There are 9 red gum balls, 5 green gum balls, 8 yellow gum balls, and 8 blue gum balls in a machine. Find P(green, then yellow)
The answer is 40/870.
P(Green) = 5/30 = 1/6
P(yellow after green) = 8/29 (as one has been taken away)
Probability of both of these happening - so green and yellow afterwards is multiply the two probabilities together together
(1/6)*(8/29) = 8/174
Which is the same as Tia said if simplified
Prob(green, then yellow)
= (5/30)(8/29)
= 4/87
Oh yeah, I forgot I could simplify too. Thanks Reiny!
To find the probability of drawing a green gum ball, then a yellow gum ball, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, let's calculate the total number of gum balls in the machine:
Total gum balls = 9 (red) + 5 (green) + 8 (yellow) + 8 (blue) = 30 gum balls
Now, let's calculate the number of favorable outcomes for drawing a green gum ball, then a yellow gum ball:
Number of green gum balls = 5
Number of yellow gum balls = 8
To calculate the probability, we divide the number of favorable outcomes by the total number of possible outcomes:
P(green, then yellow) = (Number of green gum balls / Total gum balls) * (Number of yellow gum balls / Total gum balls)
P(green, then yellow) = (5/30) * (8/30) = 40/900 = 4/90
Simplifying further, P(green, then yellow) = 2/45
Therefore, the probability of drawing a green gum ball, then a yellow gum ball is 2/45.