Solveeit Logo

Question

Question: If \( \dfrac{{1 + 3p}}{3},\dfrac{{1 - p}}{4} \) and \( \dfrac{{1 - 2p}}{2} \) are mutually exclusive...

If 1+3p3,1p4\dfrac{{1 + 3p}}{3},\dfrac{{1 - p}}{4} and 12p2\dfrac{{1 - 2p}}{2} are mutually exclusive events. Then, range of pp is
(A)13p12\left( A \right)\dfrac{1}{3} \leqslant p \leqslant \dfrac{1}{2}
(B)12p12\left( B \right)\dfrac{1}{2} \leqslant p \leqslant \dfrac{1}{2}
(C)13p23\left( C \right)\dfrac{1}{3} \leqslant p \leqslant \dfrac{2}{3}
(D)13p25\left( D \right)\dfrac{1}{3} \leqslant p \leqslant \dfrac{2}{5}

Explanation

Solution

Hint : When two events AA and BB are Mutually Exclusive it is impossible for them to happen together: The probability of AA and BB together equals 00 (impossible).
But, for Mutually Exclusive events, the probability of AA and BB is the sum of the individual probabilities: The probability of AA and BB equals the probability of AA plus the probability of BB .

Complete step-by-step answer :
To find the range of pp ,
Since, the probability lies between 00 and 11 ,
00 \leqslant 1+3p31,01p41,0\dfrac{{1 + 3p}}{3} \leqslant 1,0 \leqslant \dfrac{{1 - p}}{4} \leqslant 1,0 \leqslant 12p21\dfrac{{1 - 2p}}{2} \leqslant 1
That gives,
01+3p3,01p4,012p2\Rightarrow 0 \leqslant 1 + 3p \leqslant 3,0 \leqslant 1 - p \leqslant 4,0 \leqslant 1 - 2p \leqslant 2
13p23,3p1,12p12\Rightarrow - \dfrac{1}{3} \leqslant p \leqslant \dfrac{2}{3}, - 3 \leqslant p \leqslant 1, - \dfrac{1}{2} \leqslant p \leqslant \dfrac{1}{2} _ _ _ _ _ _ _ _ _ _ (1)\left( 1 \right)
But, when the events are mutually exclusive,
01+3p3+1p4+12p21\Rightarrow 0 \leqslant \dfrac{{1 + 3p}}{3} + \dfrac{{1 - p}}{4} + \dfrac{{1 - 2p}}{2} \leqslant 1
0133p133\Rightarrow 0 \leqslant 13 - 3p \leqslant \dfrac{{13}}{3} _ _ _ _ _ _ _ _ _ _ (2)\left( 2 \right)
From equations (1)\left( 1 \right) and (2)\left( 2 \right) ,
\max \left\\{ { - \dfrac{1}{3}, - 3, - \dfrac{1}{2},\dfrac{1}{3}} \right\\} \leqslant p \leqslant \min \left\\{ {\dfrac{2}{3},1,\dfrac{1}{2},\dfrac{{13}}{3}} \right\\}
13p12.\Rightarrow \dfrac{1}{3} \leqslant p \leqslant \dfrac{1}{2}.
Therefore, the range of pp is (A)13p12\left( A \right)\dfrac{1}{3} \leqslant p \leqslant \dfrac{1}{2} .
So, the correct answer is “Option A”.

Note : \Rightarrow In logic and probability theory, two events are mutually exclusive or disjoint if they cannot both occur at the same time.
\Rightarrow A clear example is the set of outcomes of a single coin toss, which can result in either heads or tails, but not both.