JEE Advanced 2023

LEVELJEE Main

Animated Solution for Mathematics - Probability: Let $X$ be the set of all five digit numbers formed using $1, 2, 2, 2, 4, 4, 0$ . For example, $22240$ is in $X$ while $02244$ and $44422$ are not in $X$ . Suppose that each element of $X$ has an equal chance of being chosen. Let $p$ be the conditional probability that an element chosen at random is a multiple of $20$ given that it is a multiple of $5$ . Then the value of $38 p$ is equal to

Q: Wait, why do we reject the possibility of the last digit being 5?

That is a great observation! The divisibility rule for $5$ states that a number must end in $0$ or $5$. However, we must look at our available pool of digits: $\{1, 2, 2, 2, 4, 4, 0\}$. Since there is no $5$ in our set, it is physically impossible for any number formed from these digits to end in $5$. Therefore, the last digit is forced to be $0$.

Q: Why use complementary counting for the multiple of 20 condition? Can't I just count the endings 20 and 40?

You absolutely can! Counting the endings $20$ and $40$ is a perfectly valid method. However, complementary counting is often faster and less prone to errors. By identifying that the only 'bad' ending (one that is not a multiple of $20$) is $10$, we only have to perform one calculation instead of two. It is a powerful tool in your JEE arsenal to save time and reduce the chance of missing a case.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the five-digit number as a sequence of five slots. The condition of being a multiple of

5

is the most restrictive constraint. Since our set

{1, 2, 2, 2, 4, 4, 0}

contains no

5

, what must the last digit be to satisfy the divisibility rule for

5

Hint 2

Hint 2 (The Arsenal): Once the last digit is fixed, you are left with four slots to fill using the remaining digits

{1, 2, 2, 2, 4, 4}

. You need to calculate the total number of ways to arrange these digits. Use the formula for permutations of a multiset:

\frac{n !}{n _{1} ! n _{2} ! \dots n _{k} !}

, where

n_{i}

are the frequencies of repeating digits. You must sum the arrangements for all possible selections of four digits.

Hint 3

Hint 3 (The Execution): To find the number of multiples of

20

, consider the condition that the last two digits must be divisible by

20

. Since the last digit is

0

, the second-to-last digit must be

2

4

. Instead of calculating these two cases, use the complement: subtract the number of arrangements ending in

10

from the total sample space of

38

Visualized Solution

Condition for Multiple of $5$

Digits available: ${1, 2, 2, 2, 4, 4, 0}$
The $5$ -digit number must be a multiple of $5$ .
A number is divisible by $5$ if it ends in $0$ or $5$ .
Since we only have $0$ , the $5$ th digit is fixed as $0$ .

Setting up the Sample Space

Remaining empty slots: $4$
Remaining available digits: ${1, 2, 2, 2, 4, 4}$
We need to find all possible ways to select and arrange $4$ digits.
This total count will be our sample space denominator.

Permutations with Three $2$ s

Case 1: Selecting ${1, 2, 2, 2}$
Arrangements $= \frac{4 !}{3 !} = 4$ ways
Case 2: Selecting ${2, 2, 2, 4}$
Arrangements $= \frac{4 !}{3 !} = 4$ ways

Permutations with Two $2$ s

Case 3: Selecting ${1, 2, 2, 4}$
Arrangements $= \frac{4 !}{2 !} = 12$ ways
Case 4: Selecting ${1, 2, 4, 4}$
Arrangements $= \frac{4 !}{2 !} = 12$ ways
Case 5: Selecting ${2, 2, 4, 4}$
Arrangements $= \frac{4 !}{2 ! 2 !} = 6$ ways

Total Multiples of $5$

Total multiples of $5 = 4 + 4 + 12 + 12 + 6$
Total $= 38$
This $38$ is the denominator $n (S)$ for our conditional probability.

Condition for Multiple of $20$

For a number to be a multiple of $20$ , its last two digits must be divisible by $20$ .
Possible endings: $00, 20, 40, 60, 80$ .
Since we only have one $0$ , $00$ is impossible.
Available tens digits: $1, 2, 4$ .
The number must end in $20$ or $40$ .

The Complementary Counting Trick

Instead of counting endings $20$ and $40$ , we use Complementary Counting.
The only available tens digit that makes it NOT a multiple of $20$ is $1$ .
We will count the numbers ending in $10$ .
Then subtract this count from the total ( $38$ ).

Numbers Ending in $10$

Last two digits are fixed as $10$ .
Remaining $3$ slots to be filled from ${2, 2, 2, 4, 4}$ .
Possible selections for these $3$ slots:
$∙ {2, 2, 2}$
$∙ {2, 2, 4}$
$∙ {2, 4, 4}$

Permutations for Ending in $10$

$∙ {2, 2, 2} ⟹ \frac{3 !}{3 !} = 1$ way
$∙ {2, 2, 4} ⟹ \frac{3 !}{2 !} = 3$ ways
$∙ {2, 4, 4} ⟹ \frac{3 !}{2 !} = 3$ ways
Total numbers ending in $10 = 1 + 3 + 3 = 7$

Total Multiples of $20$

Multiples of $20 = (Total multiples of 5) - (Numbers ending in 10)$
Multiples of $20 = 38 - 7 = 31$
This $31$ is our numerator $n (E)$ .

