JEE Advanced 1996

LEVELJEE Main

Animated Solution for Mathematics - Probability: For the three events $A, B$ , and $C, P$ (exactly one of the events $A$ or $B$ occurs) $= P$ (exactly one of the events $B$ or $C$ occurs) $= P$ (exactly one of the events $C$ or $A$ occurs) $= p$ and $P$ (all the three events occur simultaneously) $= p^{2}$ , where $0 < p < 1/2$ . Then the probability of at least one of the three events $A, B$ and $C$ occurring is

Q: Wait, why do we subtract $2P(A \cap B)$ instead of just $P(A \cap B)$?

That is a fantastic question! When you add $P(A)$ and $P(B)$, the intersection region $A \cap B$ is included in both $P(A)$ and $P(B)$. Therefore, it is counted twice. To remove it entirely so that we only count the regions where *exactly* one event occurs, we must subtract it twice. It is a surgical removal of the overlap!

Q: Is there a trick to solving these symmetric probability problems?

Symmetry is your best friend in JEE Advanced! Whenever you see cyclic conditions (like A-B, B-C, C-A), do not try to solve for individual probabilities. Instead, sum the equations. The symmetry will often cause terms to group together into the exact form you need for the Inclusion-Exclusion Principle.

Q: Why is the Inclusion-Exclusion Principle necessary here?

Because simple addition overcounts the overlaps. If you just add $P(A) + P(B) + P(C)$, you are counting the pairwise intersections twice and the triple intersection three times. The Inclusion-Exclusion Principle is the correction mechanism that balances the books, ensuring every region is counted exactly once.

Select Answer:

Visualized Solution

$E v e n t s A, B, an d C$

Let the three events be $A$ , $B$ , and $C$ .
We represent them using a Venn diagram.

$E x a c tl y O n e o f A or B$

The probability that exactly one of $A$ or $B$ occurs means either $A$ happens or $B$ happens, but not both.
Mathematically, this is the symmetric difference: $P (A Δ B)$ .

$E q u a t i o n f or A an d B$

The probability of exactly one of $A$ or $B$ is:
$P (A) + P (B) - 2 P (A \cap B)$
Given this equals $p$ :
$P (A) + P (B) - 2 P (A \cap B) = p$

$E q u a t i o n s f or O t h er P ai r s$

By symmetry, we can write the equations for the other pairs.
For $B$ and $C$ :
$P (B) + P (C) - 2 P (B \cap C) = p$
For $C$ and $A$ :
$P (C) + P (A) - 2 P (C \cap A) = p$

$S u mmin g t h e E q u a t i o n s$

Let's add all three equations together.
Left side:
$2 P (A) + 2 P (B) + 2 P (C) - 2 P (A \cap B) - 2 P (B \cap C) - 2 P (C \cap A)$
Right side:
$p + p + p = 3 p$

$S im pl i f y in g t h e S u m$

Factor out the $2$ on the left side:
$2 [P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A)] = 3 p$
Divide by $2$ :
$P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) = \frac{3 p}{2}$

$I n t er sec t i o n o f A l l T h r ee$

The problem also states that all three events occur simultaneously with probability $p^{2}$ .
Mathematically:
$P (A \cap B \cap C) = p^{2}$

$T h e G o a l : A t L e a s t O n e E v e n t$

We need to find the probability of at least one of the events occurring.
This is the union of all three events:
$P (A \cup B \cup C)$

$I n c l u s i o n - E x c l u s i o n P r in c i pl e$

The formula for the union of three events is:
$P (A \cup B \cup C) = P (A) + P (B) + P (C)$
$- P (A \cap B) - P (B \cap C) - P (C \cap A)$
$+ P (A \cap B \cap C)$

$S u b s t i t u t in g t h e V a l u es$

Notice that the first part of our formula is exactly what we calculated earlier!
$P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) = \frac{3 p}{2}$
And the last term is:
$P (A \cap B \cap C) = p^{2}$

$F ina l A dd i t i o n$

Substitute these values into the union formula:
$P (A \cup B \cup C) = \frac{3 p}{2} + p^{2}$
Take a common denominator of $2$ :
$P (A \cup B \cup C) = \frac{3 p + 2 p ^{2}}{2}$

$C o n c l u s i o n$

The probability of at least one event occurring is $\frac{3 p + 2 p ^{2}}{2}$ .
Comparing this with the given options, we find our answer.
Correct Option: (A)

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The Sigma Insight: Addition and Multiplication Theorems

Solution Diagram

Analyzing the Setup

We are tasked with finding the probability of the union of three events,

A

B

, and

C

, which represents the probability that at least one of these events occurs. We are given that the probability of "exactly one" of any pair occurring is

p

For any two events, say

A

and

B

, the region where exactly one occurs is defined as the set

(A \cup B) ∖ (A \cap B)

. Algebraically, the probability of this region is expressed as:

P (A) + P (B) - 2 P (A \cap B) = p

The factor of

2

is necessary because when we sum

P (A)

and

P (B)

, the intersection

P (A \cap B)

is counted twice. To isolate the region where only one event occurs, we must subtract the intersection twice.

