Question

There are 10 colored balls in a box ( 5 red, 3 blue, 2 green ). What is the probability of picking out a red ball and then a blue ball ( If I do not replace the red ball )?

Answer 1

At the beginning you have 10 balls, 5 of which are red. So, the chance of picking a red ball with the first pick is

$\dfrac{5}{10}=\dfrac{1}{2}$

When it comes to the second pick, you have 9 balls (because you didn't replace the first red ball), 3 of which are blue. So, you pick a blue ball with the second pick with probability

$\dfrac{3}{9}=\dfrac{1}{3}$

The two events will occur one after the other with the product of the probabilities:

$\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}$