JEE Main 2021 (26 Aug Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Probability: Let $A$ and $B$ be independent events such that $P (A) = p, P (B) = 2 p$ . The largest value of $p$ , for which $P (exactly one of A, B occurs) = 5/9$ , is :

Q: Wait, why do we subtract $2P(A \cap B)$? Shouldn't it just be $P(A) + P(B) - P(A \cap B)$?

That is a classic trap! $P(A) + P(B)$ counts the intersection region twice—once inside $A$ and once inside $B$. If we want the probability of 'exactly one' event, we need to remove the intersection entirely. Subtracting it once only removes one 'layer' of the intersection, leaving it counted once. To remove it completely, we must subtract it twice. Hence, $P(A) + P(B) - 2P(A \cap B)$ is the correct way to isolate the regions where only $A$ or only $B$ occurs.

Select Answer:

Visualized Solution

Define Probabilities

Let's visualize the sample space with two events $A$ and $B$ .
We are given the probability of event $A$ : $P (A) = p$ .
We are given the probability of event $B$ : $P (B) = 2 p$ .

Independence of Events

The problem states that events $A$ and $B$ are independent.
For independent events, the probability of their intersection is the product of their individual probabilities.
Formula: $P (A \cap B) = P (A) \cdot P (B)$

Calculate $P (A \cap B)$

Substitute the given values into the intersection formula.
$P (A \cap B) = p \cdot (2 p)$
$P (A \cap B) = 2 p^{2}$

Exactly One Event Occurs

We need the probability that exactly one of $A$ or $B$ occurs.
This corresponds to the regions where only $A$ happens, or only $B$ happens.
In set notation, this is the symmetric difference: $(A \cup B) - (A \cap B)$ .

Formula for Exactly One Event

The formula for the probability of exactly one event occurring is:
$P (exactly one) = P (A) + P (B) - 2 P (A \cap B)$
Why subtract twice? Adding $P (A)$ and $P (B)$ counts the intersection twice, so we must subtract it twice to remove it completely.

Substitute into the Formula

We are given that $P (exactly one) = \frac{5}{9}$ .
Substitute the known expressions:
$p + 2 p - 2 (2 p^{2}) = \frac{5}{9}$

Simplify the Equation

Combine the linear terms: $p + 2 p = 3 p$ .
Multiply the squared term: $- 2 (2 p^{2}) = - 4 p^{2}$ .
The simplified equation becomes:
$3 p - 4 p^{2} = \frac{5}{9}$

Form the Quadratic Equation

To eliminate the fraction, multiply the entire equation by $9$ :
$9 (3 p - 4 p^{2}) = 5$
$27 p - 36 p^{2} = 5$
Rearrange all terms to one side to form a standard quadratic equation:
$36 p^{2} - 27 p + 5 = 0$

Factorize the Quadratic

We need two numbers that multiply to $36 \times 5 = 180$ and add to $- 27$ .
These numbers are $- 15$ and $- 12$ .
Split the middle term:
$36 p^{2} - 15 p - 12 p + 5 = 0$
Factor by grouping:
$3 p (12 p - 5) - 1 (12 p - 5) = 0$
$(12 p - 5) (3 p - 1) = 0$

Find the Roots

Set each factor to zero to solve for $p$ :
$12 p - 5 = 0 ⟹ p = \frac{5}{12}$
$3 p - 1 = 0 ⟹ p = \frac{1}{3}$
Both values are valid probabilities (between $0$ and $1$ ).

Select the Largest Value

The question asks for the largest value of $p$ .
Compare $p = \frac{5}{12}$ and $p = \frac{1}{3}$ .
Convert $\frac{1}{3}$ to a common denominator: $\frac{1}{3} = \frac{4}{12}$ .
Since $\frac{5}{12} > \frac{4}{12}$ , the largest value is $\frac{5}{12}$ .
Final Answer: $\frac{5}{12}$

00:00 / 00:00

The Sigma Insight: Addition and Multiplication Theorems

Solution Diagram

Analyzing the Setup

In this probability space, we are given two independent events,

A

and

B

. The probabilities are defined as

P (A) = p

and

P (B) = 2 p

Because the events are independent, the occurrence of one does not influence the other. This allows us to define the probability of their intersection as the product of their individual probabilities.

