Let there be 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 'p' denote the probability of none of events occurs that satisfies 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, Probability of occurrence of $E_1$ / Probability of occurrence of $E_3$ is equal to :

### Analyzing the Setup When dealing with independent events $E_1, E_2,$ and $E_3$, we define their individual probabilities as $P_1 = P(E_1)$, $P_2 = P(E_2)$, and $P_3 = P(E_3)$. The problem defines specific regions: $\alpha$ (only $E_1$ occurs), $\beta$ (only $E_2$ occurs), $\gamma$ (only $E_3$ occurs), and $p$ (none of the events occur). Because the events are independent, we can express these as: $$\alpha = P_1(1-P_2)(1-P_3)$$ $$p = (1-P_1)(1-P_2)(1-P_3)$$ ### The Epiphany: The Ratio Trick By dividing $\alpha$ by $p$, the common terms $(1-P_2)$ and $(1-P_3)$ cancel out. This yields the elegant relationship: $$\frac{\alpha}{p} = \frac{P_1}{1-P_1}$$ By symmetry, we can derive similar expressions for the other events: $$\frac{\beta}{p} = \frac{P_2}{1-P_2} \quad \text{and} \quad \frac{\gamma}{p} = \frac{P_3}{1-P_3}$$ ### Taming the Equations We are given the equations $(\alpha - 2\beta)p = \alpha\beta$ and $(\beta - 3\gamma)p = 2\beta\gamma$. Dividing the first equation by $p^2$ transforms it into: $$\frac{\alpha}{p} - 2\frac{\beta}{p} = \frac{\alpha}{p} \cdot \frac{\beta}{p}$$ Substituting our ratio expressions, we obtain: $$\frac{P_1}{1-P_1} - 2\frac{P_2}{1-P_2} = \left( \frac{P_1}{1-P_1} \right) \left( \frac{P_2}{1-P_2} \right)$$ Multiplying by $(1-P_1)(1-P_2)$ to clear the denominators results in $P_1(1-P_2) - 2P_2(1-P_1) = P_1P_2$. Expanding this yields $P_1 - P_1P_2 - 2P_2 + 2P_1P_2 = P_1P_2$, which simplifies beautifully to: $$P_1 = 2P_2$$ ### The Final Synthesis Applying the same logic to the second equation $(\beta - 3\gamma)p = 2\beta\gamma$, we divide by $p^2$ to get: $$\frac{\beta}{p} - 3\frac{\gamma}{p} = 2\left( \frac{\beta}{p} \cdot \frac{\gamma}{p} \right)$$ Following the algebraic reduction used previously, this simplifies to: $$P_2 = 3P_3$$ Combining these results, we find $P_1 = 2(3P_3) = 6P_3$. Therefore, the ratio of the probability of $E_1$ to $E_3$ is: $$\mathbf{\frac{P_1}{P_3} = 6}$$

JEE Main 2021 (March)

LEVELJEE Main

Animated Solution for Mathematics - Probability: Let there be three independent events $E_{1}, E_{2}$ and $E_{3}$ . The probability that only $E_{1}$ occurs is $α$ , only $E_{2}$ occurs is $β$ and only $E_{3}$ occurs is $γ$ . Let 'p' denote the probability of none of events occurs that satisfies the equations $(α - 2 β) p = α β$ and $(β - 3 γ) p = 2 β γ$ . All the given probabilities are assumed to lie in the interval $(0, 1)$ . Then, Probability of occurrence of $E_{1}$ / Probability of occurrence of $E_{3}$ is equal to :

Q: Can we assume $P_1, P_2, P_3$ are independent? The question says the events are independent.

Yes, absolutely. The independence of events $E_1, E_2, E_3$ is the bedrock of this entire solution. It allows us to write the probability of the intersection as the product of individual probabilities, which is exactly how we derived the expressions for $\alpha, \beta, \gamma,$ and $p$.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z e t h e I n d e p e n d e n tE v e n t s$

Let $P (E_{1}) = P_{1}$ , $P (E_{2}) = P_{2}$ , and $P (E_{3}) = P_{3}$
Events $E_{1}, E_{2}, E_{3}$ are independent.

