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 )?

1/6
1/2
3/20
4/5

(5/10)(3/9)=

(1/2)(1/3)= ?

4/5?

To find the probability of picking a red ball and then a blue ball, without replacement, you need to consider the total number of ways the balls can be picked and the number of favorable outcomes.

Step 1: Find the total number of ways to pick both balls. Since there are 10 balls in total, the first ball can be any of the 10 balls. Once a red ball is picked and not replaced, there are 9 balls left, and the second ball can be any of those 9 balls. So, the total number of ways to pick both balls is 10 * 9 = 90.

Step 2: Find the number of favorable outcomes. The first ball needs to be red, and there are 5 red balls, so there are 5 ways to pick the first ball. After picking a red ball, there are 3 blue balls remaining, so there are 3 ways to pick the second ball. Therefore, there are 5 * 3 = 15 favorable outcomes.

Step 3: Calculate the probability. The probability of an event is given by favorable outcomes divided by total outcomes. So, the probability of picking a red ball and then a blue ball is 15/90 which simplifies to 1/6.

Therefore, the correct answer is 1/6.