P (A \cap B) = P (A) \cdot P (B) = p \cdot 2 p = 2 p^{2}

The Exclusion Principle

We are tasked with finding the probability that exactly one of these events occurs. Geometrically, this corresponds to the regions in a Venn diagram where

A

occurs without

B

, and

B

occurs without

A

To calculate this, we sum the individual probabilities and subtract the intersection twice (once for each event's inclusion of the overlap). The formula is:

P (exactly one) = P (A) + P (B) - 2 P (A \cap B)

The Algebraic Battle

Substituting the known values into our formula, we are given that the probability is

\frac{5}{9}

. This yields the following equation:

p + 2 p - 2 (2 p^{2}) = \frac{5}{9}

Simplifying the expression, we obtain:

3 p - 4 p^{2} = \frac{5}{9}

Multiplying the entire equation by

9

to clear the fraction, we get:

27 p - 36 p^{2} = 5

Rearranging this into the standard quadratic form

a x^{2} + b x + c = 0

, we have:

36 p^{2} - 27 p + 5 = 0

Solving the Quadratic

To solve for

p

, we factor the quadratic equation. We seek two numbers that multiply to

36 \times 5 = 180

and add to

- 27

. These numbers are

- 15

and

- 12

Splitting the middle term:

36 p^{2} - 15 p - 12 p + 5 = 0

Factoring by grouping:

3 p (12 p - 5) - 1 (12 p - 5) = 0

(12 p - 5) (3 p - 1) = 0

Final Selection

Solving the factors gives us two potential candidates for

p

p = \frac{5}{12} or p = \frac{1}{3}

Comparing the two values, where

\frac{1}{3} = \frac{4}{12}

, we identify the larger value. The final answer is:

p = \frac{5}{12}

Common Traps & FAQs

Wait, why do we subtract

2 P (A \cap B)

? Shouldn't it just be

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

That is a classic trap!

P (A) + P (B)

counts the intersection region twice—once inside

A

and once inside

B

. If we want the probability of 'exactly one' event, we need to remove the intersection entirely. Subtracting it once only removes one 'layer' of the intersection, leaving it counted once. To remove it completely, we must subtract it twice. Hence,

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

is the correct way to isolate the regions where only

A

or only

B

occurs.

How do I know if I should use the quadratic formula or factoring?

Factoring is usually faster if the roots are clean, like in this problem. If you look at

36 p^{2} - 27 p + 5 = 0

, you need two numbers that multiply to

180

and add to

- 27

. Once you spot

- 15

and

- 12

, factoring becomes straightforward. If you ever feel stuck or the numbers look messy, the quadratic formula is your reliable safety net—it works every single time!

Wait, both

p = \frac{5}{12}

and

p = \frac{1}{3}

are valid. How do I know which one to pick?

Always re-read the question carefully! It asks for the 'largest' value of

p

. Both values are valid probabilities (since they are between

0

and

1

), but they are not the same. Comparing

\frac{5}{12}

and

\frac{1}{3}

(which is

\frac{4}{12}

), we see that

\frac{5}{12}

is clearly larger. In competitive exams, catching these small details is what separates the top rankers.

Animated Solution for Mathematics - Probability: Let $A$ and $B$ be independent events such that $P (A) = p, P (B) = 2 p$ . The largest value of $p$ , for which $P (exactly one of A, B occurs) = 5/9$ , is :

Select Answer:

Visualized Solution

Define Probabilities

Independence of Events

Calculate $P (A \cap B)$

Exactly One Event Occurs

Formula for Exactly One Event

Substitute into the Formula

Simplify the Equation

Form the Quadratic Equation

Factorize the Quadratic

Find the Roots

Select the Largest Value

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Exclusion Principle

The Algebraic Battle

Solving the Quadratic

Final Selection

Similar Questions

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

An experiment has 10 equally likely outcomes. Let $A$ and $B$ be non-empty events of the experiment. If $A$ consists of 4 outcomes, the number of outcomes that $B$ must have so that $A$ and $B$ are independent, 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 $P (A) = 0.3$ and $P (A \cup B) = 0.8$ . If $A$ and $B$ are independent events then $P (B) = .........$

Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $11/25$ and the probability of none of them occurring is $2/25$ . If $P (T)$ denotes the probability of occurrence of the event $T$ , then

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

Select Answer:

Visualized Solution

Define Probabilities

Independence of Events

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

Exactly One Event Occurs

Formula for Exactly One Event

Substitute into the Formula

Simplify the Equation

Form the Quadratic Equation

Factorize the Quadratic

Find the Roots

Select the Largest Value

The Sigma Insight: Addition and Multiplication Theorems

Analyzing the Setup

The Exclusion Principle

The Algebraic Battle

Solving the Quadratic

Final Selection

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)=0.3 and P(A∪B)=0.8. If A and B are independent events then P(B)=.........

.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

Calculate $P (A \cap B)$

Let $A$ and $B$ be two events such that $P (A) = 0.3$ and $P (A \cup B) = 0.8$ . If $A$ and $B$ are independent events then $P (B) = .........$

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

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