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

Answer:

$Probability = \frac{1}{11}$

Step-by-step 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:

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

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

$Probability = \frac{n(Red)}{Total} * \frac{n(Green)}{Total - 1}$

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

$Probability = \frac{2}{12} * \frac{6}{12- 1}$

$Probability = \frac{2}{12} * \frac{6}{11}$

$Probability = \frac{1}{6} * \frac{6}{11}$

$Probability = \frac{1}{11}$

Hence, the required probability is 1/11