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A problem in mathematics is given to three students whose chances of solving it are respectively $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$. What is the probability that the problem will be solved ?
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Let A denotes the problem is solved by first student.
$\mathrm{B}$ denotes the problem is solved by second student.
$\mathrm{C}$ denotes the problem is solved by third student.
Clerly, A,B,C are independent events.
We have $P(A)=\frac{1}{2}$
$P(B)=\frac{1}{3}$,and
$P(C)=\frac{1}{4}$
Now, $\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})$
$=1-\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}})$
$=1-\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}})$
$=1-\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\overline{\mathrm{B}}) \mathrm{P}(\overline{\mathrm{C}})$ (Since, A,B,C are independent events, so their
compllements also)
\begin{aligned} &=-1\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \\ &=1-\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}=\frac{3}{4} \end{aligned}
by Diamond (50,375 points)

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