of solving the problem, then the probability of not solving the problem is calculated as $$\begin{aligned} & \Rightarrow P\left( {\bar{E}} \right)=P\left( {\bar{A}} \right).P\left( {\bar{B}} \right).P\left( {\bar{C}} \right).P\left( {\bar{D}} \right) \\\ & \Rightarrow P\left( {\bar{E}} \right)=\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6} \\\ & \Rightarrow P\left( {\bar{E}} \right)=\dfrac{1}{3} \\\ \end{aligned}$$ Now we can calculate the probability of solving the problem as follows $$\begin{aligned} & \Rightarrow P\left( E \right)=1-P\left( {\bar{E}} \right) \\\ & \Rightarrow P\left( E \right)=1-\dfrac{1}{3} \\\ & \Rightarrow P\left( E \right)=\dfrac{2}{3} \\\ \end{aligned}$$ Therefore the probability of solving the problem is$$\dfrac{2}{3}$$. **So, the correct answer is “Option B”.** **Note:** Students may do mistake while taking the probability of solving the problem as $$P\left( E \right)=P\left( A \right).P\left( B \right).P\left( C \right).P\left( D \right)$$. This is the wrong method where we have only solving problems by all students. But solving problems by a single student can be considered as the problem is solved. So, that is the wrong method.