Final Probability Calculation

Conditional probability $p = \frac{n ( E )}{n ( S )} = \frac{31}{38}$
We need to find the value of $38 p$ .
$38 p = 38 \times \frac{31}{38} = 31$
Final Answer: 31

00:00 / 00:00

The Sigma Insight: Conditional Probability

Solution Diagram

Analyzing the Setup

We are given the set of digits

{1, 2, 2, 2, 4, 4, 0}

and tasked with forming five-digit numbers. We seek the conditional probability that a number is a multiple of

20

, given that it is a multiple of

5

A number is divisible by

5

if its last digit is

0

5

. Since our set contains no

5

, the last digit must be

0

With the last slot fixed as

0

, we must fill the remaining four slots using the set

{1, 2, 2, 2, 4, 4}

. This set of six digits forms our sample space.

The Combinatorial Dance

To find the total number of five-digit numbers divisible by

5

, we calculate the permutations of four digits chosen from

{1, 2, 2, 2, 4, 4}

. We categorize these by the selection of digits:

1. Selection

{1, 2, 2, 2}

\frac{4 !}{3 !} = 4 ways

2. Selection

{2, 2, 2, 4}

\frac{4 !}{3 !} = 4 ways

3. Selection

{1, 2, 2, 4}

\frac{4 !}{2 !} = 12 ways

4. Selection

{1, 2, 4, 4}

\frac{4 !}{2 !} = 12 ways

5. Selection

{2, 2, 4, 4}

\frac{4 !}{2 ! 2 !} = 6 ways

Summing these, the total number of favorable outcomes for the condition (divisibility by

5

) is

n (S) = 4 + 4 + 12 + 12 + 6 = 38

The Multiple of 20 Trap

A number is divisible by

20

if its last two digits are divisible by

20

. Since the last digit is

0

, the possible two-digit endings are

20

and

40

We use complementary counting. The only other possible ending for a number ending in

0

10

. We calculate the number of ways to form a number ending in

10

using three digits from

{2, 2, 2, 4, 4}

1. Selection

{2, 2, 2}

\frac{3 !}{3 !} = 1 way

2. Selection

{2, 2, 4}

\frac{3 !}{2 !} = 3 ways

3. Selection

{2, 4, 4}

\frac{3 !}{2 !} = 3 ways

The total number of arrangements ending in

10

1 + 3 + 3 = 7

. Thus, the number of multiples of

20

38 - 7 = 31

Final Calculation

The conditional probability

p

is the ratio of the number of multiples of

20

to the total number of multiples of

5

p = \frac{31}{38}

The problem asks for the value of

38 p

. Substituting our value for

p

38 \times (\frac{31}{38}) = 31

The final answer is 31.

Common Traps & FAQs

Wait, why do we reject the possibility of the last digit being 5?

That is a great observation! The divisibility rule for

5

states that a number must end in

0

5

. However, we must look at our available pool of digits:

{1, 2, 2, 2, 4, 4, 0}

. Since there is no

5

in our set, it is physically impossible for any number formed from these digits to end in

5

. Therefore, the last digit is forced to be

0

Why is the denominator

38

and not just the total permutations of the digits?

The denominator represents the size of our sample space, which is the set of all five-digit numbers that are multiples of

5

. We already established that the last digit must be

0

. Thus, we are only counting the permutations of the remaining four digits chosen from

{1, 2, 2, 2, 4, 4}

. If we just took the total permutations of all digits, we would be including numbers that are not multiples of

5

, which would violate the condition of the conditional probability.

Why use complementary counting for the multiple of 20 condition? Can't I just count the endings 20 and 40?

You absolutely can! Counting the endings

20

and

40

is a perfectly valid method. However, complementary counting is often faster and less prone to errors. By identifying that the only 'bad' ending (one that is not a multiple of

20

) is

10

, we only have to perform one calculation instead of two. It is a powerful tool in your JEE arsenal to save time and reduce the chance of missing a case.

Enter Numerical Value:

Visualized Solution

Condition for Multiple of $5$

Setting up the Sample Space

Permutations with Three $2$ s

Permutations with Two $2$ s

Total Multiples of $5$

Condition for Multiple of $20$

The Complementary Counting Trick

Numbers Ending in $10$

Permutations for Ending in $10$

Total Multiples of $20$

Final Probability Calculation

The Sigma Insight: Conditional Probability

Analyzing the Setup

The Combinatorial Dance

The Multiple of 20 Trap

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Condition for Multiple of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5

Setting up the Sample Space

Permutations with Three .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2s

Permutations with Two .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2s

Total Multiples of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5

Condition for Multiple of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 20

The Complementary Counting Trick

Numbers Ending in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 10

Permutations for Ending in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 10

Total Multiples of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 20

Final Probability Calculation

The Sigma Insight: Conditional Probability

Analyzing the Setup

The Combinatorial Dance

The Multiple of 20 Trap

Final Calculation

Similar Questions

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Permutations with Three $2$ s

Permutations with Two $2$ s

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Condition for Multiple of $20$

Numbers Ending in $10$

Permutations for Ending in $10$

Total Multiples of $20$

Two integers are selected at random from the set ${1, 2, \dots, 11}$ . Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :

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