Question

The probability, that in a randomly selected 3-digit number at least two digits are odd, is

• A. \(\frac {19}{36}\)
• B. \(\frac {15}{36}\)
• C. \(\frac {13}{36}\)
• D. \(\frac {23}{36}\)

Answer 1

The Correct Option is A

To determine the probability that a randomly selected 3-digit number has at least two digits that are odd, we can proceed with the following steps:

1. Identify the total number of 3-digit numbers:
   A 3-digit number ranges from 100 to 999. Therefore, the total number of 3-digit numbers is: 999 - 100 + 1 = 900.
2. Calculate the probability of odd digits:
   The odd digits in our numeral system are 1, 3, 5, 7, and 9. Therefore, there are 5 possible odd digits.
3. Calculate the probability of even digits:
   The even digits are 0, 2, 4, 6, and 8, which gives us another set of 5 possible digits.
4. Compute the number of numbers with fewer than two odd digits:
   • No odd digits:
     All three digits must be even. The hundreds digit can be 2, 4, 6, or 8 (4 choices, since 0 cannot be the hundreds digit). The tens and units digits can also be 0, 2, 4, 6, or 8 (each with 5 choices). Thus, the number of numbers with no odd digits is: 4 \times 5 \times 5 = 100.
   • One odd digit:
     Select which position contains the odd digit (hundreds, tens, or units), and then fill the remaining positions with even digits.
     • Odd digit in hundreds place: 5 \times 5 \times 5 = 125
     • Odd digit in tens place: 4 \times 5 \times 5 = 100
     • Odd digit in units place: 4 \times 5 \times 5 = 100
     Total numbers with one odd digit: 125 + 100 + 100 = 325.
5. Total numbers with at least two odd digits:
   Subtract the above amounts from the total number of 3-digit numbers:
   900 - 100 - 325 = 475.
6. Calculate the required probability:
   The probability that a randomly selected 3-digit number has at least two odd digits is: \frac{475}{900} = \frac{19}{36}.

Therefore, the probability that in a randomly selected 3-digit number at least two digits are odd is \(\frac{19}{36}\).