Question

A bag contains 4 g r e e n marbles, 8 b l a c k marbles, and 7 b l u e marbles. If a marble is drawn from the bag, replaced, and another marble is drawn, what is the probability of drawing first a g r e e n marble and then a b l u e marble

Answer 1

Answer:

$P(Green\ and\ Blue) = 0.0776$

Step-by-step explanation:

Given

$Green = 4$

$Black = 8$

$Blue=7$

Required

Determine the probability of selecting green then blue marble

This probability is represented as:

$P(Green\ and\ Blue)$

And it is calculated using:

$P(Green\ and\ Blue) = P(Green) * P(Blue)$

$P(Green\ and\ Blue) = \frac{n(Green)}{Total} * \frac{n(Blue)}{Total}$

Total marble is:

$Total = 4 + 8 + 7$

$Total = 19$

So, we have:

$P(Green\ and\ Blue) = \frac{4}{19} * \frac{7}{19}$

$P(Green\ and\ Blue) = \frac{4*7}{19*19}$

$P(Green\ and\ Blue) = \frac{28}{361}$

$P(Green\ and\ Blue) = 0.0776$ -- approximated

Hence, the required probability is approximately 0.0776