JEE Advanced 1994

LEVELJEE Main

Animated Solution for Mathematics - Probability: Let $A, B, C$ be three mutually independent events. Consider the two statements $S_{1}$ and $S_{2}$ \\ $S_{1} : A$ and $B \cup C$ are independent \\ $S_{2} : A$ and $B \cap C$ are independent \\ Then,

Q: Wait, why do we need to check $P(A \cap B \cap C) = P(A)P(B)P(C)$? Isn't pairwise independence enough?

That is a classic trap! Pairwise independence (where $P(A \cap B) = P(A)P(B)$, etc.) does NOT guarantee mutual independence. You can have three events where any two are independent, but the third one is completely determined by the other two. For mutual independence, we need the stronger condition that the probability of the intersection of any combination of these events equals the product of their individual probabilities.

Q: Why did we subtract $P(A \cap B \cap C)$ in the expansion of $P(A \cap (B \cup C))$?

Great catch! We used the general addition rule: $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$. Here, we set $X = (A \cap B)$ and $Y = (A \cap C)$. The intersection of $X$ and $Y$ is $(A \cap B) \cap (A \cap C)$, which simplifies to $A \cap B \cap C$. That is why we subtract the triple intersection—to avoid double-counting the region where all three events occur.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the sample space as three overlapping circles. Mutual independence is a powerful, restrictive condition. It implies that the occurrence of any subset of these events provides no information about the others. Think beyond just pairwise independence; consider the triple intersection

P (A \cap B \cap C) = P (A) P (B) P (C)

Hint 2

Hint 2 (The Arsenal): To test independence for

S_{1}

, you must verify if

P (A \cap (B \cup C)) = P (A) P (B \cup C)

. Use the distributive law of sets:

A \cap (B \cup C) = (A \cap B) \cup (A \cap C)

. Then, apply the addition rule for probability to this union.

Hint 3

Hint 3 (The Execution): Start your proof for

S_{1}

by expanding

P (A \cap (B \cup C))

into

P (A \cap B) + P (A \cap C) - P (A \cap B \cap C)

. Now, substitute the independence conditions

P (A \cap B) = P (A) P (B)

P (A \cap C) = P (A) P (C)

, and

P (A \cap B \cap C) = P (A) P (B) P (C)

to see if it factors into

P (A) P (B \cup C)

Visualized Solution

The Setup: Mutually Independent Events

Given: $A, B, C$ are mutually independent events.
This means they are pairwise independent:
$P (A \cap B) = P (A) P (B)$
$P (B \cap C) = P (B) P (C)$
$P (C \cap A) = P (C) P (A)$
And all three are independent:
$P (A \cap B \cap C) = P (A) P (B) P (C)$

Analyzing Statement $S_{1}$

Statement $S_{1}$ : $A$ and $B \cup C$ are independent.
To prove this, we must show:
$P (A \cap (B \cup C)) = P (A) \cdot P (B \cup C)$

Applying Distributive Law

Let's expand the left-hand side: $P (A \cap (B \cup C))$
Using set theory distributive law:
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

Probability Addition Rule

Now apply the addition rule for probability: $P (X \cup Y) = P (X) + P (Y) - P (X \cap Y)$
Let $X = (A \cap B)$ and $Y = (A \cap C)$
$P ((A \cap B) \cup (A \cap C)) = P (A \cap B) + P (A \cap C) - P ((A \cap B) \cap (A \cap C))$
Notice that $(A \cap B) \cap (A \cap C) = A \cap B \cap C$

Substituting Independence Conditions

Substitute the mutual independence formulas:
$P (A \cap B) = P (A) P (B)$
$P (A \cap C) = P (A) P (C)$
$P (A \cap B \cap C) = P (A) P (B) P (C)$
Expression becomes:
$P (A) P (B) + P (A) P (C) - P (A) P (B) P (C)$

Factoring and Simplifying

Take $P (A)$ common from all terms:
$= P (A) [P (B) + P (C) - P (B) P (C)]$
Recall that $P (B) P (C) = P (B \cap C)$
So the bracket is: $[P (B) + P (C) - P (B \cap C)]$

Conclusion for Statement $S_{1}$

The term inside the bracket is the expansion of $P (B \cup C)$
$[P (B) + P (C) - P (B \cap C)] = P (B \cup C)$
Therefore, $P (A \cap (B \cup C)) = P (A) \cdot P (B \cup C)$
Statement $S_{1}$ is True.

