JEE Advanced 2003

LEVELJEE Main

Animated Solution for Mathematics - Probability: $A$ is targeting to $B, B$ and $C$ are targeting to $A$ . Probability of hitting the target by $A, B$ and $C$ are $2/3, 1/2$ and $1/3$ respectively. If $A$ is hit then find the probability that $B$ hits the target and $C$ does not.

Q: Why can't we just add $P(B \text{ hits}) + P(C \text{ hits})$ to get the probability of $A$ being hit?

That is a classic trap! If you simply add $P(B) + P(C)$, you are double-counting the scenario where both $B$ and $C$ hit $A$ simultaneously. The correct way to handle the union of two events is $P(B \cup C) = P(B) + P(C) - P(B \cap C)$. However, using the complement rule—$1 - P(\text{neither hits})$—is much faster and less prone to arithmetic errors.

Progressive Hints

Hint 1

The first step to mastering any JEE probability problem is to filter the noise. Look at the problem statement again: does

A

's ability to hit

B

have any impact on whether

A

itself gets hit? Identify the relevant shooters and discard the irrelevant data.

Hint 2

You are dealing with conditional probability. Recall the fundamental definition:

P (X ∣ Y) = \frac{P ( X \cap Y )}{P ( Y )}

. Here,

Y

is the event '

A

is hit' and

X

is the event '

B

hits and

C

misses'. How do you calculate the probability of a union of independent events?

Hint 3

To find

P (A is hit)

, it is mathematically elegant to use the complement rule:

P (A is hit) = 1 - P (B misses) \times P (C misses)

. Once you have this denominator, calculate the numerator by finding the joint probability of

B

hitting and

C

missing. Then, divide the two.

Visualized Solution

Visualizing the Scenario

Three players: $A$ , $B$ , and $C$ .
Targeting: $B \to A$ and $C \to A$ .
Goal: Find $P (B hits and C misses ∣ A is hit)$ .

The Extra Information

The problem states $A$ is targeting $B$ with probability $\frac{2}{3}$ .
However, the condition is " $A$ is hit".
This only depends on $B$ and $C$ shooting at $A$ .
$A$ 's shot is irrelevant to the event " $A$ is hit".

Defining Events

Let $E_{B}$ : Event that $B$ hits $A$ .
Let $E_{C}$ : Event that $C$ hits $A$ .
Given: $P (E_{B}) = \frac{1}{2}$ and $P (E_{C}) = \frac{1}{3}$ .
$E_{B}$ and $E_{C}$ are independent events.

The Condition: $A$ is Hit

Condition: $A$ is hit.
This happens if $B$ hits $A$ , OR $C$ hits $A$ , OR both.
In set notation: $E_{B} \cup E_{C}$ .
We need to calculate $P (E_{B} \cup E_{C})$ .

Calculating $P (A is hit)$

$P (A is hit) = P (E_{B} \cup E_{C})$
Using complement rule: $P (E_{B} \cup E_{C}) = 1 - P (\overset{ˉ}{E}_{B} \cap \overset{ˉ}{E}_{C})$
Since independent: $1 - P (\overset{ˉ}{E}_{B}) P (\overset{ˉ}{E}_{C})$

Substituting Values

$P (\overset{ˉ}{E}_{B}) = 1 - \frac{1}{2} = \frac{1}{2}$
$P (\overset{ˉ}{E}_{C}) = 1 - \frac{1}{3} = \frac{2}{3}$
Substitute: $P (E_{B} \cup E_{C}) = 1 - (\frac{1}{2}) (\frac{2}{3})$

Evaluating the Condition

$P (E_{B} \cup E_{C}) = 1 - (\frac{1}{2}) (\frac{2}{3})$
$P (E_{B} \cup E_{C}) = 1 - \frac{1}{3}$
$P (E_{B} \cup E_{C}) = \frac{2}{3}$

Defining the Target Event

Target Event: $B$ hits and $C$ misses.
In set notation: $E_{B} \cap \overset{ˉ}{E}_{C}$ .
This is the region where only $B$ is successful.

