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A problem in statistics is given to three students $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. Their probabilities of solving the problem are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively, If all of them try independently, then the probability, that problem is solved, is
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$\frac{3}{4}$
$\begin{aligned} \mathrm{P}(\mathrm{A}) & =\frac{1}{2} \\ \therefore \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right) & =1-\frac{1}{2}=\frac{1}{2} \\ \mathrm{P}(\mathrm{B}) & =\frac{1}{3}\end{aligned}$
$\begin{array}{ll}\therefore \quad & \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\frac{1}{3}=\frac{2}{3} \\ & \mathrm{P}(\mathrm{C})=\frac{1}{4} \\ \therefore \quad & \mathrm{P}\left(\mathrm{C}^{\prime}\right)=1-\frac{1}{4}=\frac{3}{4} \\ \therefore \quad & \mathrm{P}(\mathrm{Problem} \text { is not solve }) \\ & =\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime} \cap \mathrm{C}^{\prime}\right) \\ & =\mathrm{P}\left(\mathrm{A}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{C}^{\prime}\right) \\ & \quad \ldots\left[\because \mathrm{A}^{\prime}, \mathrm{B}^{\prime}, \mathrm{C}^{\prime} \text { are independent }\right]\end{array}$
$\begin{aligned} & =\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \\ & =\frac{1}{4}\end{aligned}$
$\begin{aligned} \therefore \quad & \text { P(the problem will be solve) } \\ & =1-\mathrm{P} \text { (Problem is not solved) } \\ & =1-\frac{1}{4} \\ & =\frac{3}{4}\end{aligned}$
$\begin{array}{ll}\therefore \quad & \mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\frac{1}{3}=\frac{2}{3} \\ & \mathrm{P}(\mathrm{C})=\frac{1}{4} \\ \therefore \quad & \mathrm{P}\left(\mathrm{C}^{\prime}\right)=1-\frac{1}{4}=\frac{3}{4} \\ \therefore \quad & \mathrm{P}(\mathrm{Problem} \text { is not solve }) \\ & =\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime} \cap \mathrm{C}^{\prime}\right) \\ & =\mathrm{P}\left(\mathrm{A}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{C}^{\prime}\right) \\ & \quad \ldots\left[\because \mathrm{A}^{\prime}, \mathrm{B}^{\prime}, \mathrm{C}^{\prime} \text { are independent }\right]\end{array}$
$\begin{aligned} & =\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \\ & =\frac{1}{4}\end{aligned}$
$\begin{aligned} \therefore \quad & \text { P(the problem will be solve) } \\ & =1-\mathrm{P} \text { (Problem is not solved) } \\ & =1-\frac{1}{4} \\ & =\frac{3}{4}\end{aligned}$
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