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Probability — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSProbability1 Mark
MCQ
If $\frac{(1-3\text{P})}{2},\frac{(1+4\text{P})}{3},\frac{(1+\text{P})}{6}$ are the probabilities of three mutually exclusive and exhaustive events, then the set of all values of p is:
  • A
    $(0,1)$
  • ✓
    $\Big(\frac{-1}{4},\frac{1}{3}\Big)$
  • C
    $\big(0,\frac{1}{3}\big)$
  • D
    $(0,\infty)$
✓

Answer

Correct option: B.
$\Big(\frac{-1}{4},\frac{1}{3}\Big)$
$\text{P(A)}=\frac{(1-3\text{P})}{2}$
$\text{P(B)}=\frac{(1+4\text{P})}{3}$
$\text{P()}=\frac{(1+\text{P})}{6}$
The events are mutually exclusive and exhaustive.
$\therefore\text{P}(\text{A}\cup\text{B }\cup\text{C})=\text{P(A)}+\text{P(B)}+\text{P(C)}=1$
$\Rightarrow0\leq\text{P(A)}\leq1,0\leq\text{P(B)}\leq1,0\leq\text{P(C)}\leq1$
$\Rightarrow0\leq\frac{1-3\text{P}}{2}\leq1$, $0\leq\frac{1-4\text{P}}{3}\leq1$, $0\leq\frac{1-\text{P}}{6}\leq1$
$\Rightarrow\frac{-1}{3}\leq\text{P}\leq\frac{1}{3}\ ...(1)$
$\frac{-1}{4}\leq\text{P}\leq\frac{1}{2}\ ...(2)$
$\text{and }{-1}\leq\text{P}\leq{5}\ ...(3)$
The common solution of (1), (2), and (3) is $\frac{-1}{4}\leq\text{P}\leq\frac{1}{3}$
$\therefore\text{The set values of P are}\Big(\frac{-1}{4},\frac{1}{3}\Big)$

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