Calculating $P (B \cap \overset{ˉ}{C})$

$P (E_{B} \cap \overset{ˉ}{E}_{C}) = P (E_{B}) \times P (\overset{ˉ}{E}_{C})$
Substitute: $\frac{1}{2} \times (1 - \frac{1}{3})$
Calculate: $\frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$

Applying Conditional Probability

Formula: $P (X ∣ Y) = \frac{P ( X \cap Y )}{P ( Y )}$
Here, $X = E_{B} \cap \overset{ˉ}{E}_{C}$ and $Y = E_{B} \cup E_{C}$ .
Since $(E_{B} \cap \overset{ˉ}{E}_{C})$ is a subset of $(E_{B} \cup E_{C})$ , their intersection is just $E_{B} \cap \overset{ˉ}{E}_{C}$ .
Required Probability = $\frac{P ( E _{B} \cap E ˉ _{C} )}{P ( E _{B} \cup E _{C} )}$

Final Calculation

Substitute values: $\frac{\frac{1}{3}}{\frac{2}{3}}$
Simplify: $\frac{1}{3} \times \frac{3}{2}$
Final Answer: $\frac{1}{2}$

00:00 / 00:00

The Sigma Insight: Conditional Probability

Solution Diagram

Analyzing the Setup

The problem presents a scenario with three players:

A

B

, and

C

. While the problem mentions

A

targeting

B

with a probability of

\frac{2}{3}

, we must recognize this as a distractor.

In probability, we must define our universe based strictly on the relevant events. The only events that influence whether

A

is hit are

E_{B}

(the event that

B

hits

A

) and

E_{C}

(the event that

C

hits

A

Defining the Universe

We are given the independent probabilities:

P (E_{B}) = \frac{1}{2}

P (E_{C}) = \frac{1}{3}

To find the probability that

A

is hit, we define the event

Y = E_{B} \cup E_{C}

. Using the complement rule,

A

is hit if at least one shooter succeeds, which is the complement of both shooters missing.

P (Y) = 1 - P (\overset{ˉ}{E}_{B}) \times P (\overset{ˉ}{E}_{C})

Substituting the values:

P (\overset{ˉ}{E}_{B}) = 1 - \frac{1}{2} = \frac{1}{2}

P (\overset{ˉ}{E}_{C}) = 1 - \frac{1}{3} = \frac{2}{3}

P (Y) = 1 - (\frac{1}{2} \times \frac{2}{3}) = 1 - \frac{1}{3} = \frac{2}{3}

The Target Event

We seek the probability that

B

hits and

C

misses, given that

A

is hit. Let this event be

X = E_{B} \cap \overset{ˉ}{E}_{C}

Since

B

and

C

are independent, we calculate:

P (X) = P (E_{B}) \times P (\overset{ˉ}{E}_{C}) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}

Final Calculation

We require the conditional probability

P (X ∣ Y)

. By the definition of conditional probability:

P (X ∣ Y) = \frac{P ( X \cap Y )}{P ( Y )}

Because the event

X

(B hits and C misses) is a subset of

Y

(A is hit), the intersection

P (X \cap Y)

is simply

P (X)

P (X ∣ Y) = \frac{1/3}{2/3} = \frac{1}{2}

The final probability that

B

hits

A

given that

A

is hit is $\frac{1}{2}$ .

Common Traps & FAQs

Why do we ignore the fact that

A

targets

B

? Isn't that part of the problem?

In JEE Advanced, examiners love to add 'noise' to test your conceptual clarity. The condition given is '

A

is hit'. Since

A

cannot hit itself, the only events that determine whether

A

is hit are

B

shooting at

A

and

C

shooting at

A

A

's own accuracy is completely irrelevant to the sample space defined by the condition. Always ask yourself: 'Does this variable actually influence the outcome I am measuring?' If not, discard it.

Why can't we just add

P (B hits) + P (C hits)

to get the probability of

A

being hit?

That is a classic trap! If you simply add

P (B) + P (C)

, you are double-counting the scenario where both

B

and

C

hit

A

simultaneously. The correct way to handle the union of two events is

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

. However, using the complement rule—

1 - P (neither hits)

—is much faster and less prone to arithmetic errors.

