Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability tha

Q: Why do we have to split the cases for the thousands place? Can't we just use simple permutations?

That is a common trap! Permutations assume all positions are identical, but in 4-digit numbers, the thousands place is special—it cannot be 0. By splitting the cases based on the position of the 7, we handle this restriction naturally without overcounting or violating the 'exactly one 7' rule.

Q: Why does the last digit being 7 change the calculation compared to the last digit being 2?

Because of the 'exactly one 7' rule! If the last digit is 7, you have already used your 'quota' of sevens. You cannot place another 7 anywhere else. If the last digit is 2, you still have the freedom to place the 7 in the thousands, hundreds, or tens place. The constraint dictates the available choices for the remaining slots.

Q: Is there a faster way to calculate the total sample space?

You could use the complement method (Total 4-digit numbers - numbers with no 7s - numbers with two or more 7s), but that is often more prone to error in high-pressure exams. The direct case-by-case method we used is the safest, most robust path to accuracy in JEE.

Question

Let $A$ be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of $A$ leaves remainder 2 when divided by 5 is:

Options:

(A) $\frac{1}{5}$

(B) $\frac{2}{9}$

(C) $\frac{97}{297}$

(D) $\frac{122}{297}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To solve this problem, we define a 4-digit number as a sequence of four slots. We are constrained by the condition that exactly one digit must be a $7$.

We partition the total sample space into two mutually exclusive cases based on the position of the $7$.

### Defining the Universe

**Case 1: The $7$ is at the thousands place.**
The first digit is fixed as $7$. The remaining three slots can be any digit from $\{0, 1, 2, 3, 4, 5, 6, 8, 9\}$, which provides $9$ choices for each slot.
$$n_1 = 1 	imes 9 	imes 9 	imes 9 = 729$$

**Case 2: The $7$ is not at the thousands place.**
The $7$ can occupy one of the three remaining positions (hundreds, tens, or units). The thousands place cannot be $0$ or $7$, leaving $8$ choices. The other two slots, excluding the $7$, have $9$ choices each.
$$n_2 = 3 	imes (8 	imes 9 	imes 9) = 1944$$

The total sample space is the sum of these cases:
$$n(A) = 729 + 1944 = 2673$$

### The Divisibility Trap

We require the number to leave a remainder of $2$ when divided by $5$. According to the divisibility rule for $5$, the last digit must be either $2$ or $7$.

**Sub-case 1: The last digit is $7$.**
Since the number must contain exactly one $7$, and it is fixed at the units place, no other digit can be $7$. The thousands place has $8$ choices (excluding $0$ and $7$), while the hundreds and tens places have $9$ choices each.
$$n(E_1) = 8 	imes 9 	imes 9 	imes 1 = 648$$

**Sub-case 2: The last digit is $2$.**
Here, the $7$ must occupy one of the other three positions. If the $7$ is at the thousands place, we have $1 	imes 9 	imes 9 	imes 1 = 81$ ways. If the $7$ is at the hundreds or tens place, we have $2 	imes 8 	imes 9 	imes 1 = 144$ ways.
$$n(E_2) = 81 + 144 = 225$$

### Final Calculation

The total number of favorable outcomes is the sum of the sub-cases:
$$n(E) = 648 + 225 = 873$$

The probability $P(E)$ is the ratio of favorable outcomes to the total sample space:
$$P(E) = \frac{873}{2673}$$

By dividing both the numerator and the denominator by their greatest common divisor, $9$, we obtain the final result:
**$$P(E) = \frac{97}{297}$$**

Animated Solution for Mathematics - Probability: Let $A$ be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of $A$ leaves remainder 2 when divided by 5 is:

Select Answer:

Visualized Solution

Defining the Total Sample Space $n (A)$

Case 1: 7 at the Thousands Place

Case 2: 7 is Not at the Thousands Place

Calculating Total Elements $n (A)$

Defining Favorable Condition

Favorable Sub-case 1: Last Digit is 7

Favorable Sub-case 2a: Last Digit is 2, 7 at Thousands

Favorable Sub-case 2b: Last Digit is 2, 7 at Hundreds/Tens

Summing Favorable Cases $n (E)$

Final Probability Calculation

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

Defining the Universe

The Divisibility Trap

Final Calculation

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