A ten-digit number is selected at random from the set of all possible ten-digit numbers. The probability that the sum of the ten digits in this selected number is odd, is nearest to

A ten-digit number has its first digit from 1-9 and the remaining nine digits from 0-9.

For the first digit:

• Odd digits: 1, 3, 5, 7, 9 (5 options)
• Even digits: 2, 4, 6, 8 (4 options)
• Total: 9 options

For each of the remaining 9 digits:

• Odd digits: 1, 3, 5, 7, 9 (5 options)
• Even digits: 0, 2, 4, 6, 8 (5 options)
• Total: 10 options

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The sum of all ten digits is odd if and only if there is an odd number of odd digits in the number.

For the last 9 positions (digits 2 through 10):

Each digit independently has probability $\frac{5}{10} = \frac{1}{2}$ of being odd.

By symmetry, the sum of these 9 digits has:

• Probability $\frac{1}{2}$ of being odd
• Probability $\frac{1}{2}$ of being even

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For the first digit:

Probability of being odd $= \frac{5}{9}$

Probability of being even $= \frac{4}{9}$

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The total sum of all 10 digits is odd when:

• First digit is odd AND sum of remaining 9 digits is even, OR
• First digit is even AND sum of remaining 9 digits is odd

$P(\text{sum is odd}) = P(\text{first odd}) \times P(\text{remaining even}) + P(\text{first even}) \times P(\text{remaining odd})$

$= \frac{5}{9} \times \frac{1}{2} + \frac{4}{9} \times \frac{1}{2}$

$= \frac{5}{18} + \frac{4}{18}$

$= \frac{9}{18}$

$= \frac{1}{2}$

$= 0.5$

Therefore, the probability is nearest to 0.5.