JEE Advanced 2014

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Animated Solution for Mathematics - Probability: Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is

Q: Wait, why do we treat the boys and girls as distinct? Doesn't the problem imply they are identical?

In probability problems involving arrangements, we treat all individuals as distinct to ensure each outcome in our sample space is equally likely. Even if we treated them as identical, the ratio would remain the same, but treating them as distinct is the safest, most robust method to avoid errors.

Q: Why is $b_2 = y - 2$?

Great question! At position $y$, there are $y-1$ people ahead of the second girl. One of those people is the first girl. So, the number of boys ahead of her is the total people ahead ($y-1$) minus the one girl ahead ($1$), which gives $y-2$.

Q: Can we use symmetry to solve this?

While symmetry is powerful, this specific constraint is asymmetric because the condition for the second girl is stricter than the first. Direct counting is the most reliable path here.

Select Answer:

Progressive Hints

Hint 1

The problem asks for a specific condition regarding the number of boys ahead of each girl. Instead of looking at the whole queue, focus on the two girls. Let's label them

G_{1}

and

G_{2}

, where

G_{1}

is ahead of

G_{2}

. What does the condition

b_{i} \geq g_{i} + 1

imply for each girl individually?

Hint 2

Think of the queue as

5

slots,

1

through

5

. If

G_{1}

is at position

x

and

G_{2}

is at position

y

, how many boys are ahead of each? For

G_{1}

, the number of boys ahead is

b_{1} = x - 1

. For

G_{2}

, the number of boys ahead is

b_{2} = y - 2

. Now, substitute these into the given inequalities.

Hint 3

You have

b_{1} \geq 1

and

b_{2} \geq 2

. This translates to

x - 1 \geq 1

(so

x \geq 2

) and

y - 2 \geq 2

(so

y \geq 4

). With the constraint

1 \leq x < y \leq 5

, list all valid pairs

(x, y)

that satisfy these conditions. Once you have the valid positions, the rest is just counting arrangements.

Visualized Solution

Understanding the Queue Setup

We have $3$ boys and $2$ girls, making a total of $5$ people.
Let's represent the queue positions as slots $1, 2, 3, 4, 5$ from front to back.

Calculating Total Arrangements

The total number of ways to arrange $5$ distinct people in $5$ slots is given by $5!$ .
$Total Arrangements = 5! = 120$ .

Decoding the Mathematical Condition

Let $G_{1}$ and $G_{2}$ be the two girls, where $G_{1}$ is ahead of $G_{2}$ .
Let $b_{i}$ be the number of boys ahead of girl $i$ , and $g_{i}$ be the number of girls ahead of girl $i$ .
The condition states: $b_{i} \geq g_{i} + 1$ for every girl.
For the first girl ( $G_{1}$ ): since she is the first girl in the queue, $g_{1} = 0$ .
Therefore, the condition for $G_{1}$ becomes: $b_{1} \geq 0 + 1 ⟹ b_{1} \geq 1$ .

Condition for the Second Girl

For the second girl ( $G_{2}$ ): since she is behind $G_{1}$ , $g_{2} = 1$ .
Therefore, the condition for $G_{2}$ becomes: $b_{2} \geq 1 + 1 ⟹ b_{2} \geq 2$ .
So, there must be at least $2$ boys ahead of the second girl.

Finding Valid Positions for the Girls

Let the position of the first girl $G_{1}$ be $x$ , and the second girl $G_{2}$ be $y$ , where $1 \leq x < y \leq 5$ .
Since $b_{1} \geq 1$ , the first girl cannot be at position $1$ (as that would mean $b_{1} = 0$ ).
Thus, the position of the first girl must satisfy $x \geq 2$ .
Since $b_{2} \geq 2$ , and there is $1$ girl ( $G_{1}$ ) ahead of $G_{2}$ , the number of boys ahead of $G_{2}$ is $b_{2} = y - 2$ .
Thus, $y - 2 \geq 2 ⟹ y \geq 4$ .

Listing the Valid Position Pairs $(x, y)$

We have the constraints: $2 \leq x < y \leq 5$ and $y \geq 4$ .
Let's systematically find all valid pairs $(x, y)$ :
If $x = 2$ : $y$ can be $4$ or $5$ . (Pairs: $(2, 4)$ , $(2, 5)$ )
If $x = 3$ : $y$ can be $4$ or $5$ . (Pairs: $(3, 4)$ , $(3, 5)$ )
If $x = 4$ : $y$ can only be $5$ . (Pair: $(4, 5)$ )
Total number of valid position pairs for the girls is $5$ .

Calculating Favorable Arrangements

For each of the $5$ valid position pairs, we must arrange the actual individuals.
The $2$ girls can be arranged in the $2$ chosen slots in $2! = 2$ ways.
The $3$ boys can be arranged in the remaining $3$ slots in $3! = 6$ ways.
Favorable arrangements for each pair = $2! \times 3! = 12$ .
Total favorable arrangements = $5 \times 12 = 60$ .

