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