Analyzing Statement $S_{2}$

Statement $S_{2}$ : $A$ and $B \cap C$ are independent.
To prove this, we must show:
$P (A \cap (B \cap C)) = P (A) \cdot P (B \cap C)$

Proving Statement $S_{2}$

Start with the left-hand side: $P (A \cap B \cap C)$
Since $A, B, C$ are mutually independent:
$P (A \cap B \cap C) = P (A) \cdot P (B) \cdot P (C)$
Group $P (B)$ and $P (C)$ :
$= P (A) \cdot [P (B) \cdot P (C)]$

Conclusion for Statement $S_{2}$

Since $B$ and $C$ are independent, $P (B) \cdot P (C) = P (B \cap C)$
Substitute this back:
$= P (A) \cdot P (B \cap C)$
Therefore, $P (A \cap (B \cap C)) = P (A) \cdot P (B \cap C)$
Statement $S_{2}$ is True.

Final Conclusion

Both $S_{1}$ and $S_{2}$ are true.
General Rule: If events are mutually independent, any one event is independent of any combination of the others formed by unions, intersections, or complements.
Correct Option: (0) Both $S_{1}$ and $S_{2}$ are true

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The Sigma Insight: Addition and Multiplication Theorems

Solution Diagram

Analyzing the Setup

Welcome, future engineers! Today, we are diving into the heart of probability. We have three events,

A

B

, and

C

, and we are told they are mutually independent.

When we say events are mutually independent, we are saying that the occurrence of one event tells us absolutely nothing about the occurrence of the others. It is the ultimate state of "non-interference."

The Master Keys

Before we touch the statements, let us lay out our tools. Mutual independence gives us four golden equations:

P (A \cap B) = P (A) P (B)

P (B \cap C) = P (B) P (C)

P (C \cap A) = P (C) P (A)

P (A \cap B \cap C) = P (A) P (B) P (C)

These are the keys to the kingdom. If you ever feel lost in a probability problem, return to these definitions.

Decoding Statement $S_{1}$

Statement

S_{1}

asks if

A

is independent of

B \cup C

. To prove this, we must show that

P (A \cap (B \cup C)) = P (A) \cdot P (B \cup C)

Using the distributive law of sets, we can rewrite

A \cap (B \cup C)

(A \cap B) \cup (A \cap C)

. Now, we apply the addition rule for probability:

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

Notice that the intersection of

(A \cap B)

and

(A \cap C)

is simply

A \cap B \cap C

. Substituting our independence keys, the expression becomes:

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

Factoring out

P (A)

, we get

P (A) [P (B) + P (C) - P (B) P (C)]

. Since

B

and

C

are independent,

P (B) P (C) = P (B \cap C)

Thus, the bracket is

P (B) + P (C) - P (B \cap C)

, which is exactly

P (B \cup C)

. We have proven

P (A \cap (B \cup C)) = P (A) P (B \cup C)

, so $S_{1}$ is true.

Decoding Statement $S_{2}$

Now for

S_{2}

: is

A

independent of

B \cap C

? We need to show

P (A \cap (B \cap C)) = P (A) \cdot P (B \cap C)

The left-hand side is

P (A \cap B \cap C)

. By the definition of mutual independence, this is

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

We can group this as

P (A) \cdot [P (B) P (C)]

. Since

B

and

C

are independent,

P (B) P (C) = P (B \cap C)

Therefore,

P (A \cap B \cap C) = P (A) \cdot P (B \cap C)

. $S_{2}$ is also true.

The Golden Rule

We have arrived at the conclusion: both statements are true. This is a general, beautiful rule in probability: if a set of events is mutually independent, any event is independent of any combination of the others.

Whether you use unions, intersections, or complements, the independence holds. Keep this in your toolkit, and you will conquer any probability problem the JEE throws at you.

Common Traps & FAQs

Wait, why do we need to check

P (A \cap B \cap C) = P (A) P (B) P (C)

? Isn't pairwise independence enough?

That is a classic trap! Pairwise independence (where

P (A \cap B) = P (A) P (B)

, etc.) does NOT guarantee mutual independence. You can have three events where any two are independent, but the third one is completely determined by the other two. For mutual independence, we need the stronger condition that the probability of the intersection of any combination of these events equals the product of their individual probabilities.

