Abag contains only 3 blue disks, 3 green disks, and 4 orange disks. if 3 disks are selected at random from the bag, what is the probability that 1 of the disks will be green and 2 of the disks will be orangea) 1/30
b) 1/20
c) 3/40
d) 1/10
e) 3/20

E= 3/20

Step-by-step explanation:

Number of blue disk= 3

Number of green disk= 3

Number of orange disk= 4

Total number of disk = 3+3+34

= 10

Let B represent blue disk

Let G represent green disk

Let O represent orange disk

If three disk are selected at random, the possible outcome of one green disk and two orange disks are

OOG, OGO, GOO

Pr(OOG) = Pr(O1) * Pr(O2) * Pr(G3)

= 4/10 * 3/9 * 3/8

= 36/720

= 1/20

Pr(OGO) = Pr(O1) * Pr(G2) * Pr(O3)

= 4/10 * 3/9 * 3/8

= 36/720

= 1/20

Pr(GOOG) = Pr(G1) * Pr(O2) * Pr(O3)

= 4/10 * 4/9 * 3/8

= 36/720

= 1/20

Pr(total) = Pr(OOG) + Pr(OGO) + Pr(GOO)

= 1/20 + 1/20 + 1/20

= 3/20

the answer is vanilla

step-by-step explanation:

can someone answer this question plz

step-by-step explanation: