Question

If $\dfrac{{1 + 4p}}{4},\dfrac{{1 + p}}{4},\dfrac{{1 - 2p}}{2}$ are probabilities of three mutually exclusive events, then which of the following holds,
A) $\dfrac{1}{3} \leqslant p \leqslant \dfrac{1}{2}$
B) $0 \leqslant p \leqslant \dfrac{2}{3}$
C) $\dfrac{1}{6} \leqslant p \leqslant \dfrac{1}{2}$
D) $0 \leqslant p \leqslant \dfrac{1}{2}$

Answer 1

Two events are said to be mutually exclusive events when both cannot occur at the same time.
The probability of any event lies between 0 and 1.
Using this condition, we will find the range of$p$.

_Complete step-by-step answer: _
It is given that, $\dfrac{{1 + 4p}}{4},\dfrac{{1 + p}}{4},\dfrac{{1 - 2p}}{2}$ are probabilities of three mutually exclusive events.
We know that, probability of any event lies between 0 and 1.
Let us consider, $P$ be the probability of any event.
That is, $0 \leqslant P \leqslant 1$
So, we have,
$0 \leqslant \dfrac{{1 + 4p}}{4} \leqslant 1$
From the above inequality we can simplify and get corresponding inequality values,
Let us multiply the inequality by 4 initially,
$0 \leqslant 1 + 4p \leqslant 4$
Let us subtract the inequality by 1 then we get,
$- 1 \leqslant 4p \leqslant 3$
Now let us again divide the inequality by 4, so that we get the range of p,
$\dfrac{{ - 1}}{4} \leqslant p \leqslant \dfrac{3}{4}$… (1)
Similarly, we have,
$0 \leqslant \dfrac{{1 + p}}{4} \leqslant 1$
Let us multiply the inequality by 4 we get,
$0 \leqslant 1 + p \leqslant 4$
Now let us subtract the inequality by 1 we get,
$- 1 \leqslant p \leqslant 3$ … (2)
Similarly, we have,
$0 \leqslant \dfrac{{1 - 2p}}{2} \leqslant 1$
Let us multiply the inequality by 2 we get,
$0 \leqslant 1 - 2p \leqslant 2$
Now let us subtract the inequality by 1 we get,
$- 1 \leqslant - 2p \leqslant 1$
Let us again divide the inequality by -2
$\dfrac{{ - 1}}{2} \leqslant p \leqslant \dfrac{1}{2}$… (3)
Again we have,
$0 \leqslant \dfrac{{1 + 4p}}{4} + \dfrac{{1 + p}}{4} + \dfrac{{1 - 2p}}{2} \leqslant 1$
By solving the inequality we get,
$0 \leqslant 1 + \dfrac{p}{4} \leqslant 1$
Now let us subtract the inequality by 1 we get,
$- 1 \leqslant \dfrac{p}{4} \leqslant 0$
On multiplying the inequality by 4 we get,
$- 4 \leqslant p \leqslant 0$ … (4)
From the expression (1), (2), (3), (4) we have,
We have, max$\\{ \dfrac{{ - 1}}{4}, - 1,\dfrac{{ - 1}}{2}, - 1\\} \leqslant p \leqslant$min $\\{ \dfrac{3}{4},3,\dfrac{1}{2},0\\}$
So, the value of $p$lies between $\dfrac{{ - 1}}{2}$ and$0$.
By taking the range to the positive bound we have the value of p lies in 0 and$\dfrac{1}{2}$

Hence, the correct option is (D) $0 \leqslant p \leqslant \dfrac{1}{2}$

Note:
As per the condition we have, $p$lies between $\dfrac{{ - 1}}{2}$ and$0$ but the suitable range of p is$0 \leqslant p \leqslant \dfrac{1}{2}$.
While operating operations in inequalities it should be done in all terms of the inequality.
While multiplying and dividing the inequality the inequality by negative terms the inequality changes from less than to greater than or vice versa.
In the sum we have used the above mentioned property in expression (3) .