Given that the probability of three events are $$P(A) = {{3x + 1} \over 3} \Rightarrow P(B) = {{1 - x} \over 4} \Rightarrow P(C) = {{1 - 2x} \over 2}$$

As we know that probability of event lies between 0 and 1.

This can be represented as $$0 \le P(x) \le 1$$. So let us use this theorem for each probability one by one.

For given event A, we have $$0 \le P(A) \le 1$$

Let us now substitute the value given

$$0 \le \left[ {P(A) = {{3x + 1} \over 3}} \right] < 1$$ ..... (i)

We get the interval of x from Eq. (i),

$$0 \le {{3x + 1} \over 3} \le 1$$

$$ \Rightarrow {{ - 1} \over 3} \le x \le {2 \over 3}$$ ..... (ii)

Similarly, for the given event B, we have

$$0 \le P(B) \le 1$$

Let us now substitute the value given.

Let us simplify further to find the interval of x

$$0 \le {{1 - x} \over 4} \le 1$$

$$ - 3 \le x \le 1$$ ..... (iii)

For event C

$$0 \le p(c) \le 1$$

$$0 \le {{1 - 2x} \over 2} \le 1$$

$$\Rightarrow 1-2x\le 2$$

$$\Rightarrow -\frac{1}{2}\le x \le\frac{1}{2}$$ ..... (iv)

Now, three mutually exclusive events are

$$0\le P(A \cup B \cup C) \le 1$$

$$\therefore$$ [Probability must be 0 to 1]

$$P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C) -P(C \cap A)+P(A \cap B \cap C)$$

So, probability of intersection any of these event is 0

$$P(A \cap B)=P(B \cap C) = P(C \cap A)$$

$$=P(A\cap B \cap C)=0$$

So, the union of events become

$$P(A \cup B \cup C)=P(A)+P(B)+P(C)-0-0-0$$

$$\Rightarrow P(A \cup B \cup C)=P(A)+P(B)+P(C)$$

$$0<\frac{3x+1}{3}+\frac{1-x}{4}+\frac{1-2x}{2}\le 1$$

$$0\le13-3x\le12$$

$$\frac{1}{3}\le x\le\frac{13}{3}$$ ..... (v)

Finally, we have four inequalities

$$ - {1 \over 3} \le x \le {2 \over 3}...$$ [from Eq. (ii)]

$$ - 3 \le x \le 1...$$ [from Eq. (iii)]

$$ - {1 \over 2} \le x \le {1 \over 2}...$$ [from Eq. (iv)]

$${1 \over 3} \le x \le {{13} \over 3}...$$ [from Eq. (v)]

Hence, the common of each inequality is $${1 \over 3} \le x \le {1 \over 2}$$.

Probability

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