Animated Solution for Mathematics - Probability: The probabilities of three events $A,B$ and $C$ are given by $P(A)=0.6,P(B)=0.4$ and $P(C)=0.5$ If $P(A\cup B)=0.8,P(A\cap C)=0.3,P(A\cap B\cap C)=0.2,P(B\cap C)=\beta$ and $P(A\cup B\cup C)=\alpha$, where $0.85\le \alpha \le 0.95$, then $\beta$ lies in the interval :

Unlocking the Venn Diagram

Welcome, future engineers! Today, we are going to peel back the layers of a beautiful probability problem. Often, when we see a question involving three events—$A$, $B$, and $C$—our minds immediately jump to the Inclusion-Exclusion Principle.

But before we dive into the heavy algebra, let's pause and visualize. Imagine three circles on a piece of paper, overlapping in a complex, beautiful dance.

The probability of their union, $\alpha$, is simply the total area covered by these circles. Our mission is to find the range of $\beta$, which represents the intersection of events $B$ and $C$.

The Two-Event Warmup

Before we tackle the three-circle beast, let's look at the relationship between $A$ and $B$. We are given $P(A)=0.6$, $P(B)=0.4$, and $P(A\cup B)=0.8$.

The addition theorem for two events is our best friend here:

Why do we subtract $P(A\cap B)$? Because when we add the area of circle $A$ and circle $B$, the overlapping region is counted twice. We subtract it once to correct that double-counting.

Plugging in our values, we get $0.8=0.6+0.4-P(A\cap B)$. A quick calculation shows that $P(A\cap B)=0.2$. Keep this number safe; it is a vital piece of our larger puzzle.

The Inclusion-Exclusion Principle

Now, let's expand our horizon to all three events. The Inclusion-Exclusion Principle is the backbone of this problem. It states that the probability of the union of three events is given by:

This formula is elegant. We sum the individual probabilities, subtract the pairwise intersections (because they were double-counted), and then add back the triple intersection $P(A\cap B\cap C)$ because it was subtracted one too many times during the pairwise step. It is a perfect balancing act.

Let's substitute our known values: $\alpha =0.6+0.4+0.5-(0.2+\beta +0.3)+0.2$. Notice how we used our previously calculated $P(A\cap B)=0.2$ here.

Simplifying this, we sum the positive terms: $0.6+0.4+0.5+0.2=1.7$. Then, we combine the negative terms: $-(0.2+0.3)=-0.5$. The equation simplifies beautifully to $\alpha =1.2-\beta$.

The Inequality Dance

We are almost there! The problem provides a constraint: $0.85\le \alpha \le 0.95$. Since we know $\alpha =1.2-\beta$, we can substitute this into our inequality:

To isolate $\beta$, we first subtract $1.2$ from all parts of the inequality. This gives us $-0.35\le -\beta \le -0.25$.

Now, here is the moment of truth. To get $\beta$ by itself, we multiply the entire inequality by $-1$. Remember the golden rule: when you multiply by a negative number, the inequality signs must flip!

Thus, we get $0.35\ge \beta \ge 0.25$. Rearranging this into standard interval notation, we find that $\beta \in [0.25,0.35]$.

Look at what we have achieved! We navigated the complexities of the Inclusion-Exclusion Principle and handled the inequality constraints with precision.

Probability is not just about formulas; it is about understanding the geometry of overlapping possibilities. Keep practicing, keep visualizing, and most importantly, keep falling in love with the logic behind the math. You've got this!