JEE Advanced 2007

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Animated Solution for Mathematics - Probability: One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is

Q: Wait, why do we treat the Indian couple as two separate people in the denominator $n(E_2)$?

This is the most common trap! The condition $E_2$ only mandates that the American couples must sit together. It imposes absolutely no restriction on the Indian couple. In probability, if a condition doesn't force a constraint, you must assume the entities are free to move independently. Therefore, in the denominator, the Indian man and his wife are two distinct, free-roaming individuals, not a single block.

Q: Why do we multiply by $2^n$ in the final calculation?

Great observation! That $2^n$ represents the internal arrangement of the couples. Even if a couple is 'glued' together as a block, the husband can sit to the left of the wife, or the wife can sit to the left of the husband. That is $2!$ (which is $2$) ways for each couple. Since we have multiple couples, we raise $2$ to the power of the number of couples. It is a crucial step that students often forget!

Select Answer:

Progressive Hints

Hint 1

Think of the circular table as a stage where the rules of the game change based on the 'given' condition. When we say 'given that each American man is seated adjacent to his wife,' we are effectively shrinking our universe. We are no longer looking at all possible seating arrangements; we are only looking at the subset where the American couples are already 'glued' together as single units. Your first task is to visualize this restricted sample space.

Hint 2

Remember your toolkit for circular permutations: if you have

n

distinct items, there are

(n - 1)!

ways to arrange them in a circle. But do not forget the 'internal degrees of freedom'! Each couple is a block, and within that block, the husband and wife can swap places in

2!

ways. This internal arrangement is a multiplier that applies to every couple block you create.

Hint 3

Start by writing the ratio:

P (E_{1} ∣ E_{2}) = \frac{n ( E _{1} \cap E _{2} )}{n ( E _{2} )}

. For the denominator

n (E_{2})

, treat the 4 American couples as 4 blocks and the 2 Indian individuals as 2 separate units. For the numerator

n (E_{1} \cap E_{2})

, treat the 4 American couples as 4 blocks and the 1 Indian couple as 1 block. Now, apply the circular permutation formula

(n - 1)!

to these sets and multiply by the internal arrangements

(2!)^{k}

for each case.

Visualized Solution

The Circular Setup

Total people: $10$
$1$ Indian couple + $4$ American couples
Seated randomly around a circular table.

Defining the Events

Let $E_{1}$ : Indian man is adjacent to his wife.
Let $E_{2}$ : Each American man is adjacent to his wife.
We need to find $P (E_{1} ∣ E_{2})$ .

Conditional Probability Formula

Formula: $P (E_{1} ∣ E_{2}) = \frac{P ( E _{1} \cap E _{2} )}{P ( E _{2} )}$
Using sample space: $P (E_{1} ∣ E_{2}) = \frac{n ( E _{1} \cap E _{2} )}{n ( E _{2} )}$

Analyzing Condition $E_{2}$

Focus on $E_{2}$ : Each American couple sits together.
Block Method: Treat each American couple as a single unit.

Counting Units for $E_{2}$

$4$ American units (couples).
$2$ Indian individuals (not grouped yet).
Total units for circular arrangement = $4 + 2 = 6$ .

Circular Permutations for $E_{2}$

Arranging $n$ units in a circle: $(n - 1)!$
For $6$ units: $(6 - 1)! = 5!$ ways.

Internal Arrangements for $E_{2}$

Each American couple can swap seats internally.
$2!$ ways for each of the $4$ couples.
Total $n (E_{2}) = 5! \times (2!)^{4} = 5! \times 2^{4}$ .

Analyzing Intersection $E_{1} \cap E_{2}$

Now consider $E_{1} \cap E_{2}$ : All couples sit together.
The Indian couple must also be treated as a single unit.

Counting Units for $E_{1} \cap E_{2}$

$4$ American units + $1$ Indian unit.
Total units = $5$ couple blocks.

Circular Permutations for $E_{1} \cap E_{2}$

Arranging $5$ units in a circle.
Ways = $(5 - 1)! = 4!$

Internal Arrangements for $E_{1} \cap E_{2}$

Each of the $5$ couples can swap seats internally.
$2!$ ways for each couple.
Total $n (E_{1} \cap E_{2}) = 4! \times (2!)^{5} = 4! \times 2^{5}$ .

Setting Up the Final Ratio

Substitute into $P (E_{1} ∣ E_{2}) = \frac{n ( E _{1} \cap E _{2} )}{n ( E _{2} )}$
$P (E_{1} ∣ E_{2}) = \frac{4 ! \times 2 ^{5}}{5 ! \times 2 ^{4}}$

Simplifying the Expression

Simplify powers of $2$ : $\frac{2 ^{5}}{2 ^{4}} = 2$
Simplify factorials: $\frac{4 !}{5 !} = \frac{4 !}{5 \times 4 !} = \frac{1}{5}$

The Final Answer

Multiply the simplified parts: $\frac{1}{5} \times 2 = \frac{2}{5}$
The conditional probability is $\frac{2}{5}$ .