Calculating the Final Probability

The required probability is: $P = \frac{Favorable Arrangements}{Total Arrangements}$
$P = \frac{60}{120} = \frac{1}{2}$
Thus, the correct option is 1/2.

00:00 / 00:00

The Sigma Insight: Classical Definition of Probability

Solution Diagram

The Universe of Possibilities

Before we impose any rules, let us look at the total number of ways to arrange these five distinct individuals. Since we have

5

distinct people, the total number of linear arrangements is simply:

5! = 120

This is our sample space, the denominator of our probability fraction. It represents every possible way these five people could stand in line.

Decoding the Constraint

Now, let us look at the condition. For each girl, the number of boys ahead (

b_{i}

) must be at least one more than the number of girls ahead (

g_{i}

). That is:

b_{i} \geq g_{i} + 1

Let us label our girls

G_{1}

and

G_{2}

, with

G_{1}

standing closer to the front. For

G_{1}

, there are no girls ahead, so

g_{1} = 0

. The condition becomes

b_{1} \geq 0 + 1

, or

b_{1} \geq 1

. This means at least one boy must stand before

G_{1}

For

G_{2}

, there is one girl ahead (

G_{1}

), so

g_{2} = 1

. The condition becomes

b_{2} \geq 1 + 1

, or

b_{2} \geq 2

. This means at least two boys must stand before

G_{2}

The Geometry of the Queue

Let us map these conditions to positions

x

and

y

for

G_{1}

and

G_{2}

, where

1 \leq x < y \leq 5

. The number of boys ahead of

G_{1}

x - 1

. Thus:

x - 1 \geq 1 ⟹ x \geq 2

The number of boys ahead of

G_{2}

y - 2

. Thus:

y - 2 \geq 2 ⟹ y \geq 4

We are looking for pairs

(x, y)

such that

2 \leq x < y \leq 5

and

y \geq 4

. Let us list them:

x = 2

y

can be

4

5

x = 3

y

can be

4

5

x = 4

y

must be

5

That gives us exactly

5

valid position pairs.

The Final Tally

For each of these

5

valid position pairs, we must arrange the individuals. The

2

girls can be arranged in

2! = 2

ways, and the

3

boys can be arranged in

3! = 6

ways.

So, for each pair, there are

2 \times 6 = 12

arrangements. With

5

valid pairs, the total number of favorable arrangements is:

5 \times 12 = 60

The probability is then:

\frac{60}{120} = \frac{1}{2}

It is a clean, elegant result, emerging from the constraints we carefully mapped. The final probability is $1/2$ .

Common Traps & FAQs

Wait, why do we treat the boys and girls as distinct? Doesn't the problem imply they are identical?

In probability problems involving arrangements, we treat all individuals as distinct to ensure each outcome in our sample space is equally likely. Even if we treated them as identical, the ratio would remain the same, but treating them as distinct is the safest, most robust method to avoid errors.

Why is

b_{2} = y - 2

Great question! At position

y

, there are

y - 1

people ahead of the second girl. One of those people is the first girl. So, the number of boys ahead of her is the total people ahead (

y - 1

) minus the one girl ahead (

1

), which gives

y - 2

Can we use symmetry to solve this?

While symmetry is powerful, this specific constraint is asymmetric because the condition for the second girl is stricter than the first. Direct counting is the most reliable path here.

Animated Solution for Mathematics - Probability: Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is

Select Answer:

Visualized Solution

Understanding the Queue Setup

Calculating Total Arrangements

Decoding the Mathematical Condition

Condition for the Second Girl

Finding Valid Positions for the Girls

Listing the Valid Position Pairs $(x, y)$

Calculating Favorable Arrangements

Calculating the Final Probability

The Sigma Insight: Classical Definition of Probability

The Universe of Possibilities

Decoding the Constraint

The Geometry of the Queue

The Final Tally

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Animated Solution for Mathematics - Probability: Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is

Select Answer:

Visualized Solution

Understanding the Queue Setup

Calculating Total Arrangements

Decoding the Mathematical Condition

Condition for the Second Girl

Finding Valid Positions for the Girls

Listing the Valid Position Pairs .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x,y)

Calculating Favorable Arrangements

Calculating the Final Probability

The Sigma Insight: Classical Definition of Probability

The Universe of Possibilities

Decoding the Constraint

The Geometry of the Queue

The Final Tally

Similar Questions

Six boys and six girls sit in a row randomly. Find the probability that (i) the six girls sit together (ii) the boys and girls sit alternately.

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is

Balls are drawn one-by-one without replacement from a box containing 2 black, 4 white and 3 red balls till all the balls are drawn. Find the probability that the balls drawn are in the order 2 black, 4 white and 3 red.

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

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