Is the conditional probability formula always necessary, or can I just use intuition?

Intuition is a great starting point, but in the heat of an exam, it can betray you. The conditional probability formula

P (X ∣ Y) = \frac{P ( X \cap Y )}{P ( Y )}

is your safety net. It mathematically restricts your 'universe' to only those cases where

A

is hit. By using the formula, you ensure that you are not accidentally including cases where

A

remains safe.

Animated Solution for Mathematics - Probability: $A$ is targeting to $B, B$ and $C$ are targeting to $A$ . Probability of hitting the target by $A, B$ and $C$ are $2/3, 1/2$ and $1/3$ respectively. If $A$ is hit then find the probability that $B$ hits the target and $C$ does not.

Visualized Solution

Visualizing the Scenario

The Extra Information

Defining Events

The Condition: $A$ is Hit

Calculating $P (A is hit)$

Substituting Values

Evaluating the Condition

Defining the Target Event

Calculating $P (B \cap \overset{ˉ}{C})$

Applying Conditional Probability

Final Calculation

The Sigma Insight: Conditional Probability

Analyzing the Setup

Defining the Universe

The Target Event

Final Calculation

Similar Questions

It is given that the events $A$ and $B$ are such that $P (A) = 1/4, P (A ∣ B) = 1/2$ and $P (B ∣ A) = 2/3$ . Then $P (B)$ is

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 :

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 $A$ and $B$ be two independent events such that $P (A) = \frac{1}{3}$ and $P (B) = \frac{1}{6}$ . Then, which of the following is TRUE ?

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

Let $A$ and $B$ be two events such that $P (B ∣ A) = \frac{2}{5}$ , $P (A ∣ B) = \frac{1}{7}$ and $P (A \cap B) = \frac{1}{9}$ . Consider (S1) $P (A^{'} \cup B) = \frac{5}{6}$ , (S2) $P (A^{'} \cap B^{'}) = \frac{1}{18}$ . Then

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:-

Let $A$ and $B$ be two independent events such that $P (A) = 1/3$ and $P (B) = 1/6$ . Then, which of the following is TRUE?

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

Visualized Solution

Visualizing the Scenario

The Extra Information

Defining Events

The Condition: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A is Hit

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

Substituting Values

Evaluating the Condition

Defining the Target Event

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

Applying Conditional Probability

Final Calculation

The Sigma Insight: Conditional Probability

Analyzing the Setup

Defining the Universe

The Target Event

Final Calculation

Similar Questions

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

.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 A and B be two independent events such that P(A)=31​ and P(B)=61​. Then, which of the following is TRUE ?

.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; } Let A and B be two events such that P(B∣A)=52​, P(A∣B)=71​ and P(A∩B)=91​. Consider (S1) P(A′∪B)=65​, (S2) P(A′∩B′)=181​. 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)=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; } Let A and B be two independent events such that P(A)=1/3 and P(B)=1/6. Then, which of the following is TRUE?

.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

The Condition: $A$ is Hit

Calculating $P (A is hit)$

Calculating $P (B \cap \overset{ˉ}{C})$

It is given that the events $A$ and $B$ are such that $P (A) = 1/4, P (A ∣ B) = 1/2$ and $P (B ∣ A) = 2/3$ . Then $P (B)$ is

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 $A$ and $B$ be two independent events such that $P (A) = \frac{1}{3}$ and $P (B) = \frac{1}{6}$ . Then, which of the following is TRUE ?

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

Let $A$ and $B$ be two events such that $P (B ∣ A) = \frac{2}{5}$ , $P (A ∣ B) = \frac{1}{7}$ and $P (A \cap B) = \frac{1}{9}$ . Consider (S1) $P (A^{'} \cup B) = \frac{5}{6}$ , (S2) $P (A^{'} \cap B^{'}) = \frac{1}{18}$ . Then

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:-

Let $A$ and $B$ be two independent events such that $P (A) = 1/3$ and $P (B) = 1/6$ . Then, which of the following is TRUE?

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