Animated Solution for Mathematics - Probability: Let $E_1,E_2,E_3$ be three mutually exclusive events such that $P(E_1)=\frac{2+3p}{6}$, $P(E_2)=\frac{2-p}{8}$ and $P(E_3)=\frac{1-p}{2}$. If the maximum and minimum values of $p$ are $p_1$ and $p_2$, then $(p_1+p_2)$ is equal to :

Welcome, fellow traveler on the path to JEE excellence. Today, we aren't just solving a math problem; we are building a logical fortress.

When we look at probability, we often think of dice or cards, but at its core, probability is a game of boundaries. We are given three events, $E_1,E_2,E_3$, and a parameter $p$ that dictates their likelihood. Our mission is to find the 'safe zone' for $p$ where these events can coexist without violating the fundamental laws of the universe.

Before we look at the group, we must look at the individuals. The most sacred rule in probability is that for any event $E$, the probability $P(E)$ must satisfy $0\le P(E)\le 1$.

If we violate this, we are essentially saying an event is 'more than certain' or 'less than impossible,' both of which are mathematical nonsense.

For $E_1$, we have $P(E_1)=\frac{2+3p}{6}$. Setting this between 0 and 1 gives us:

Multiplying by 6 and subtracting 2, we find $3p$ is between $-2$ and $4$, leading to $p\in [-\frac{2}{3},\frac{4}{3}]$. This is our first constraint.

We repeat this for $E_2$ and $E_3$. For $E_2$, we find $p\in [-6,2]$, and for $E_3$, we find $p\in [-1,1]$. These are the 'survival ranges' for each event.

Now, here is where the plot thickens. We are told these events are 'mutually exclusive.' This is a powerful geometric constraint.

It means these events occupy distinct, non-overlapping regions in the sample space. Because they don't overlap, the probability of their union is simply the sum of their individual probabilities.

Since the total probability of the sample space is 1, we must have:

This is the constraint that catches most students off guard. It is not enough for the events to be individually valid; they must be collectively valid. Substituting our expressions, we get:

To solve this, we find the common denominator, which is 24. Multiplying through, we get:

Expanding this, we see the $p$ terms dance: $8+12p+6-3p+12-12p\le 24$. Simplifying, we arrive at $26-3p\le 24$, which elegantly simplifies to $3p\ge 2$, or $p\ge \frac{2}{3}$.

We now have four separate constraints on $p$. To find the valid range, we must find the intersection of all these sets.

We need $p$ to be in $[-\frac{2}{3},\frac{4}{3}]$, AND $[-6,2]$, AND $[-1,1]$, AND $[\frac{2}{3},\infty )$.

Imagine these as four different filters. Only the values of $p$ that pass through all four filters will survive. The intersection of these intervals is $[\frac{2}{3},1]$.

This is the 'Golden Interval.' Any value of $p$ within this range keeps our probability model perfectly balanced.

We have identified our boundaries: the minimum value $p_2=\frac{2}{3}$ and the maximum value $p_1=1$. The problem asks for the sum $p_1+p_2$.

Look at that! Through careful, systematic application of the axioms of probability, we have navigated the constraints and arrived at the solution.

Remember, in the JEE, the math is rarely the enemy—it is the lack of structure. Keep your constraints organized, respect the boundaries, and the answer will always reveal itself.