If (1+3p)/3, (1-p)/4 and (1-2p)/2 are the probabilities of three mutually exclusive events, then the

Q: Wait, why do we need $P_i \ge 0$ and $P_i \le 1$?

That is the bedrock of probability theory! By definition, probability is a measure of likelihood, and it is normalized to be between $0$ (impossible) and $1$ (certain). If any of your expressions resulted in a value outside this range, that specific value of $p$ would be physically meaningless.

Q: Why does the sum have to be $\le 1$?

Because the events are mutually exclusive! If events are mutually exclusive, the probability of their union is the sum of their individual probabilities. Since the probability of the union of any events cannot exceed $1$, the sum of these probabilities must also be $\le 1$.

Q: What if I forget to flip the inequality sign when dividing by a negative number?

That is a classic trap! Whenever you multiply or divide an inequality by a negative number, the direction of the inequality must reverse. If you miss this, your entire interval will be wrong. Always double-check your signs when isolating $p$!

Question

If $\frac{1+3p}{3}, \frac{1-p}{4}$ and $\frac{1-2p}{2}$ are the probabilities of three mutually exclusive events, then the set of all values of $p$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the threshold of a decision, where the outcome is not certain but governed by the elegant laws of probability. In this problem, we are given three expressions for the probabilities of three mutually exclusive events:

$$P_1 = \frac{1+3p}{3}, \quad P_2 = \frac{1-p}{4}, \quad P_3 = \frac{1-2p}{2}$$

Our mission is to find the range of $p$ that keeps these probabilities valid. This is not just algebra; it is the art of defining the boundaries of possibility.

### The First Phase: Individual Constraints

Before we consider the events together, we must ensure each event is individually valid. The first axiom of probability dictates that for any event $E$, $0 \le P(E) \le 1$.

For $P_1$:
$$0 \le \frac{1+3p}{3} \le 1 \Rightarrow 0 \le 1+3p \le 3 \Rightarrow -1 \le 3p \le 2 \Rightarrow -\frac{1}{3} \le p \le \frac{2}{3}$$

For $P_2$:
$$0 \le \frac{1-p}{4} \le 1 \Rightarrow 0 \le 1-p \le 4 \Rightarrow -1 \le -p \le 3 \Rightarrow -3 \le p \le 1$$

For $P_3$:
$$0 \le \frac{1-2p}{2} \le 1 \Rightarrow 0 \le 1-2p \le 2 \Rightarrow -1 \le -2p \le 1 \Rightarrow -\frac{1}{2} \le p \le \frac{1}{2}$$

### The Second Phase: The Collective Constraint

Now, we invoke the power of mutual exclusivity. Because these events cannot happen at the same time, the sum of their probabilities must satisfy $P_1 + P_2 + P_3 \le 1$.

Substituting our expressions:
$$\frac{1+3p}{3} + \frac{1-p}{4} + \frac{1-2p}{2} \le 1$$

To solve this, we find a common denominator of $12$:
$$\frac{4(1+3p) + 3(1-p) + 6(1-2p)}{12} \le 1$$

Multiplying by $12$ and expanding the brackets:
$$4 + 12p + 3 - 3p + 6 - 12p \le 12$$

Simplifying the expression:
$$13 - 3p \le 12 \Rightarrow 1 \le 3p \Rightarrow p \ge \frac{1}{3}$$

### The Final Synthesis: Finding the Intersection

We now have four conditions for $p$:
1. $p \in [-\frac{1}{3}, \frac{2}{3}]$
2. $p \in [-3, 1]$
3. $p \in [-\frac{1}{2}, \frac{1}{2}]$
4. $p \in [\frac{1}{3}, \infty)$

To find the valid set of $p$, we must find the intersection of these intervals. The lower bound is determined by the most restrictive condition, which is $\frac{1}{3}$. The upper bound is determined by the most restrictive condition, which is $\frac{1}{2}$.

Thus, the valid range for $p$ is:
$$\frac{1}{3} \le p \le \frac{1}{2}$$

You have successfully navigated the constraints and found the truth hidden within the algebra!

Animated Solution for Mathematics - Probability: If $\frac{1 + 3 p}{3}, \frac{1 - p}{4}$ and $\frac{1 - 2 p}{2}$ are the probabilities of three mutually exclusive events, then the set of all values of $p$ is .........

Visualized Solution

The Given Probabilities

Axioms of Probability

Bounding $P_{1}$

Solving for $p$ in $P_{1}$

Bounding $P_{2}$

Solving for $p$ in $P_{2}$

Bounding $P_{3}$

Solving for $p$ in $P_{3}$

The Sum Constraint

Simplifying the Sum

Expanding the Numerator

Final Constraint on $p$

Finding the Intersection

The Final Answer

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The First Phase

The Second Phase

The Final Synthesis

Similar Questions

Let $E_{1}, E_{2}, E_{3}$ be three mutually exclusive events such that $P (E_{1}) = \frac{2 + 3 p}{6}$ , $P (E_{2}) = \frac{2 - p}{8}$ 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 :

Three dice are rolled. If the probability of getting different numbers on the three dice is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then $q - p$ is equal to

Comprehension Passage

The value of $\frac{625}{4} p_{1}$ is

The value of $\frac{125}{4} p_{2}$ is

The probability that two randomly selected subsets of the set ${1, 2, 3, 4, 5}$ have exactly two elements in their intersection, is:

$A$ is a set containing $n$ elements. A subset $P$ of $A$ is chosen at random. The set $A$ is reconstructed by replacing the elements of $P$ . A subset $Q$ of $A$ is again chosen at random. Find the probability that $P$ and $Q$ have no common elements.

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is

A die is thrown. Let $A$ be the event that the number obtained is greater than 3. Let $B$ be the event that the number obtained is less than 5. Then $P (A \cup B)$ is

Let $S = {E_{1}, E_{2}, ..., E_{8}}$ be a sample space of random experiment such that $P (E_{n}) = \frac{n}{36}$ for every $n = 1, 2, ..., 8$ . Then the number of elements in the set ${A \subset S : P (A) \geq \frac{4}{5}}$ is ____.

If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is