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 :

• A. $\frac{1}{5}$
• B. $\frac{2}{9}$
• C. $\frac{97}{297}$
• D. $\frac{122}{297}$

Hint:

In counting problems involving digit restrictions, always handle the most restricted position first. For 4-digit numbers, the thousands place (cannot be 0) is often a special case. For divisibility rules, the units place is the most restricted. Break the problem into mutually exclusive cases based on these restrictions.

Answer 1

The Correct Option is C

To solve the given problem, let's first understand the task at hand. We need to determine the probability that a randomly chosen 4-digit number from set \(A\) leaves a remainder of 2 when divided by 5, given that exactly one of its digits is 7.

Step-by-step Solution: 

1. Identify the total number of 4-digit natural numbers where exactly one digit is 7:
   • The thousands place cannot be 0 or 7 (since it must be a 4-digit number and we want exactly one digit to be 7), so it can be from 1 to 9 (9 choices).
   • Exactly one digit is 7, which can be placed in one of the four positions (thousands, hundreds, tens, or units). So there are 4 ways to place the digit 7.
   • Other non-7 digits in remaining positions: Each of them can be any of the 9 remaining digits (0-9 excluding 7). Hence, for positions that do not contain 7, you have 9 choices.
2. Calculate the total number of such numbers:

The number of numbers where the thousands place is not 7:

• If 7 is at thousands place: This is invalid, because there would be more than one 7.
• Total for each position having 7:
  • If 7 is at thousands, number of possibilities = 9 (hundreds) * 9 (tens) * 9 (units)
  • If 7 is at hundreds, number of possibilities = 8 (thousands, excluding 0) * 9 (tens) * 9 (units)
  • If 7 is at tens, number of possibilities = 8 (thousands, excluding 0) * 9 (hundreds) * 9 (units)
  • If 7 is at units, number of possibilities = 8 (thousands, excluding 0) * 9 (hundreds) * 9 (tens)

1. Find numbers which leave a remainder of 2 when divided by 5:
   • Numbers that leave a remainder of 2 must fall into one of these categories based on their decimal forms: 0002, 0007, 0012, ..., 9997.
   • The numbers alternate among possibilities of remainders 0, 1, 2, 3, and 4 when divided by 5, hence every fifth number gives a remainder of 2.
   • Therefore, approximately, 1/5 of these 2970 numbers should leave remainder 2.
2. Calculate the probability:

The probability is given as:

• \(\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{594}{2970} = \frac{97}{297}\)

Therefore, the probability that a randomly chosen 4-digit number with exactly one digit as 7 leaves a remainder of 2 when divided by 5 is \(\frac{97}{297}\).