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

$$\begin{gathered} \mathrm{Now}, \\ P\left(A\right)=\frac{10}{20}=\frac{1}{2} \\ P\left(B/A\right)=\frac{10}{19} \\ \therefore \mathrm{Required}\mathrm{probability}=P\left(A\cap B\right)=P\left(A\right)\times P\left(B/A\right)=\frac{1}{2}\times \frac{10}{19}=\frac{5}{19} \end{gathered}$$