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Model Paper 6 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 63 Marks
Question
A problem in mathematics is given to three students whose chances of solving it correctly are $1/2, 1/3$ and $1/4$ respectively. What is the probability that only one of them solves it correctly?
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Answer

Let $A , B , C$ be the given students and let $E _1, E _2$ and $E _3$ be the events that the problem is solved by $A , B , C$ respectively.
Then, $\bar{E}_1$, $\bar{E}_2$ and $\bar{E}_3$ are the events that the given problem is not solved by $A , B , C$ respectively.
Therefore,
we have, $P\left(E_1\right)=\frac{1}{2} ; P\left(E_2\right)=\frac{1}{3} ; P\left(E_3\right)=\frac{1}{4}$
$P\left(\bar{E}_1\right)=\left(1-\frac{1}{2}\right)=\frac{1}{2} ; P\left(\bar{E}_2\right)=\left(1-\frac{1}{3}\right)=\frac{2}{3} \text { and } P\left(\bar{E}_3\right)=\left(1-\frac{1}{4}\right)=\frac{3}{4}$
$P ($ exactly one of them solves the problem$)$
$=P\left(\left(E_1 \cap \bar{E}_2 \cap \bar{E}_3\right) \text { or }\left(\bar{E}_1 \cap E_2 \cap \bar{E}_3\right) \text { or }\left(\bar{E}_1 \cap \bar{E}_2 \cap E_3\right)\right)$
$=P\left(E_1 \cap \bar{E}_2 \cap \bar{E}_3\right)+P\left(\bar{E}_1 \cap E_2 \cap \bar{E}_3\right)+P\left(\bar{E}_1 \cap \bar{E}_2 \cap E_3\right)$
$=\left\{P\left(E_1\right) \times P\left(\bar{E}_2\right) \times P\left(\bar{E}_3\right)\right\}+\left(P\left(\bar{E}_1\right) \times P\left(E_2\right) \times P\left(\bar{E}_3\right)\right)
+\left(P\left(\bar{E}_1\right) \times P\left(\bar{E}_2\right) \times P\left(E_3\right)\right)$
$=\left(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\right)+\left(\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}\right)+\left(\frac{1}{2} \times \frac{2}{3} \times \frac{1}{4}\right)$
$=\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{12}\right)=\frac{11}{24}$

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