Question

If $\dfrac{1+4p}{4},\dfrac{1-p}{4},\dfrac{1-2p}{2}$ are probabilities of three mutually exclusive events, then
\[\begin{align}
  & a.\dfrac{1}{3}\le p\le \dfrac{1}{2} \\
 & b.\dfrac{1}{3}\le p\le \dfrac{2}{3} \\
 & c.\dfrac{1}{6}\le p\le \dfrac{1}{2} \\
 & d.\text{None of these} \\
\end{align}\]

Answer 1

Hint: We are given probabilities of 3 mutually exclusive events in terms of a variable ‘p’. We will solve this question by using the concept of mutually exclusive events, that is, P (A or B) = P (A) + P (B) and we will consider the fact that the probability of any event always lies in the range [0,1].

Complete step-by-step solution:
Here, we have been given the probabilities of 3 mutually exclusive events as $\dfrac{1+4p}{4},\dfrac{1-p}{4},\dfrac{1-2p}{2}$.
In order to solve this question, we need to understand the concept of mutually exclusive events. So, in probability theory, two events are said to be mutually exclusive if they cannot occur at the same time. In such a case, we use the formula,
P (A or B) = P (A) + P (B)
Here, P (A or B) means that the probability that the events A or B occur. P (A) is the probability of occurrence of event A and P (B) is the probability of occurrence of event B.
Also, we know that the probability of any event, P (E) will always lie in the range [0,1].
So, we can write,
$0\le \dfrac{1+4p}{4}+\dfrac{1-p}{4}+\dfrac{1-2p}{2}\le 1.........(i)$
Also, the events have to individually satisfy the condition that P (E) belong in the range [0,1].
So, we get,
$\begin{align}
  & 0\le \dfrac{1+4p}{4}\le 1.........(ii) \\
 & 0\le \dfrac{1-p}{4}\le 1.........(iii) \\
 & 0\le \dfrac{1-2p}{2}\le 1.........(iv) \\
\end{align}$
By solving (i), (ii), (iii) and (iv), we will get,
From (i), $\dfrac{1}{2}\le p\le \dfrac{5}{2}$
From (ii), $-\dfrac{1}{4}\le p\le \dfrac{3}{4}$
From (iii), $-1\le p\le 1$
From (iv), $-\dfrac{1}{2}\le p\le \dfrac{1}{2}$
Thus, we consider,
$\max \left( -\dfrac{1}{4},-\dfrac{1}{2},-1,\dfrac{1}{2} \right)\le p\le \min \left( \dfrac{3}{4},1,\dfrac{1}{2},\dfrac{5}{2} \right)$
So, we get,
$\dfrac{-1}{4}\le p\le \dfrac{1}{2}$
So, we get the value of p as $p=\dfrac{1}{2}$.
Therefore, option (d) is the correct answer.

Note: After finding the probability, to check whether it is a reasonable value or not, the students can check if the value of the probability lies between 0 and 1. If so it is correct. So, the students must always remember that the value of the probability will lie between 0 and 1 only. Also, students should not get confused between mutually exclusive and mutually inclusive events. In mutually inclusive events, two events cannot occur independently and it is given as P (A or B) = P (A) + P (B) – P (A and B).