A yellow bag contains 3 green marbles and 7 yellow marbles. Find the probability of drawing a green marble, keeping it, and drawing another green marble.

Question

Is my answer key wrong? It says 1/15. But I'm just not seeing how. 

Answer 1

The probability of drawing a green marble is 3/10

We keep that green marble, so now there are only 2 green marbles and 7 yellow marbles in the bag (a total of 9)

The probability of drawing another green marble is 2/9

 

We multiply these two probabilities together:

(3/10)*(2/9) = 6/90

=1/15

 

Answer 2

Hi Jaymie, the probability of drawing the first green marble is 3 / 10. The probability of drawing a second green marble is 2 / 9.

 

The probability of drawing two green marbles without replacement is:

 

p = 3 / 10 * 2 / 9

 

p = 6 / 90

 

Reduce to lowest terms:

 

p = 1 / 15

 

Questions?

Answer 3

Hi Jaymie,

 

The probability of first drawing a green marble is 3/10. Now, by keeping the green marble, only 9 total marbles remain (2 of which are now green). So the probability of then drawing another green marble is 2/9. To find the probability of this entire event happening, we multiply the two probabilities above to get:

 

(3/10)*(2/9)=(6/90), which we can reduce to (1/15)  since 6 goes into both the numerator and denominator evenly. Hope this helps!

 

 

-Sam