$D e f inin g t h e R e g i o n s$

$α = P_{1} (1 - P_{2}) (1 - P_{3})$
$β = (1 - P_{1}) P_{2} (1 - P_{3})$
$γ = (1 - P_{1}) (1 - P_{2}) P_{3}$
$p = (1 - P_{1}) (1 - P_{2}) (1 - P_{3})$

$M a t h e ma t i c a l T r an s l a t i o n$

Since events are independent, we multiply probabilities.
Only $E_{1}$ means $E_{1}$ occurs, $E_{2}$ does not, $E_{3}$ does not.

$T h e R a t i o T r i c k : \frac{α}{p}$

Dividing $α$ by $p$ :
$\frac{α}{p} = \frac{P _{1} ( 1 - P _{2} ) ( 1 - P _{3} )}{( 1 - P _{1} ) ( 1 - P _{2} ) ( 1 - P _{3} )} = \frac{P _{1}}{1 - P _{1}}$

$E x t e n d in g t h e R a t i o T r i c k$

Similarly for $β$ and $γ$ :
$\frac{β}{p} = \frac{P _{2}}{1 - P _{2}}$
$\frac{γ}{p} = \frac{P _{3}}{1 - P _{3}}$

$T r an s f or min g t h e F i r s tE q u a t i o n$

Given: $(α - 2 β) p = α β$
Divide by $p^{2}$ :
$\frac{α}{p} - 2 \frac{β}{p} = \frac{α}{p} \cdot \frac{β}{p}$

$S u b s t i t u t in g in t o t h e F i r s tE q u a t i o n$

Substitute the ratios:
$\frac{P _{1}}{1 - P _{1}} - 2 \frac{P _{2}}{1 - P _{2}} = \frac{P _{1}}{1 - P _{1}} \cdot \frac{P _{2}}{1 - P _{2}}$

$S im pl i f y in g t h e F i r s tE q u a t i o n$

Multiply by $(1 - P_{1}) (1 - P_{2})$ :
$P_{1} (1 - P_{2}) - 2 P_{2} (1 - P_{1}) = P_{1} P_{2}$

$S o l v in g f or P_{1} an d P_{2}$

Expand the brackets:
$P_{1} - P_{1} P_{2} - 2 P_{2} + 2 P_{1} P_{2} = P_{1} P_{2}$
$P_{1} - 2 P_{2} = 0 ⟹ P_{1} = 2 P_{2}$

$T r an s f or min g t h e S eco n d E q u a t i o n$

Given: $(β - 3 γ) p = 2 β γ$
Divide by $p^{2}$ :
$\frac{β}{p} - 3 \frac{γ}{p} = 2 \frac{β}{p} \cdot \frac{γ}{p}$

$S u b s t i t u t in g in t o t h e S eco n d E q u a t i o n$

Substitute the ratios:
$\frac{P _{2}}{1 - P _{2}} - 3 \frac{P _{3}}{1 - P _{3}} = 2 \frac{P _{2}}{1 - P _{2}} \cdot \frac{P _{3}}{1 - P _{3}}$

$S im pl i f y in g t h e S eco n d E q u a t i o n$

Multiply by $(1 - P_{2}) (1 - P_{3})$ :
$P_{2} (1 - P_{3}) - 3 P_{3} (1 - P_{2}) = 2 P_{2} P_{3}$

$S o l v in g f or P_{2} an d P_{3}$

Expand the brackets:
$P_{2} - P_{2} P_{3} - 3 P_{3} + 3 P_{2} P_{3} = 2 P_{2} P_{3}$
$P_{2} - 3 P_{3} = 0 ⟹ P_{2} = 3 P_{3}$

$T h e F ina l C o nn ec t i o n$

We have two relations:
$P_{1} = 2 P_{2}$
$P_{2} = 3 P_{3}$
We need to find the ratio $\frac{P _{1}}{P _{3}}$

$T h e F ina l A n s w er$

Substitute $P_{2}$ into the first relation:
$P_{1} = 2 (3 P_{3}) = 6 P_{3}$
$\frac{P _{1}}{P _{3}} = 6$

00:00 / 00:00

The Sigma Insight: Addition and Multiplication Theorems

Solution Diagram

Analyzing the Setup

When dealing with independent events

E_{1}, E_{2},

and

E_{3}

, we define their individual probabilities as

P_{1} = P (E_{1})

P_{2} = P (E_{2})

, and

P_{3} = P (E_{3})

The problem defines specific regions:

