Question

A jar contains 26 marbles. It has 10 red, 8 black, and 8 green marbles. Two marbles are drawn; the first is not returned before the second one is drawn. What is the probability that both marbles are green?
P(Both green) = 13/325
P(Both green) = 28/325
P(Both green) = 32/325
P(Both green) = 14/169

Answer 1

a) what it he probability the first one is green?

Answer 2

This is "sampling without replacement."
How many (total) marbles are in the jar?
Of these, how many are green?
What is the ratio of (number of green marbles) to (total number of marbles)?

We're not going to replace that green marble.
How many (total) marbles are now in the jar?
Of those, how many are green?
Write the same ratio (as before, using the new, different total number of marbles and the new, different count of green marbles)

Any conclusions?

What do you need to do next to finish this problem solution?

Answer 3

i know there is a formula

Answer 4

but i dont understand what i need to plug in to the formula

Answer 5

P(A|B)= P(A ∩ B) / P(B)

Answer 6

i need help with pluggin in the numbers and solving

Answer 7

If I remember correctly it's something along the lines of the following -
\[\frac{ 8 }{ 26 } \times \frac{ 7 }{ 25 }\]

Answer 8

This is because you have an 8 out of 26 chance to draw a green marble. Since you're not putting the marble back, you have one less green marble and one less marble in total.