JEE Main 2018 (16 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Probability: Let A, B and C be three events, which are pair-wise independent and $\overset{ˉ}{E}$ denotes the complement of an event E. If $P (A \cap B \cap C) = 0$ and $P (C) > 0$ , then $P [(\overset{ˉ}{A} \cap \overset{ˉ}{B}) ∣ C]$ is equal to :

Q: Why does $P(A \cap B \cap C) = 0$ simplify the Inclusion-Exclusion formula so much?

The Inclusion-Exclusion Principle for two sets $X$ and $Y$ is $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$. Here, $X = A \cap C$ and $Y = B \cap C$. The intersection $X \cap Y$ is $(A \cap C) \cap (B \cap C)$, which is exactly $A \cap B \cap C$. Since this is given as zero, the subtraction term vanishes, leaving us with a clean sum of probabilities.

Select Answer:

Visualized Solution

Visualize the Events $A, B, C$

Events $A, B, C$ are pair-wise independent.
$P (A \cap B) = P (A) P (B)$
$P (B \cap C) = P (B) P (C)$
$P (A \cap C) = P (A) P (C)$

The Empty Intersection

We are given a crucial condition:
$P (A \cap B \cap C) = 0$

Define the Target Probability

We need to find: $P [(\overset{ˉ}{A} \cap \overset{ˉ}{B}) ∣ C]$
Using the definition of Conditional Probability:
$P [(\overset{ˉ}{A} \cap \overset{ˉ}{B}) ∣ C] = \frac{P ( A ˉ \cap B ˉ \cap C )}{P ( C )}$

Apply De Morgan's Law

Focus on the numerator: $P (\overset{ˉ}{A} \cap \overset{ˉ}{B} \cap C)$
Using De Morgan's Law:
$\overset{ˉ}{A} \cap \overset{ˉ}{B} = \overline{A \cup B}$
The numerator becomes: $P (\overline{A \cup B} \cap C)$

Express as Set Subtraction

From set theory properties:
$P (\overline{A \cup B} \cap C) = P (C) - P ((A \cup B) \cap C)$

Apply Distributive Law

Expand the subtracted term using the Distributive Law:
$(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$

Inclusion-Exclusion Principle

Apply Inclusion-Exclusion to the union:
$P ((A \cap C) \cup (B \cap C)) = P (A \cap C) + P (B \cap C) - P (A \cap B \cap C)$

Substitute Known Values

Substitute $P (A \cap B \cap C) = 0$ :
$P ((A \cup B) \cap C) = P (A \cap C) + P (B \cap C) - 0$

Use Pair-wise Independence

Substitute Pair-wise Independence conditions:
$P (A \cap C) = P (A) P (C)$
$P (B \cap C) = P (B) P (C)$
$P ((A \cup B) \cap C) = P (A) P (C) + P (B) P (C)$

Simplify the Numerator

Substitute back into the numerator expression:
$Numerator = P (C) - [P (A) P (C) + P (B) P (C)]$
Factoring out $P (C)$ :
$Numerator = P (C) [1 - P (A) - P (B)]$

Final Calculation

Substitute the numerator back into the conditional formula:
$P [(\overset{ˉ}{A} \cap \overset{ˉ}{B}) ∣ C] = \frac{P ( C ) [ 1 - P ( A ) - P ( B )]}{P ( C )}$
Canceling $P (C)$ (since $P (C) > 0$ ):
$= 1 - P (A) - P (B)$

Conclusion

Since $1 - P (A) = P (\overset{ˉ}{A})$ :
$= P (\overset{ˉ}{A}) - P (B)$
This matches Option 3.

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The Sigma Insight: Conditional Probability

Solution Diagram

Analyzing the Setup

We are given three events,

A

B

, and

C

, which are pair-wise independent. This implies the following relationships:

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

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

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

Furthermore, we are given the constraint that the intersection of all three events is empty:

P (A \cap B \cap C) = 0

The Conditional Lens

We aim to calculate the conditional probability

P [(\overset{ˉ}{A} \cap \overset{ˉ}{B}) ∣ C]

. By the definition of conditional probability, we have:

P [(\overset{ˉ}{A} \cap \overset{ˉ}{B}) ∣ C] = \frac{P ( A ˉ \cap B ˉ \cap C )}{P ( C )}

Our primary objective is to simplify the numerator,

P (\overset{ˉ}{A} \cap \overset{ˉ}{B} \cap C)

De Morgan's Magic

Using De Morgan's Law, we recognize that

\overset{ˉ}{A} \cap \overset{ˉ}{B} = \overline{A \cup B}

. Consequently, the numerator can be expressed as:

P (\overline{A \cup B} \cap C) = P (C) - P ((A \cup B) \cap C)

This represents the region within

C

that does not overlap with the union of

A

and

B

The Distributive Dance

We apply the Distributive Law to the term

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

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

Applying the Inclusion-Exclusion Principle to this union, we obtain:

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

The Final Calculation

Given that

P (A \cap B \cap C) = 0

, the expression simplifies significantly. Substituting the pair-wise independence conditions, we get:

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

Substituting this back into our numerator expression:

P (\overset{ˉ}{A} \cap \overset{ˉ}{B} \cap C) = P (C) - [P (A) P (C) + P (B) P (C)]

P (\overset{ˉ}{A} \cap \overset{ˉ}{B} \cap C) = P (C) [1 - P (A) - P (B)]

Finally, dividing by

P (C)

to complete the conditional probability calculation:

P [(\overset{ˉ}{A} \cap \overset{ˉ}{B}) ∣ C] = \frac{P ( C ) [ 1 - P ( A ) - P ( B )]}{P ( C )}

The final result is $1 - P (A) - P (B)$ .

Common Traps & FAQs

Wait, why can't we just say

P (\overset{ˉ}{A} \cap \overset{ˉ}{B}) = P (\overset{ˉ}{A}) P (\overset{ˉ}{B})

? Isn't that the definition of independence?

That is a common trap! The problem states the events are pair-wise independent, meaning

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

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

, and

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

. However, this does not automatically imply that the complements

\overset{ˉ}{A}

and

\overset{ˉ}{B}

are independent in the context of the conditional space

C

. Always stick to the definition of conditional probability first.

Why did we use

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

? Is there a simpler way to think about this?

Think of it geometrically. We are looking for the probability of being in

C

but not in

A

B

. The set

C

can be partitioned into two parts: the part that overlaps with

(A \cup B)

and the part that does not. Therefore, the probability of the part that does not overlap is simply the total probability of

C

minus the probability of the overlap.

Why does

P (A \cap B \cap C) = 0

simplify the Inclusion-Exclusion formula so much?

The Inclusion-Exclusion Principle for two sets

X

and

Y

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

. Here,

X = A \cap C

and

Y = B \cap C

. The intersection

X \cap Y

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

, which is exactly

A \cap B \cap C

. Since this is given as zero, the subtraction term vanishes, leaving us with a clean sum of probabilities.

Animated Solution for Mathematics - Probability: Let A, B and C be three events, which are pair-wise independent and $\overset{ˉ}{E}$ denotes the complement of an event E. If $P (A \cap B \cap C) = 0$ and $P (C) > 0$ , then $P [(\overset{ˉ}{A} \cap \overset{ˉ}{B}) ∣ C]$ is equal to :

Select Answer:

Visualized Solution

Visualize the Events $A, B, C$

The Empty Intersection

Define the Target Probability

Apply De Morgan's Law

Express as Set Subtraction

Apply Distributive Law

Inclusion-Exclusion Principle

Substitute Known Values

Use Pair-wise Independence

Simplify the Numerator

Final Calculation

Conclusion

The Sigma Insight: Conditional Probability

Analyzing the Setup

The Conditional Lens

De Morgan's Magic

The Distributive Dance

The Final Calculation

Similar Questions

Let $E^{c}$ denote the complement of an event $E$ . Let $E, F, G$ be pairwise independent events with $P (G) > 0$ and $P (E \cap F \cap G) = 0$ . Then $P (E^{c} \cap F^{c} ∣ G)$ equals

Let $E^{C}$ denote the complement of an event $E$ . Let $E_{1}, E_{2}$ and $E_{3}$ be any pairwise independent events with $P (E_{1}) > 0$ and $P (E_{1} \cap E_{2} \cap E_{3}) = 0$ . Then $P (E_{2}^{C} \cap E_{3}^{C} / E_{1})$ is equal to :

If $A$ and $B$ are two events such that $P (A) = 0.7$ , $P (B) = 0.4$ and $P (A \cap \overset{ˉ}{B}) = 0.5$ , where $\overset{ˉ}{B}$ denotes the complement of $B$ , then $P (B ∣ (A \cup \overset{ˉ}{B}))$ is equal:-

If $\overset{ˉ}{E}$ and $\overset{ˉ}{F}$ are the complementary events of events $E$ and $F$ respectively and if $0 < P (F) < 1$ , then

If $A$ and $B$ are two events such that $P (A) = \frac{1}{3}$ , $P (B) = \frac{1}{5}$ and $P (A \cup B) = \frac{1}{2}$ , then $P (A ∣ B^{'}) + P (B ∣ A^{'})$ is equal to