Establishing the System of Equations

Given the symmetry of the problem, we can write the same condition for all three pairs:

(A, B)

(B, C)

, and

(C, A)

. This yields the following system:

P (A) + P (B) - 2 P (A \cap B) = p

P (B) + P (C) - 2 P (B \cap C) = p

P (C) + P (A) - 2 P (C \cap A) = p

Summing these three equations allows us to observe a structural pattern. Adding them together, we obtain:

2 [P (A) + P (B) + P (C)] - 2 [P (A \cap B) + P (B \cap C) + P (C \cap A)] = 3 p

The Master Equation

Dividing the entire equation by

2

, we isolate the core component of the Inclusion-Exclusion Principle:

P (A) + P (B) + P (C) - [P (A \cap B) + P (B \cap C) + P (C \cap A)] = \frac{3 p}{2}

The Inclusion-Exclusion Principle for three events is defined as:

P (A \cup B \cup C) = P (A) + P (B) + P (C) - [P (A \cap B) + P (B \cap C) + P (C \cap A)] + P (A \cap B \cap C)

Final Calculation

We have already determined that the first part of the Inclusion-Exclusion formula equals

\frac{3 p}{2}

. The problem further specifies that the intersection of all three events,

P (A \cap B \cap C)

, is equal to

p^{2}

Substituting these values into the formula, we find:

P (A \cup B \cup C) = \frac{3 p}{2} + p^{2}

Combining these terms into a single fraction, the final probability is:

\frac{3p + 2 p ^{2}}{2}

Common Traps & FAQs

Wait, why do we subtract

2 P (A \cap B)

instead of just

P (A \cap B)

That is a fantastic question! When you add

P (A)

and

P (B)

, the intersection region

A \cap B

is included in both

P (A)

and

P (B)

. Therefore, it is counted twice. To remove it entirely so that we only count the regions where *exactly* one event occurs, we must subtract it twice. It is a surgical removal of the overlap!

Is there a trick to solving these symmetric probability problems?

Symmetry is your best friend in JEE Advanced! Whenever you see cyclic conditions (like A-B, B-C, C-A), do not try to solve for individual probabilities. Instead, sum the equations. The symmetry will often cause terms to group together into the exact form you need for the Inclusion-Exclusion Principle.

Why is the Inclusion-Exclusion Principle necessary here?

Because simple addition overcounts the overlaps. If you just add

P (A) + P (B) + P (C)

, you are counting the pairwise intersections twice and the triple intersection three times. The Inclusion-Exclusion Principle is the correction mechanism that balances the books, ensuring every region is counted exactly once.

Select Answer:

Visualized Solution

$E v e n t s A, B, an d C$

$E x a c tl y O n e o f A or B$

$E q u a t i o n f or A an d B$

$E q u a t i o n s f or O t h er P ai r s$

$S u mmin g t h e E q u a t i o n s$

$S im pl i f y in g t h e S u m$

$I n t er sec t i o n o f A l l T h r ee$

$T h e G o a l : A t L e a s t O n e E v e n t$

$I n c l u s i o n - E x c l u s i o n P r in c i pl e$

$S u b s t i t u t in g t h e V a l u es$

$F ina l A dd i t i o n$

$C o n c l u s i o n$

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

Establishing the System of Equations

The Master Equation

Final Calculation

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:

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

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

Events $A, B, C$ are mutually exclusive events such that $P (A) = \frac{3 x + 1}{3}, P (B) = \frac{1 - x}{4}$ and $P (C) = \frac{1 - 2 x}{2}$ . The set of possible values of $x$ are in the interval.

Let $A$ and $B$ be two events such that the probability that exactly one of them occurs is $2/5$ and the probability that $A$ or $B$ occurs is $1/2$ , then the probability of both of them occur together is

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

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Events A, B, and C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Exactly One of A or B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation for A and B

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Goal: At Least One Event

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

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

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

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

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

Establishing the System of Equations

The Master Equation

Final Calculation

Similar Questions

.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

.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; } Events A,B,C are mutually exclusive events such that P(A)=33x+1​,P(B)=41−x​ and P(C)=21−2x​. The set of possible values of x are in the interval.

.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

$E v e n t s A, B, an d C$

$E x a c tl y O n e o f A or B$

$E q u a t i o n f or A an d B$

$E q u a t i o n s f or O t h er P ai r s$

$S u mmin g t h e E q u a t i o n s$

$S im pl i f y in g t h e S u m$

$I n t er sec t i o n o f A l l T h r ee$

$T h e G o a l : A t L e a s t O n e E v e n t$

$I n c l u s i o n - E x c l u s i o n P r in c i pl e$

$S u b s t i t u t in g t h e V a l u es$

$F ina l A dd i t i o n$

$C o n c l u s i o n$

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

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

Events $A, B, C$ are mutually exclusive events such that $P (A) = \frac{3 x + 1}{3}, P (B) = \frac{1 - x}{4}$ and $P (C) = \frac{1 - 2 x}{2}$ . The set of possible values of $x$ are in the interval.

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