JEE Main 2021 (24 February Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Probability: Let $B_{i} (i = 1, 2, 3)$ be three independent events in a sample space. The probability that only $B_{1}$ occur is $α$ , only $B_{2}$ occurs is $β$ and only $B_{3}$ occurs is $γ$ . Let $p$ be the probability that none of the events $B_{i}$ occurs and these 4 probabilities satisfy the equations $(α - 2 β) p = α β$ and $(β - 3 γ) p = 2 β γ$ (All the probabilities are assumed to lie in the interval $(0, 1)$ ). Then $\frac{P ( B _{1} )}{P ( B _{3} )}$ is equal to

Q: Why does the term $(1-z)$ cancel out in the first equation but $(1-x)$ cancels in the second?

It comes down to the definitions of $\alpha, \beta,$ and $\gamma$. In the first equation, we are dealing with $\alpha$ and $\beta$, both of which contain the term $(1-z)$. In the second equation, we deal with $\beta$ and $\gamma$, both of which contain $(1-x)$. The math is telling you exactly which variables are 'active' in that specific relationship. Trust the algebra; it is guiding you to the solution.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Visualize the sample space as a Venn diagram. Since the events

B_{1}, B_{2}, B_{3}

are independent, the probability of 'only

B_{1}

occurring' is not just

P (B_{1})

, but

P (B_{1}) \cdot P (B_{1}^{c}) \cdot P (B_{1}^{c})

. Express

α, β, γ,

and

p

in terms of

x, y,

and

z

Hint 2

Hint 2: Do not rush to expand the algebraic expressions immediately. Look for common factors like

(1 - x), (1 - y),

(1 - z)

that appear on both sides of the given equations. These are your keys to simplifying the problem without getting lost in a sea of variables.

Hint 3

Hint 3: Start by substituting the expressions into the first equation:

[x (1 - y) (1 - z) - 2 y (1 - x) (1 - z)] (1 - x) (1 - y) (1 - z) = x (1 - y) (1 - z) \cdot y (1 - x) (1 - z)

. Divide both sides by the common factor

(1 - x) (1 - y) (1 - z)^{2}

to reveal the hidden relationship between

x

and

y

Visualized Solution

$D e f inin g E v e n t s an d P r o babi l i t i es$

Let $P (B_{1}) = x$ , $P (B_{2}) = y$ , and $P (B_{3}) = z$ .
The events $B_{1}, B_{2}, B_{3}$ are independent.
Sample space $S$ contains all possible outcomes.

$F or m u l a t in g R e g i o n P r o babi l i t i es$

Probability of only $B_{1}$ : $α = x (1 - y) (1 - z)$
Probability of only $B_{2}$ : $β = y (1 - x) (1 - z)$
Probability of only $B_{3}$ : $γ = z (1 - x) (1 - y)$
Probability of none: $p = (1 - x) (1 - y) (1 - z)$

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

Given equation: $(α - 2 β) p = α β$
Substitute the expressions for $α, β$ , and $p$ .
$[x (1 - y) (1 - z) - 2 y (1 - x) (1 - z)] (1 - x) (1 - y) (1 - z) = x (1 - y) (1 - z) \cdot y (1 - x) (1 - z)$

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

Notice the common factor $(1 - z)$ in the bracket.
Divide both sides by $(1 - x) (1 - y) (1 - z)^{2}$ .
Result: $x (1 - y) - 2 y (1 - x) = x y$

$S o l v in g f or x an d y$

Expand the terms: $x - x y - 2 y + 2 x y = x y$
Combine like terms: $x - 2 y + x y = x y$
Subtract $x y$ from both sides: $x - 2 y = 0$
Conclusion: $x = 2 y$

$S e tt in g U pt h e S eco n d E q u a t i o n$

Given equation: $(β - 3 γ) p = 2 β γ$
Substitute the expressions for $β, γ$ , and $p$ .
$[y (1 - x) (1 - z) - 3 z (1 - x) (1 - y)] (1 - x) (1 - y) (1 - z) = 2 [y (1 - x) (1 - z) \cdot z (1 - x) (1 - y)]$

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

Notice the common factor $(1 - x)$ in the bracket.
Divide both sides by $(1 - x)^{2} (1 - y) (1 - z)$ .
Result: $y (1 - z) - 3 z (1 - y) = 2 y z$

$S o l v in g f or y an d z$

Expand the terms: $y - y z - 3 z + 3 y z = 2 y z$
Combine like terms: $y - 3 z + 2 y z = 2 y z$
Subtract $2 y z$ from both sides: $y - 3 z = 0$
Conclusion: $y = 3 z$

$C a l c u l a t in g t h e F ina l R a t i o$

We need to find the ratio $\frac{P ( B _{1} )}{P ( B _{3} )} = \frac{x}{z}$ .
We know $x = 2 y$ and $y = 3 z$ .
Substitute $y$ into the first equation: $x = 2 (3 z) = 6 z$ .
Therefore, $\frac{x}{z} = 6$ .

00:00 / 00:00

The Sigma Insight: Addition and Multiplication Theorems

Solution Diagram

Analyzing the Setup

When we say events

B_{1}, B_{2},

and

B_{3}

are independent, we are saying that the occurrence of one tells us absolutely nothing about the occurrence of the others. They are free spirits in the sample space.

