Question

Events A, B, C are mutually exclusive events such that $P\left( A \right) = \dfrac{{3x + 1}}{3}$ , $P\left( B \right) = \dfrac{{1 - x}}{4}$ and $P\left( C \right) = \dfrac{{1 - 2x}}{2}$ . Then the set of possible values of x are in the interval.

$$A.{\text{ }}\left[ {\dfrac{1}{3},\dfrac{2}{3}} \right] \\\ B.{\text{ }}\left[ {\dfrac{1}{3},\dfrac{{13}}{3}} \right] \\\ C.{\text{ }}\left[ {0,1} \right] \\\ D.{\text{ }}\left[ {\dfrac{1}{3},\dfrac{1}{2}} \right] \\\$$

Answer 1

Hint:In order to solve the given problem we will use the concept that the probability of any event lies between 0 and 1. With the use of this concept on all the given probabilities we will get 3 inequality. Further we will find the union of all the three events A, B and C and by the use of formula for mutually exclusive events and the above concept we will find fourth inequality. Further we will find the intersection of each of the inequality we will find the final interval of x.

Complete step-by-step answer:
Given that the probability of three events are:

$$P\left( A \right) = \dfrac{{3x + 1}}{3} \\\ P\left( B \right) = \dfrac{{1 - x}}{4} \\\ P\left( C \right) = \dfrac{{1 - 2x}}{2} \\\$$

We have to find the interval of x.
As we know that the probability of the event lies between 0 and 1.
This can be represented as:
$0 \leqslant P\left( x \right) \leqslant 1$
So let us use this theorem for each probability one by one.
For the given event A, we have:
$0 \leqslant P\left( A \right) \leqslant 1$
Let us now substitute the value given:
$0 \leqslant \left[ {P\left( A \right) = \dfrac{{3x + 1}}{3}} \right] \leqslant 1$
Let us simplify further to find the interval of x.

$$\Rightarrow 0 \leqslant \dfrac{{3x + 1}}{3} \leqslant 1 \\\ \Rightarrow 0 \leqslant 3x + 1 \leqslant 3 \\\ \Rightarrow - 1 \leqslant 3x \leqslant 2 \\\ \Rightarrow \dfrac{{ - 1}}{3} \leqslant x \leqslant \dfrac{2}{3} \\\$$

So we have the inequality as:
$\Rightarrow \dfrac{{ - 1}}{3} \leqslant x \leqslant \dfrac{2}{3}............(1)$
Similarly for the given event B, we have:
$0 \leqslant P\left( B \right) \leqslant 1$
Let us now substitute the value given:
$0 \leqslant \left[ {P\left( B \right) = \dfrac{{1 - x}}{4}} \right] \leqslant 1$
Let us simplify further to find the interval of x.

$$\Rightarrow 0 \leqslant \dfrac{{1 - x}}{4} \leqslant 1 \\\ \Rightarrow 0 \leqslant 1 - x \leqslant 4 \\\ \Rightarrow - 4 \leqslant x - 1 \leqslant 0 \\\ \Rightarrow - 3 \leqslant x \leqslant 1 \\\$$

So we have the inequality as:
$\Rightarrow - 3 \leqslant x \leqslant 1............(2)$
Also for the given event C, we have:
$0 \leqslant P\left( C \right) \leqslant 1$
Let us now substitute the value given:
$0 \leqslant \left[ {P\left( C \right) = \dfrac{{1 - 2x}}{2}} \right] \leqslant 1$
Let us simplify further to find the interval of x.

$$\Rightarrow 0 \leqslant \dfrac{{1 - 2x}}{2} \leqslant 1 \\\ \Rightarrow 0 \leqslant 1 - 2x \leqslant 2 \\\ \Rightarrow - 2 \leqslant 2x - 1 \leqslant 0 \\\ \Rightarrow - 1 \leqslant 2x \leqslant 1 \\\ \Rightarrow \dfrac{{ - 1}}{2} \leqslant x \leqslant \dfrac{1}{2} \\\$$

So we have the inequality as:
$\Rightarrow \dfrac{{ - 1}}{2} \leqslant x \leqslant \dfrac{1}{2}............(3)$
We have got the three inequalities from the above three equations which represents the interval of x.
Now let us proceed to find the probability of the union of three mutually exclusive events.
Union of three events is represented as:
$P\left( {A \cup B \cup C} \right)$
Using the above theorem we have:
For the given union of events A, B and C we have:
$0 \leqslant P\left( {A \cup B \cup C} \right) \leqslant 1$
As we know that union of the events can be further simplified as:
$P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right) + P\left( {A \cap B \cap C} \right)$
As the events A, B and C are mutually exclusive so the probability of intersection of any of these events is 0. So, we have:
$P\left( {A \cap B} \right) = P\left( {B \cap C} \right) = P\left( {C \cap A} \right) = P\left( {A \cap B \cap C} \right) = 0$
So, the union of the events becomes:

$$P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) - 0 - 0 - 0 + 0 \\\ P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) \\\$$

Hence our inequality becomes:

$$\because 0 \leqslant P\left( {A \cup B \cup C} \right) \leqslant 1 \\\ \Rightarrow 0 \leqslant P\left( A \right) + P\left( B \right) + P\left( C \right) \leqslant 1 \\\$$

Let us now substitute the value given:

$$\because 0 \leqslant P\left( A \right) + P\left( B \right) + P\left( C \right) \leqslant 1 \\\ \Rightarrow 0 \leqslant \dfrac{{3x + 1}}{3} + \dfrac{{1 - x}}{4} + \dfrac{{1 - 2x}}{2} \leqslant 1 \\\ \Rightarrow 0 \leqslant \dfrac{{13 - 3x}}{{12}} \leqslant 1 \\\$$

Let us simplify further to find the interval of x.

$$\Rightarrow 0 \leqslant \dfrac{{13 - 3x}}{{12}} \leqslant 1 \\\ \Rightarrow 0 \leqslant 13 - 3x \leqslant 12 \\\ \Rightarrow - 12 \leqslant 3x - 13 \leqslant 0 \\\ \Rightarrow 1 \leqslant 3x \leqslant 13 \\\ \Rightarrow \dfrac{1}{3} \leqslant x \leqslant \dfrac{{13}}{3} \\\$$

So we have the inequality as:
$\Rightarrow \dfrac{1}{3} \leqslant x \leqslant \dfrac{{13}}{3}............(4)$
So finally we have four equations of inequality.

$$\Rightarrow \dfrac{{ - 1}}{3} \leqslant x \leqslant \dfrac{2}{3}............(1) \\\ \Rightarrow - 3 \leqslant x \leqslant 1............(2) \\\ \Rightarrow \dfrac{{ - 1}}{2} \leqslant x \leqslant \dfrac{1}{2}............(3) \\\ \Rightarrow \dfrac{1}{3} \leqslant x \leqslant \dfrac{{13}}{3}............(4) \\\$$

Let us find the intersection of these four inequalities.
Comparing all the inequality of x and taking the intersection or common of each of the inequality, we get:
$\Rightarrow \dfrac{1}{3} \leqslant x \leqslant \dfrac{1}{2}$
Hence, the set of possible values of x are in the interval $\dfrac{1}{3} \leqslant x \leqslant \dfrac{1}{2}$
So, the correct answer is option D

Note:In order to solve such problems related to finding the interval of some variables first we need to find out different inequality equations that can be formed by the help of given problem statements and other information. Further students must remember how to find the common interval. We don't take the minimum and maximum to find the interval, rather we take the least possible interval within different inequality which is found to be common between all the inequalities.