A date is picked at random on the calendar for April.
\begin{tabular}{||c|c|c|c|c|c|c|}
\hline \multicolumn{8}{|c|}{ April } \\
\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
8 & 9 & 10 & 11 & 12 & 13 & 14 \\
15 & 16 & 17 & 18 & 19 & 20 & 21 \\
22 & 23 & 24 & 25 & 26 & 27 & 28 \\
\hline 29 & 30 & & & & & \\
\hline
\end{tabular}

What is the probability that the date chosen will be a two-digit odd number?
$\frac{1}{3}$
$70 \%$
$\frac{1}{2}$
$10 \%$

Step 1 ：To solve this problem, we need to determine the total number of two-digit odd numbers in April and divide that by the total number of days in April.

Step 2 ：April has 30 days, so that is the denominator of the probability fraction.

Step 3 ：For the numerator, we count the two-digit odd numbers, which are 11, 13, 15, 17, 19, 21, 23, 25, 27, and 29.

Step 4 ：That's a total of 10 two-digit odd numbers.

Step 5 ：The probability is calculated as the number of two-digit odd numbers divided by the total number of days in April.

Step 6 ：The calculation is \( \frac{10}{30} \)

Step 7 ：Simplifying the fraction, we get \( \frac{1}{3} \)

Step 8 ：Final Answer: \(\boxed{\frac{1}{3}}\)