If $A$ and $B$ are two events such that $P (A) > 0$ , and $P (B) \neq = 1$ , then $P (\frac{A ˉ}{B ˉ})$ is equal to

Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P (E_{1} ∣ E_{2}) = \frac{1}{2}$ , $P (E_{2} ∣ E_{1}) = \frac{3}{4}$ and $P (E_{1} \cap E_{2}) = \frac{1}{8}$ . Then:

If two events $A$ and $B$ are such that $P (A^{c}) = 0.3, P (B) = 0.4$ and $P (A \cap B^{c}) = 0.5$ , then $P (B ∣ A \cup B^{c}) = .........$

Let $X$ and $Y$ be two events such that $P (X) = \frac{1}{3}$ , $P (X ∣ Y) = \frac{1}{2}$ and $P (Y ∣ X) = \frac{2}{5}$ . Then

Let $A$ and $B$ be two non-null events such that $A \subset B$ . Then, which of the following statements is always correct ?

Select Answer:

Visualized Solution

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

The Empty Intersection

Define the Target Probability

Apply De Morgan's Law

Express as Set Subtraction

Apply Distributive Law

Inclusion-Exclusion Principle

Substitute Known Values

Use Pair-wise Independence

Simplify the Numerator

Final Calculation

Conclusion

The Sigma Insight: Conditional Probability

Analyzing the Setup

The Conditional Lens

De Morgan's Magic

The Distributive Dance

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Ec denote the complement of an event E. Let E,F,G be pairwise independent events with P(G)>0 and P(E∩F∩G)=0. Then P(Ec∩Fc∣G) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A and B are two events such that P(A)=0.7, P(B)=0.4 and P(A∩Bˉ)=0.5, where Bˉ denotes the complement of B, then P(B∣(A∪Bˉ)) is equal:-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If Eˉ and Fˉ are the complementary events of events E and F respectively and if 0<P(F)<1, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A and B are two events such that P(A)=31​, P(B)=51​ and P(A∪B)=21​, then P(A∣B′)+P(B∣A′) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A and B are two events such that P(A)>0, and P(B)=1, then P(BˉAˉ​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let E1​ and E2​ be two events such that the conditional probabilities P(E1​∣E2​)=21​, P(E2​∣E1​)=43​ and P(E1​∩E2​)=81​. Then:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two events A and B are such that P(Ac)=0.3,P(B)=0.4 and P(A∩Bc)=0.5, then P(B∣A∪Bc)=.........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let X and Y be two events such that P(X)=31​, P(X∣Y)=21​ and P(Y∣X)=52​. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A and B be two non-null events such that A⊂B. Then, which of the following statements is always correct ?

Visualize the Events $A, B, C$

Let $E^{c}$ denote the complement of an event $E$ . Let $E, F, G$ be pairwise independent events with $P (G) > 0$ and $P (E \cap F \cap G) = 0$ . Then $P (E^{c} \cap F^{c} ∣ G)$ equals

If $A$ and $B$ are two events such that $P (A) = 0.7$ , $P (B) = 0.4$ and $P (A \cap \overset{ˉ}{B}) = 0.5$ , where $\overset{ˉ}{B}$ denotes the complement of $B$ , then $P (B ∣ (A \cup \overset{ˉ}{B}))$ is equal:-

If $\overset{ˉ}{E}$ and $\overset{ˉ}{F}$ are the complementary events of events $E$ and $F$ respectively and if $0 < P (F) < 1$ , then

If $A$ and $B$ are two events such that $P (A) = \frac{1}{3}$ , $P (B) = \frac{1}{5}$ and $P (A \cup B) = \frac{1}{2}$ , then $P (A ∣ B^{'}) + P (B ∣ A^{'})$ is equal to

If $A$ and $B$ are two events such that $P (A) > 0$ , and $P (B) \neq = 1$ , then $P (\frac{A ˉ}{B ˉ})$ is equal to

Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P (E_{1} ∣ E_{2}) = \frac{1}{2}$ , $P (E_{2} ∣ E_{1}) = \frac{3}{4}$ and $P (E_{1} \cap E_{2}) = \frac{1}{8}$ . Then:

If two events $A$ and $B$ are such that $P (A^{c}) = 0.3, P (B) = 0.4$ and $P (A \cap B^{c}) = 0.5$ , then $P (B ∣ A \cup B^{c}) = .........$

Let $X$ and $Y$ be two events such that $P (X) = \frac{1}{3}$ , $P (X ∣ Y) = \frac{1}{2}$ and $P (Y ∣ X) = \frac{2}{5}$ . Then

Let $A$ and $B$ be two non-null events such that $A \subset B$ . Then, which of the following statements is always correct ?