Question
Question: If \( \dfrac{{1 + 3p}}{3},\dfrac{{1 - p}}{4} \) and \( \dfrac{{1 - 2p}}{2} \) are mutually exclusive...
If 31+3p,41−p and 21−2p are mutually exclusive events. Then, range of p is
(A)31⩽p⩽21
(B)21⩽p⩽21
(C)31⩽p⩽32
(D)31⩽p⩽52
Solution
Hint : When two events A and B are Mutually Exclusive it is impossible for them to happen together: The probability of A and B together equals 0 (impossible).
But, for Mutually Exclusive events, the probability of A and B is the sum of the individual probabilities: The probability of A and B equals the probability of A plus the probability of B .
Complete step-by-step answer :
To find the range of p ,
Since, the probability lies between 0 and 1 ,
0⩽ 31+3p⩽1,0⩽41−p⩽1,0⩽ 21−2p⩽1
That gives,
⇒0⩽1+3p⩽3,0⩽1−p⩽4,0⩽1−2p⩽2
⇒−31⩽p⩽32,−3⩽p⩽1,−21⩽p⩽21 _ _ _ _ _ _ _ _ _ _ (1)
But, when the events are mutually exclusive,
⇒0⩽31+3p+41−p+21−2p⩽1
⇒0⩽13−3p⩽313 _ _ _ _ _ _ _ _ _ _ (2)
From equations (1) and (2) ,
\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\\}
⇒31⩽p⩽21.
Therefore, the range of p is (A)31⩽p⩽21 .
So, the correct answer is “Option A”.
Note : ⇒ In logic and probability theory, two events are mutually exclusive or disjoint if they cannot both occur at the same time.
⇒ A clear example is the set of outcomes of a single coin toss, which can result in either heads or tails, but not both.