JEE Advanced 1980

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Animated Solution for Mathematics - Probability: Two events $A$ and $B$ have probabilities $0.25$ and $0.50$ respectively. The probability that both $A$ and $B$ occur simultaneously is $0.14$ . Then the probability that neither $A$ nor $B$ occurs is

Q: Why do we subtract the intersection? Can't we just add $P(A)$ and $P(B)$?

Great question! If you just add $P(A)$ and $P(B)$, you are counting the overlapping region $P(A \cap B)$ twice—once as part of $A$ and once as part of $B$. To get the true area of the union, we must subtract the overlap once to correct this double-counting.

Select Answer:

Visualized Solution

Visualizing the Events $A$ and $B$

Let the rectangle represent the sample space $S$ , where $P (S) = 1$ .
Event $A$ is represented by the blue circle: $P (A) = 0.25$ .
Event $B$ is represented by the green circle: $P (B) = 0.50$ .
The intersection $A \cap B$ is the overlapping region: $P (A \cap B) = 0.14$ .

The Addition Theorem of Probability

To find the probability of either $A$ or $B$ occurring, we use the Addition Theorem.
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

Substituting the Given Values

Substitute the given probabilities into the formula:
$P (A \cup B) = 0.25 + 0.50 - 0.14$

Adding the Probabilities

First, add $P (A)$ and $P (B)$ :
$0.25 + 0.50 = 0.75$
$P (A \cup B) = 0.75 - 0.14$

Calculating $P (A \cup B)$

Subtract the intersection:
$P (A \cup B) = 0.75 - 0.14$
$P (A \cup B) = 0.61$

Applying De Morgan's Law

The event 'neither $A$ nor $B$ ' is represented as $A^{'} \cap B^{'}$ .
By De Morgan's Law: $A^{'} \cap B^{'} = (A \cup B)^{'}$
This represents the region outside the circles.

Probability of the Complement

The probability of a complement is $1 - P (Event)$ .
Therefore, $P ((A \cup B)^{'}) = 1 - P (A \cup B)$
Substitute the value: $P ((A \cup B)^{'}) = 1 - 0.61$

Final Calculation

$P (neither A nor B) = 1 - 0.61$
$P (neither A nor B) = 0.39$

Key Takeaways

Key Takeaway: $P (neither A nor B) = 1 - P (A \cup B)$ .
Formula Recap: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .
Next Challenge: What if $A$ and $B$ were mutually exclusive events? How would $P (A \cap B)$ change?

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The Sigma Insight: Addition and Multiplication Theorems

Solution Diagram

Analyzing the Setup

Probability is the language of uncertainty, a way to quantify the unknown. When we look at events

A

and

B

, we are not just looking at numbers; we are looking at the geometry of possibility.

Imagine a rectangle representing our entire sample space

S

, where the total probability is

P (S) = 1

. Inside this rectangle, events

A

and

B

are like two overlapping circles.

We are given the following values:

P (A) = 0.25

P (B) = 0.50

P (A \cap B) = 0.14

Our goal is to find the probability that neither event occurs.

The Addition Theorem

Avoiding the Double-Counting Trap

To find the probability of "neither," we must first understand the probability of "at least one." This is the union

P (A \cup B)

A common mistake is to simply add

P (A)

and

P (B)

. If you do that, you are counting the overlapping region

P (A \cap B)

twice—once as part of

A

and once as part of

B

To correct this, we use the Addition Theorem:

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

Substituting our values, we get:

P (A \cup B) = 0.25 + 0.50 - 0.14

Adding

0.25

and

0.50

gives

0.75

. Subtracting the intersection

0.14

yields

P (A \cup B) = 0.61

. This

0.61

represents the total area covered by the two circles.

De Morgan's Law

The Final Step

The question asks for the probability that neither

A

nor

B

occurs. This is the region outside both circles. In set notation, this is

A^{'} \cap B^{'}

By De Morgan's Law, this is equivalent to the complement of the union:

P ((A \cup B)^{'}) = 1 - P (A \cup B)

Since the total probability is

1

, we simply subtract our union from

1

P (neither) = 1 - 0.61 = 0.39

The beauty of this result lies in its simplicity. By carefully navigating the geometry of the Venn diagram and applying the fundamental theorems of probability, we have turned a complex problem into a clear, logical conclusion.

The final answer is $0.39$ .

Common Traps & FAQs

Why do we subtract the intersection? Can't we just add

P (A)

and

P (B)

Great question! If you just add

P (A)

and

P (B)

, you are counting the overlapping region

P (A \cap B)

twice—once as part of

A

and once as part of

B

. To get the true area of the union, we must subtract the overlap once to correct this double-counting.

What does 'neither A nor B' actually mean in terms of set notation?

In set theory, 'neither A nor B' means the event is outside both

A

and

B

. This is written as

A^{'} \cap B^{'}

. According to De Morgan's Law, this is equivalent to

(A \cup B)^{'}

, which is the complement of the union of

A

and

B

Is there a scenario where

P (A \cup B)

could be greater than 1?

No, that is impossible! The probability of any event, including the union of two events, must always be between 0 and 1. If you ever calculate a probability greater than 1, it is a clear sign that you have made a calculation error or misunderstood the relationship between the events.

Animated Solution for Mathematics - Probability: Two events $A$ and $B$ have probabilities $0.25$ and $0.50$ respectively. The probability that both $A$ and $B$ occur simultaneously is $0.14$ . Then the probability that neither $A$ nor $B$ occurs is

Select Answer:

Visualized Solution

Visualizing the Events $A$ and $B$

The Addition Theorem of Probability

Substituting the Given Values

Adding the Probabilities

Calculating $P (A \cup B)$

Applying De Morgan's Law

Probability of the Complement

Final Calculation

Key Takeaways

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Addition Theorem

De Morgan's Law

Similar Questions

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

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

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

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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

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

Select Answer:

Visualized Solution

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

The Addition Theorem of Probability

Substituting the Given Values

Adding the Probabilities

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(A∪B)

Applying De Morgan's Law

Probability of the Complement

Final Calculation

Key Takeaways

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Addition Theorem

De Morgan's Law

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and B are events such that P(A∪B)=3/4,P(A∩B)=1/4,P(Aˉ)=2/3 then P(Aˉ∩B) is

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.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

Visualizing the Events $A$ and $B$

Calculating $P (A \cup B)$

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

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