Is it always true that if

A, B, C

are independent, then

A

is independent of

B \cup C

Yes, absolutely! This is a fundamental theorem in probability. If a collection of events is mutually independent, then any event

A

is independent of any event formed by the union, intersection, or complement of the other events. It is a beautiful, elegant property that simplifies complex systems.

Why did we subtract

P (A \cap B \cap C)

in the expansion of

P (A \cap (B \cup C))

Great catch! We used the general addition rule:

P (X \cup Y) = P (X) + P (Y) - P (X \cap Y)

. Here, we set

X = (A \cap B)

and

Y = (A \cap C)

. The intersection of

X

and

Y

(A \cap B) \cap (A \cap C)

, which simplifies to

A \cap B \cap C

. That is why we subtract the triple intersection—to avoid double-counting the region where all three events occur.

Animated Solution for Mathematics - Probability: Let $A, B, C$ be three mutually independent events. Consider the two statements $S_{1}$ and $S_{2}$ \\ $S_{1} : A$ and $B \cup C$ are independent \\ $S_{2} : A$ and $B \cap C$ are independent \\ Then,

Select Answer:

Visualized Solution

The Setup: Mutually Independent Events

Analyzing Statement $S_{1}$

Applying Distributive Law

Probability Addition Rule

Substituting Independence Conditions

Factoring and Simplifying

Conclusion for Statement $S_{1}$

Analyzing Statement $S_{2}$

Proving Statement $S_{2}$

Conclusion for Statement $S_{2}$

Final Conclusion

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Master Keys

Decoding Statement $S_{1}$

Decoding Statement $S_{2}$

The Golden Rule

Similar Questions

Let $A$ and $B$ be two events such that $P (A \cup B) = 5/6, P (A \cap B) = 1/4$ and $P (\overset{ˉ}{A}) = 1/4$ , where $\overset{ˉ}{A}$ stands for the complement of the event $A$ . Then the events $A$ and $B$ are

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

$A$ and $B$ are two independent events. $C$ is event in which exactly one of $A$ or $B$ occurs. Prove that $P (C) \geq P (A \cup B) P (\overset{ˉ}{A} \cap \overset{ˉ}{B})$ .

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

Let two fair six-faced dice $A$ and $B$ be thrown simultaneously. If $E_{1}$ is the event that die $A$ shows up four, $E_{2}$ is the event that die $B$ shows up two and $E_{3}$ is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true?

Two fair dice are tossed. Let $x$ be the event that the first die shows an even number and $y$ be the event that the second die shows an odd number. The two events $x$ and $y$ are :

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

Select Answer:

Visualized Solution

The Setup: Mutually Independent Events

Analyzing Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Applying Distributive Law

Probability Addition Rule

Substituting Independence Conditions

Factoring and Simplifying

Conclusion for Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Analyzing Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Proving Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Conclusion for Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

Final Conclusion

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Master Keys

Decoding Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Decoding Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

The Golden Rule

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A and B be two events such that P(A∪B)=5/6,P(A∩B)=1/4 and P(Aˉ)=1/4, where Aˉ stands for the complement of the event A. Then the events A and B are

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and B are two independent events. C is event in which exactly one of A or B occurs. Prove that P(C)≥P(A∪B)P(Aˉ∩Bˉ).

.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; } Two fair dice are tossed. Let x be the event that the first die shows an even number and y be the event that the second die shows an odd number. The two events x and y are :

.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

Analyzing Statement $S_{1}$

Conclusion for Statement $S_{1}$

Analyzing Statement $S_{2}$

Proving Statement $S_{2}$

Conclusion for Statement $S_{2}$

Decoding Statement $S_{1}$

Decoding Statement $S_{2}$

Let $A$ and $B$ be two events such that $P (A \cup B) = 5/6, P (A \cap B) = 1/4$ and $P (\overset{ˉ}{A}) = 1/4$ , where $\overset{ˉ}{A}$ stands for the complement of the event $A$ . Then the events $A$ and $B$ are

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

$A$ and $B$ are two independent events. $C$ is event in which exactly one of $A$ or $B$ occurs. Prove that $P (C) \geq P (A \cup B) P (\overset{ˉ}{A} \cap \overset{ˉ}{B})$ .

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

Two fair dice are tossed. Let $x$ be the event that the first die shows an even number and $y$ be the event that the second die shows an odd number. The two events $x$ and $y$ are :

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