Let us define their probabilities as

P (B_{1}) = x

P (B_{2}) = y

, and

P (B_{3}) = z

The Translation

From Words to Math

Imagine a Venn diagram. The region representing 'only

B_{1}

' is the intersection of

B_{1}

with the complement of

B_{2}

and the complement of

B_{3}

. Because of independence, the probability of this intersection is simply the product of the individual probabilities:

α = x (1 - y) (1 - z)

Similarly, we define:

β = y (1 - x) (1 - z)

γ = z (1 - x) (1 - y)

The probability that none of the events occur is the intersection of all complements:

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

The First Dance

Simplifying the Complex

We are given the equation

(α - 2 β) p = α β

. Substituting our definitions, we obtain:

[x (1 - y) (1 - z) - 2 y (1 - x) (1 - z)] (1 - x) (1 - y) (1 - z) = x (1 - y) (1 - z) \cdot y (1 - x) (1 - z)

Notice the common factor

(1 - z)

and the product

(1 - x) (1 - y) (1 - z)

appearing on both sides. By dividing both sides by

(1 - x) (1 - y) (1 - z)^{2}

, the equation collapses into:

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

Expanding this gives

x - x y - 2 y + 2 x y = x y

. The

x y

terms cancel out, leaving us with

x - 2 y = 0

, or simply:

x = 2 y

The Second Act

Symmetry in Action

Now, we apply the same logic to the second equation:

(β - 3 γ) p = 2 β γ

. Substituting our expressions, we get:

[y (1 - x) (1 - z) - 3 z (1 - x) (1 - y)] (1 - x) (1 - y) (1 - z) = 2 [y (1 - x) (1 - z) \cdot z (1 - x) (1 - y)]

Again, notice the common factor

(1 - x)

. By dividing both sides by

(1 - x)^{2} (1 - y) (1 - z)

, we are left with:

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

Expanding this yields

y - y z - 3 z + 3 y z = 2 y z

. The

y z

terms cancel out beautifully, leaving

y - 3 z = 0

, or:

y = 3 z

The Final Convergence

We have arrived at the finish line. We know

x = 2 y

and

y = 3 z

. The question asks for the ratio

\frac{P ( B _{1} )}{P ( B _{3} )}

, which is

\frac{x}{z}

Substituting our findings:

x = 2 (3 z) = 6 z

Therefore, the final ratio is:

\frac{x}{z} = 6

Common Traps & FAQs

Wait, why do we assume

P (B_{1}) = x, P (B_{2}) = y, P (B_{3}) = z

? Can't we just solve for the probabilities directly?

That is a great question! In this problem, we aren't given the absolute values of the probabilities, but rather the relationships between them. By assigning variables

x, y,

and

z

, we create a mathematical framework that allows us to manipulate these relationships algebraically. We aren't solving for the values themselves, but for the ratio between them, which is exactly what the question asks for.

The algebra looks terrifyingly long. Is there a trick to avoid expanding everything?

You have hit on the most important lesson of JEE Advanced algebra! Never expand blindly. Always look for common factors first. In this problem, the structure of independent events creates repeating terms like

(1 - x), (1 - y),

and

(1 - z)

. By dividing these out early, you turn a massive polynomial equation into a simple linear one. Always look for the 'cancellation path' before you start multiplying.

Why does the term

(1 - z)

cancel out in the first equation but

(1 - x)

cancels in the second?

It comes down to the definitions of

α, β,

and

γ

. In the first equation, we are dealing with

α

and

β

, both of which contain the term

(1 - z)

. In the second equation, we deal with

β

and

γ

, both of which contain

(1 - x)

. The math is telling you exactly which variables are 'active' in that specific relationship. Trust the algebra; it is guiding you to the solution.

Enter Numerical Value:

Visualized Solution

$D e f inin g E v e n t s an d P r o babi l i t i es$

$F or m u l a t in g R e g i o n P r o babi l i t i es$

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

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

$S o l v in g f or x an d y$

$S e tt in g U pt h e S eco n d E q u a t i o n$

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

$S o l v in g f or y an d z$

$C a l c u l a t in g t h e F ina l R a t i o$

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Translation

The First Dance

The Second Act

The Final Convergence

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:

For any two events $A$ and $B$ in a sample space

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 :

For two given events $A$ and $B, P (A \cap B)$ is

Let $0 < P (A) < 1, 0 < P (B) < 1$ and $P (A \cup B) = P (A) + P (B) - P (A) P (B)$ then

If $M$ and $N$ are any two events, the probability that exactly one of them occurs is

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

Enter Numerical Value:

Visualized Solution

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

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

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

.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; } Solvingforxandy

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

.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; } Solvingforyandz

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

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Translation

The First Dance

The Second Act

The Final Convergence

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For any two events A and B in a sample space

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For two given events A and B,P(A∩B) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 0<P(A)<1,0<P(B)<1 and P(A∪B)=P(A)+P(B)−P(A)P(B) then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If M and N are any two events, the probability that exactly one of them occurs is

.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

$D e f inin g E v e n t s an d P r o babi l i t i es$

$F or m u l a t in g R e g i o n P r o babi l i t i es$

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

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

$S o l v in g f or x an d y$

$S e tt in g U pt h e S eco n d E q u a t i o n$

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

$S o l v in g f or y an d z$

$C a l c u l a t in g t h e F ina l R a t i o$

For any two events $A$ and $B$ in a sample space

For two given events $A$ and $B, P (A \cap B)$ is

Let $0 < P (A) < 1, 0 < P (B) < 1$ and $P (A \cup B) = P (A) + P (B) - P (A) P (B)$ then

If $M$ and $N$ are any two events, the probability that exactly one of them occurs is

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