A problem in statistics is given to three students A, B, and C. T…

Question

A problem in statistics is given to three students $A, B$, and $C$. Their chances of solving the problem are $1 / 3,1 / 4$, and $1 / 5$ respectively. If all of them try independently, what is the probability that,:
1.problem is solved?
2.problem is not solved?
3.exactly two students solve the problem?

✓

Answer

Let $\mathrm{A}$ be the event that student $\mathrm{A}$ can solve the problem.
$B$ be the event that student $B$ can solve the problem.
$C$ be the event that student $C$ can solve problem.
$
\begin{aligned}
& \therefore P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(C)=\frac{1}{5} \\
& \therefore P\left(A^{\prime}\right)=1-P(A)=1-\frac{1}{3}=\frac{2}{3} \\
& P\left(B^{\prime}\right)=1-P(B)=1-\frac{1}{4}=\frac{5}{4} \\
& P\left(C^{\prime}\right)=1-P(C)=1-\frac{1}{5}=\frac{4}{5}
\end{aligned}
$
Since A, B, C are independent events
$\therefore A^{\prime}, B^{\prime}, C^{\prime}$ are also independent events.:
1. Let $\mathrm{X}$ be the event that problem is solved.
Problem can be solved if at least one of the three students solves the problem.
$P(X)=P($ at least one student solves the problem)
$=1-P($ no student solved problem)
$=1-P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right)$
$=1-P\left(A^{\prime}\right) P\left(B^{\prime}\right) P\left(C^{\prime}\right)$
$=1-\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5}$
$=1-\frac{2}{5}$
$
=\frac{3}{5}
$

2.Let $Y$ be the event that problem is not solved
$
\begin{aligned}
& \therefore P(Y)=P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right) \\
& =P\left(A^{\prime}\right) P\left(B^{\prime}\right) P\left(C^{\prime}\right) \\
& =\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \\
& =\frac{2}{5}
\end{aligned}
$

3.Let $Z$ be the event that exactly two students solve the problem.
$
\begin{aligned}
& \therefore P(Z)=P\left(A \cap B \cap C^{\prime}\right) \cup P\left(A \cap B^{\prime} \cap C\right) \cup P\left(A^{\prime} \cap B \cap C\right) \\
& =P(A) \cdot P(B) \cdot P\left(C^{\prime}\right)+P(A) \cdot P\left(B^{\prime}\right) \cdot P(C)+P\left(A^{\prime}\right) \cdot P(B) \cdot P(C) \\
& =\left(\frac{1}{3} \times \frac{1}{4} \times \frac{4}{5}\right)+\left(\frac{1}{3} \times \frac{3}{4} \times \frac{1}{5}\right)+\left(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\right) \\
& =\frac{4}{60}+\frac{3}{60}+\frac{2}{60} \\
& =\frac{3}{20}
\end{aligned}
$

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