Question

You have a bag of 12 marbles. Four are blue, 6 are green, 2 are red, and the others are yellow. What is the probability of drawing a red marble and, without replacing it, then drawing a green marble?

Answer 1

`Probability = (1)/(11)`

Explanation:

Given

`Blue = 4`

`Green = 6`

`Red = 2`

`Total = 12`

Required

Determine the probability of drawing red and green without replacement

This is calculated as follows:

`Probability = P(Red\ and\ Green)`

`Probability = P(Red)\ and\ P(Green)`

`Probability = (n(Red))/(Total) * (n(Green))/(Total - 1)`

We used Total - 1 because it's a probability without replacement:

`Probability = (2)/(12) * (6)/(12- 1)`

`Probability = (2)/(12) * (6)/(11)`

`Probability = (1)/(6) * (6)/(11)`

`Probability = (1)/(11)`

Hence, the required probability is 1/11