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Probability (p-2) — Maths (commerce) STD 11 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Probability (p-2)2 Marks
Question
$A, B$ and $C$ are mutually exclusive and exhaustive events associated with the random experiment. Find $P(A)$, given that
$P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$
✓

Answer

$P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$
Since $A, B, C$ are mutually exclusive and exhaustive events,
$
\begin{array}{ll}
\therefore & \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})=1 \\
\therefore & \mathrm{P}(\mathrm{A})+\frac{3}{2} \mathrm{P}(\mathrm{A})+\frac{1}{2} \mathrm{P}(\mathrm{B})=1 \\
\therefore & \mathrm{P}(\mathrm{A})+\frac{3}{2} \mathrm{P}(\mathrm{A})+\frac{1}{2} \times \frac{3}{2} \mathrm{P}(\mathrm{A})=1 \\
\therefore & \mathrm{P}(\mathrm{A})+\frac{3}{2} \mathrm{P}(\mathrm{A})+\frac{3}{4} \mathrm{P}(\mathrm{A})=1 \\
\therefore & \mathrm{P}(\mathrm{A}) \times\left(1+\frac{3}{2}+\frac{3}{4}\right)=1 \\
\therefore & \mathrm{P}(\mathrm{A}) \times\left(\frac{13}{4}\right)=1 \\
\therefore & \mathrm{P}(\mathrm{A})=\frac{4}{13}
\end{array}
$

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