Of the three independent events E₁, E₂ and E₃, the probability that only E₁ occurs is α, only E₂ occ

Q: Wait, why can't we just assume $P(E_1) = \alpha$? Isn't that what the problem implies?

That is a classic trap! The problem states $\alpha$ is the probability that *only* $E_1$ occurs. This means $E_1$ happens AND $E_2$ does not happen AND $E_3$ does not happen. If you assume $P(E_1) = \alpha$, you are ignoring the possibility that $E_2$ or $E_3$ could also occur. Always distinguish between the marginal probability of an event and the probability of an event occurring in isolation.

Q: Why did we divide by $p^2$? That feels like a random step.

It is not random; it is strategic! When you see equations involving products of probabilities like $\alpha\beta$ and $\alpha - 2\beta$, you should immediately look for a way to normalize the terms. By dividing by $p^2$, we convert the equation into a form involving the ratios $\frac{\alpha}{p}$ and $\frac{\beta}{p}$. This reduces the degree of the equation and makes the algebra collapse beautifully. In JEE Advanced, look for these symmetries!

Q: What if $p=0$? Can we divide by it?

Great catch on the domain! The problem explicitly states that all given probabilities lie in the interval $(0, 1)$. This implies that $x, y, z$ are strictly between $0$ and $1$, which means $(1-x), (1-y), (1-z)$ are also non-zero. Therefore, $p = (1-x)(1-y)(1-z)$ is strictly greater than $0$. We are safe to divide.

Question

Of the three independent events $E_1, E_2$ and $E_3$, the probability that only $E_1$ occurs is $\alpha$, only $E_2$ occurs is $\beta$ and only $E_3$ occurs is $\gamma$. Let the probability $p$ that none of events $E_1, E_2$ or $E_3$ occurs satisfy the equations $(\alpha - 2\beta)p = \alpha\beta$ and $(\beta - 3\gamma)p = 2\beta\gamma$. All the given probabilities are assumed to lie in the interval $(0, 1)$. Then $\frac{	ext{Probability of occurrence of } E_1}{	ext{Probability of occurrence of } E_3}$ is .........

Sigma Percentile · Accepted Answer

### The Symphony of Independent Events: A Journey into Probability

Welcome, future engineer. Today, we are not just solving a probability problem; we are peeling back the layers of a system governed by the beautiful laws of independence.

When you look at a problem involving three independent events, $E_1$, $E_2$, and $E_3$, do not see them as abstract symbols. See them as three distinct circles in a Venn diagram, overlapping in a dance of chance. Our goal is to find the ratio of the probability of $E_1$ to the probability of $E_3$.

### Phase 1: Defining the Landscape

First, let us ground ourselves. We define the individual probabilities as $P(E_1) = x$, $P(E_2) = y$, and $P(E_3) = z$.

Since these are independent events, the probability of any combination of them occurring is simply the product of their individual probabilities (or their complements).

The problem defines $\alpha$ as the probability that *only* $E_1$ occurs. This is a specific region in our Venn diagram: $E_1$ happens, but $E_2$ and $E_3$ do not. Mathematically, this is expressed as:

$$\alpha = x(1-y)(1-z)$$

Similarly, for $\beta$ (only $E_2$) and $\gamma$ (only $E_3$), we have:

$$\beta = (1-x)y(1-z)$$

$$\gamma = (1-x)(1-y)z$$

And finally, $p$, the probability that none of the events occur, is the region outside all three circles:

$$p = (1-x)(1-y)(1-z)$$

### Phase 2: The Ratio Trick

Here is where the magic happens. If you try to substitute these expressions directly into the given equations, you will find yourself drowning in a sea of variables.

Notice how $\alpha$, $\beta$, and $\gamma$ are almost identical to $p$, just with one term swapped. If we divide $\alpha$ by $p$, the terms $(1-y)$ and $(1-z)$ cancel out perfectly!

We are left with:

$$\frac{\alpha}{p} = \frac{x(1-y)(1-z)}{(1-x)(1-y)(1-z)} = \frac{x}{1-x}$$

This is our golden key. By applying this logic to $\beta$ and $\gamma$, we get:

$$\frac{\beta}{p} = \frac{y}{1-y}$$

$$\frac{\gamma}{p} = \frac{z}{1-z}$$

### Phase 3: The Algebraic Collapse

Now, let us look at the first given equation: $(\alpha - 2\beta)p = \alpha\beta$. If we divide both sides by $p^2$, we get:

$$\frac{\alpha}{p} - \frac{2\beta}{p} = \frac{\alpha}{p} \cdot \frac{\beta}{p}$$

Substitute our ratio terms: let $u = \frac{x}{1-x}$ and $v = \frac{y}{1-y}$. The equation becomes $u - 2v = uv$.

Multiplying by $(1-x)(1-y)$ clears the denominators and leads us to:

$$x(1-y) - 2y(1-x) = xy$$

$$x - xy - 2y + 2xy = xy$$

$$x - 2y = 0 \implies x = 2y$$

Isn't that elegant? The complexity vanishes, leaving a simple linear relationship. We repeat this process for the second equation: $(\beta - 3\gamma)p = 2\beta\gamma$.

Dividing by $p^2$ gives:

$$\frac{\beta}{p} - \frac{3\gamma}{p} = 2 \cdot \frac{\beta}{p} \cdot \frac{\gamma}{p}$$

Substituting our ratios, we find:

$$y(1-z) - 3z(1-y) = 2yz$$

$$y - yz - 3z + 3yz = 2yz$$

$$y - 3z = 0 \implies y = 3z$$

### The Final Victory

We have arrived at the finish line. We know $x = 2y$ and $y = 3z$. To find the ratio of the probability of $E_1$ to $E_3$, we simply calculate $\frac{x}{z}$:

$$\frac{x}{z} = \frac{2y}{z} = 2 \left( \frac{y}{z} \right) = 2(3) = 6$$

The final answer is **6**.

Remember, in JEE Advanced, the math is rarely about brute force. It is about finding the symmetry, the ratio, or the substitution that makes the problem solve itself.

Enter Numerical Value:

Visualized Solution

Defining Event Probabilities

Expressing $α$ (Only $E_{1}$ )

Expressing $β$ and $γ$

Expressing $p$ (None Occur)

The Ratio Transformation

Setting up Equation 1

Solving for $x$ and $y$

Setting up Equation 2

Solving for $y$ and $z$

Final Ratio Calculation

The Sigma Insight: Addition and Multiplication Theorems

The Symphony of Independent Events

Phase 1

Phase 2

Phase 3

The Final Victory

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