α

(only

E_{1}

occurs),

β

(only

E_{2}

occurs),

γ

(only

E_{3}

occurs), and

p

(none of the events occur). Because the events are independent, we can express these as:

α = P_{1} (1 - P_{2}) (1 - P_{3})

p = (1 - P_{1}) (1 - P_{2}) (1 - P_{3})

The Epiphany

The Ratio Trick

By dividing

α

p

, the common terms

(1 - P_{2})

and

(1 - P_{3})

cancel out. This yields the elegant relationship:

\frac{α}{p} = \frac{P _{1}}{1 - P _{1}}

By symmetry, we can derive similar expressions for the other events:

\frac{β}{p} = \frac{P _{2}}{1 - P _{2}} and \frac{γ}{p} = \frac{P _{3}}{1 - P _{3}}

Taming the Equations

We are given the equations

(α - 2 β) p = α β

and

(β - 3 γ) p = 2 β γ

. Dividing the first equation by

p^{2}

transforms it into:

\frac{α}{p} - 2 \frac{β}{p} = \frac{α}{p} \cdot \frac{β}{p}

Substituting our ratio expressions, we obtain:

\frac{P _{1}}{1 - P _{1}} - 2 \frac{P _{2}}{1 - P _{2}} = (\frac{P _{1}}{1 - P _{1}}) (\frac{P _{2}}{1 - P _{2}})

Multiplying by

(1 - P_{1}) (1 - P_{2})

to clear the denominators results in

P_{1} (1 - P_{2}) - 2 P_{2} (1 - P_{1}) = P_{1} P_{2}

. Expanding this yields

P_{1} - P_{1} P_{2} - 2 P_{2} + 2 P_{1} P_{2} = P_{1} P_{2}

, which simplifies beautifully to:

P_{1} = 2 P_{2}

The Final Synthesis

Applying the same logic to the second equation

(β - 3 γ) p = 2 β γ

, we divide by

p^{2}

to get:

\frac{β}{p} - 3 \frac{γ}{p} = 2 (\frac{β}{p} \cdot \frac{γ}{p})

Following the algebraic reduction used previously, this simplifies to:

P_{2} = 3 P_{3}

Combining these results, we find

P_{1} = 2 (3 P_{3}) = 6 P_{3}

. Therefore, the ratio of the probability of

E_{1}

E_{3}

is:

\frac{P _{1}}{P _{3}} = 6

Common Traps & FAQs

Why do we divide by

p^{2}

in the equations? Isn't that just a random algebraic manipulation?

It is not random at all! In JEE Advanced problems involving independent events, we often face products of probabilities. By dividing by

p^{2}

, we transform a non-linear product equation into a linear one involving the ratios

\frac{α}{p}

and

\frac{β}{p}

. This is a classic 'linearization' technique that turns a terrifying equation into a simple, solvable one.

Wait, why do the cross terms like

P_{1} P_{2}

cancel out in the end? Is that a coincidence?

It is definitely not a coincidence! It is the hallmark of a well-constructed problem. When you expand the terms, the cross-product terms appear on both sides of the equation with the same sign. This cancellation is the universe's way of telling you that you are on the right path. Always look for this symmetry in probability problems.

Can we assume

P_{1}, P_{2}, P_{3}

are independent? The question says the events are independent.

Yes, absolutely. The independence of events

E_{1}, E_{2}, E_{3}

is the bedrock of this entire solution. It allows us to write the probability of the intersection as the product of individual probabilities, which is exactly how we derived the expressions for

α, β, γ,

and

p

Enter Numerical Value:

Visualized Solution

$V i s u a l i z e t h e I n d e p e n d e n tE v e n t s$

$D e f inin g t h e R e g i o n s$

$M a t h e ma t i c a l T r an s l a t i o n$

$T h e R a t i o T r i c k : \frac{α}{p}$

$E x t e n d in g t h e R a t i o T r i c k$

$T r an s f or min g t h e F i r s tE q u a t i o n$

$S u b s t i t u t in g in t o t h e F i r s tE q u a t i o n$

$S im pl i f y in g t h e F i r s tE q u a t i o n$

$S o l v in g f or P_{1} an d P_{2}$

$T r an s f or min g t h e S eco n d E q u a t i o n$

$S u b s t i t u t in g in t o t h e S eco n d E q u a t i o n$

$S im pl i f y in g t h e S eco n d E q u a t i o n$

$S o l v in g f or P_{2} an d P_{3}$

$T h e F ina l C o nn ec t i o n$

$T h e F ina l A n s w er$

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Epiphany

Taming the Equations

The Final Synthesis

Similar Questions

For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = $\frac{1}{4}$ and P(All the three events occur simultaneously) = $\frac{1}{16}$ . Then the probability that at least one of the events occurs, is:

