Question

A bag contains 20 tickets, numbered from 1 to 20. Two tickets are drawn without replacement. What is the probability that the first ticket has an even number and the second an odd number.

Answer 1

There are 10 even numbers and 10 odd numbers between 1 to 20.

Consider the given events.
A = An even number in the first draw
B = An odd number in the second draw

\[\text{ Now }, \]
\[P\left( A \right) = \frac{10}{20} = \frac{1}{2}\]
\[P\left( B/A \right) = \frac{10}{19}\]
\[ \therefore \text{ Required probability } = P\left( A \cap B \right) = P\left( A \right) \times P\left( B/A \right) = \frac{1}{2} \times \frac{10}{19} = \frac{5}{19}\]