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The Sigma Insight: Conditional Probability

Solution Diagram

Analyzing the Setup

We are tasked with finding the conditional probability

P (E_{1} ∣ E_{2})

, where

E_{1}

is the event that the Indian couple sits together, and

E_{2}

is the event that all four American couples sit together at a circular table of ten people.

The formula for conditional probability serves as our guide:

P (E_{1} ∣ E_{2}) = \frac{n ( E _{1} \cap E _{2} )}{n ( E _{2} )}

We do not need the total sample space of all possible arrangements, as the denominator cancels out in the ratio. We only need to count the favorable arrangements for the intersection and the given condition.

Phase 1

The Denominator - The World of

E_{2}

To calculate

n (E_{2})

, we use the Block Method. We treat each of the 4 American couples as a single, inseparable unit.

Since the Indian couple is not restricted by

E_{2}

, they remain as 2 separate individuals. This gives us a total of

4

blocks +

2

individuals =

6

units to arrange around a circle.

The number of ways to arrange

n

units in a circle is

(n - 1)!

. Thus, we have

(6 - 1)! = 5!

ways.

We must also account for the internal arrangements of the 4 American couples. Each couple can swap seats in

2!

ways, leading to a factor of

2^{4}

. Therefore:

n (E_{2}) = 5! \times 2^{4}

Phase 2

The Numerator - The Intersection

E_{1} \cap E_{2}

In the intersection

E_{1} \cap E_{2}

, both the American couples and the Indian couple must sit together. We now treat the Indian couple as a single block as well.

We have 4 American blocks + 1 Indian block =

5

units to arrange in a circle. The number of circular arrangements is

(5 - 1)! = 4!

Next, we account for the internal arrangements of all 5 couples (4 American + 1 Indian). Each couple can swap seats in

2!

ways, giving us a factor of

2^{5}

. Thus:

n (E_{1} \cap E_{2}) = 4! \times 2^{5}

Phase 3

The Elegant Cancellation

Now, we substitute our values into the conditional probability formula:

P (E_{1} ∣ E_{2}) = \frac{4 ! \times 2 ^{5}}{5 ! \times 2 ^{4}}

We simplify the expression by observing the powers of 2 and the factorials:

\frac{2 ^{5}}{2 ^{4}} = 2

\frac{4 !}{5 !} = \frac{4 !}{5 \times 4 !} = \frac{1}{5}

Multiplying these results, we find the final probability:

$P (E_{1} ∣ E_{2}) = 2 \times \frac{1}{5} = \frac{2}{5}$

The probability is $\frac{2}{5}$ . By treating the couples as blocks and respecting the circular symmetry, we have successfully navigated the constraints of the problem.

Common Traps & FAQs

Wait, why do we treat the Indian couple as two separate people in the denominator

n (E_{2})

This is the most common trap! The condition

E_{2}

only mandates that the American couples must sit together. It imposes absolutely no restriction on the Indian couple. In probability, if a condition doesn't force a constraint, you must assume the entities are free to move independently. Therefore, in the denominator, the Indian man and his wife are two distinct, free-roaming individuals, not a single block.

Why is it

(n - 1)!

and not

n!

? I always get confused with circular arrangements.

Think of it this way: in a linear row, position matters relative to the ends. In a circle, there are no ends. If everyone shifts one seat to the left, the relative arrangement is identical. To break this rotational symmetry, we 'fix' one person or one block in a specific seat. Once that one unit is fixed, the remaining

(n - 1)

units can be arranged in

(n - 1)!

ways. It is the standard way to anchor a circular system.

Why do we multiply by

2^{n}

in the final calculation?

Great observation! That

2^{n}

represents the internal arrangement of the couples. Even if a couple is 'glued' together as a block, the husband can sit to the left of the wife, or the wife can sit to the left of the husband. That is

2!

(which is

2

) ways for each couple. Since we have multiple couples, we raise

2

to the power of the number of couples. It is a crucial step that students often forget!

Animated Solution for Mathematics - Probability: One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is

Select Answer:

Visualized Solution

The Circular Setup

Defining the Events

Conditional Probability Formula

Analyzing Condition $E_{2}$

Counting Units for $E_{2}$

Circular Permutations for $E_{2}$

Internal Arrangements for $E_{2}$

Analyzing Intersection $E_{1} \cap E_{2}$

Counting Units for $E_{1} \cap E_{2}$

Circular Permutations for $E_{1} \cap E_{2}$

Internal Arrangements for $E_{1} \cap E_{2}$

Setting Up the Final Ratio

Simplifying the Expression

The Final Answer

The Sigma Insight: Conditional Probability

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

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Animated Solution for Mathematics - Probability: One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is

Select Answer:

Visualized Solution

The Circular Setup

Defining the Events

Conditional Probability Formula

Analyzing Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Counting Units for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Circular Permutations for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Internal Arrangements for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Analyzing Intersection .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​∩E2​

Counting Units for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​∩E2​

Circular Permutations for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​∩E2​

Internal Arrangements for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​∩E2​

Setting Up the Final Ratio

Simplifying the Expression

The Final Answer

The Sigma Insight: Conditional Probability

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

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Analyzing Intersection $E_{1} \cap E_{2}$

Counting Units for $E_{1} \cap E_{2}$

Circular Permutations for $E_{1} \cap E_{2}$

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