$E$ and $F$ are two independent events. The probability that both $E$ and $F$ happen is $1/12$ and the probability that neither $E$ nor $F$ happens is $1/2$ . Then,

Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $11/25$ and the probability of none of them occurring is $2/25$ . If $P (T)$ denotes the probability of occurrence of the event $T$ , then

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 $P (A \cup B) = 0.8, P (A \cap C) = 0.3, P (A \cap B \cap C) = 0.2, P (B \cap C) = β$ and $P (A \cup B \cup C) = α$ , where $0.85 \leq α \leq 0.95$ , then $β$ lies in the interval :

The probability that at least one of the events $A$ and $B$ occurs is $0.6$ . If $A$ and $B$ occur simultaneously with probability $0.2$ , then $P (\overset{ˉ}{A}) + P (\overset{ˉ}{B})$ is

If $P (B) = \frac{3}{4}, P (A \cap B \cap \overset{ˉ}{C}) = \frac{1}{3}$ and $P (\overset{ˉ}{A} \cap B \cap \overset{ˉ}{C}) = \frac{1}{3}$ , then $P (B \cap C)$ is

If $E$ and $F$ are independent events such that $0 < P (E) < 1$ and $0 < P (F) < 1$ , then

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VisualizetheIndependentEvents

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DefiningtheRegions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MathematicalTranslation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheRatioTrick:pα​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ExtendingtheRatioTrick

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TransformingtheFirstEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SubstitutingintotheFirstEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifyingtheFirstEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SolvingforP1​andP2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TransformingtheSecondEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SubstitutingintotheSecondEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifyingtheSecondEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SolvingforP2​andP3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheFinalConnection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheFinalAnswer

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Epiphany

Taming the Equations

The Final Synthesis

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E and F are two independent events. The probability that both E and F happen is 1/12 and the probability that neither E nor F happens is 1/2. Then,

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then P(Aˉ)+P(Bˉ) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P(B)=43​,P(A∩B∩Cˉ)=31​ and P(Aˉ∩B∩Cˉ)=31​, then P(B∩C) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If E and F are independent events such that 0<P(E)<1 and 0<P(F)<1, then

$V i s u a l i z e t h e I n d e p e n d e n tE v e n t s$

$D e f inin g t h e R e g i o n s$

$M a t h e ma t i c a l T r an s l a t i o n$

$T h e R a t i o T r i c k : \frac{α}{p}$

$E x t e n d in g t h e R a t i o T r i c k$

$T r an s f or min g t h e F i r s tE q u a t i o n$

$S u b s t i t u t in g in t o t h e F i r s tE q u a t i o n$

$S im pl i f y in g t h e F i r s tE q u a t i o n$

$S o l v in g f or P_{1} an d P_{2}$

$T r an s f or min g t h e S eco n d E q u a t i o n$

$S u b s t i t u t in g in t o t h e S eco n d E q u a t i o n$

$S im pl i f y in g t h e S eco n d E q u a t i o n$

$S o l v in g f or P_{2} an d P_{3}$

$T h e F ina l C o nn ec t i o n$

$T h e F ina l A n s w er$

$E$ and $F$ are two independent events. The probability that both $E$ and $F$ happen is $1/12$ and the probability that neither $E$ nor $F$ happens is $1/2$ . Then,

The probability that at least one of the events $A$ and $B$ occurs is $0.6$ . If $A$ and $B$ occur simultaneously with probability $0.2$ , then $P (\overset{ˉ}{A}) + P (\overset{ˉ}{B})$ is

If $P (B) = \frac{3}{4}, P (A \cap B \cap \overset{ˉ}{C}) = \frac{1}{3}$ and $P (\overset{ˉ}{A} \cap B \cap \overset{ˉ}{C}) = \frac{1}{3}$ , then $P (B \cap C)$ is

If $E$ and $F$ are independent events such that $0 < P (E) < 1$ and $0 < P (F) < 1$ , then