KCET · Maths · Probability
A random experiment has five outcomes \(\mathrm{w}_1, \mathrm{w}_2, \mathrm{w}_3, \mathrm{w}_4\) and \(\mathrm{w}_5\). The probabilities of the occurrence of the outcomes \(\mathrm{w}_1, \mathrm{w}_2, \mathrm{w}_3, \mathrm{w}_4\) and \(\mathrm{w}_5\) are respectively \(\frac{1}{6}, \mathrm{a}, \mathrm{b}\) and \(\frac{1}{12}\) such that \(12 \mathrm{a}+12 \mathrm{~b}-1=0\). Then the probabilities of occurrence of the outcome \(w_3\) is
- A \(\frac{2}{3}\)
- B \(\frac{1}{3}\)
- C \(\frac{1}{6}\)
- D \(\frac{1}{12}\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}\mathrm{p}\left(\mathrm{w}_1\right)=\frac{1}{6} & \\ \mathrm{p}\left(\mathrm{w}_2\right)=\mathrm{a} & \Rightarrow \frac{1}{6}+\mathrm{a}+\mathrm{b}+\mathrm{x}+\frac{1}{12}=1 \\ \mathrm{p}\left(\mathrm{w}_3\right)=\mathrm{b} & \Rightarrow 12(\mathrm{a}+\mathrm{b}+\mathrm{x})=9 \\ \mathrm{p}\left(\mathrm{w}_4\right)=\mathrm{c} & \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{x}=\frac{3}{4} \\ \mathrm{p}\left(\mathrm{w}_5\right)=\frac{1}{12} & \Rightarrow \frac{1}{12}+\mathrm{x}=\frac{3}{4} \Rightarrow \mathrm{x}=\frac{2}{3}\end{array}\)
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