HarveyKlaus wrote:
A bag contains 3 blue disks, 3 green disks and 4 orange disks. If three disks are selected random from the bag, what is the probability that 1 of the disks will be green and the other 2 of the disks will be orange?
A) 1/30
B) 1/20
C) 3/40
D) 1/10
E) 3/20
We are given that in a bag there are 3 blue disks, 3 green disks, and 4 orange disks. We need to determine the probability that from 3 selected disks, 1 will be green and 2 will be orange.
Let’s first determine the total number of ways to select 1 green disk from 3.
3C1 = 3
Next, let’s determine the number of ways to select 2 orange disks from 4.
4C2 = (4 x 3)/2! = (4 x 3)/(2 x 1) = 6 ways
So, there are 3 x 6 = 18 ways to select 1 green and 2 orange disks.
Now let’s determine the number of ways to select 3 disks from 10:
10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2) = 5 x 3 x 8 = 120
Thus, the probability of selecting 1 green disk and 2 orange disks is 18/120 = 3/20.
Alternate Solution:
Selecting 3 disks at once is equivalent to selecting them one at a time without replacement. Remember, we are determining the probability of selecting 1 green disk and 2 orange disks from 10 total disks.
If we select the green disk first, since there are 3 green disks and 10 total disks, there is a 3/10 chance that the green disk will be selected. Next, since there are 4 orange disks and 9 disks left, there is a 4/9 chance an orange disk will be selected. Similarly, for the third disk chosen, there is a 3/8 chance another orange disk will be selected. However, there are 3 different ways to select the 1 green disk and 2 orange disks:
G - O - O
O - G - O
O - O - G
Each of these 3 ways has the same probability of occurring. Thus, the total probability is:
3 x (3/10 x 4/9 x 3/8) = 3/20
Answer: E