A gumball machine contains three yellow, eight blue, and six pink gumballs. What is the probability that the next gumball that comes out will be neither yellow nor blue? show work pls tnx!

total balls in the Machine = 7+5+4 = 16

Step-by-step explanation:

METHOD-1

Probability of First ball to be Blue = 7/(7+5+4) = 7/16

Probability of Second ball to be Green = 5/(15) = 5/15 [15 balls remaining after one being taken out in Step 1]

Probability of Third ball to be red = 4/(14) = 4/14 [14 balls remaining after Two being taken out in Step 1&2]

But this can happen in any order so the orders in which it can happen = 3!

i.e. Final Probability that machine dispenses one gumball of each color = (7/16)*(5/15)*(4/14)*3! = 1/4

Option E

METHOD-2

The total ways in which three balls can be Picked out of a total of 16 balls (with orders) = 16*15*14 = 840

The ways in which three balls of different colors can be taken out (with orders) = 7*5*4*3! = 3360

Probability = Favourable Outcome / Total Outcomes

i.e Probability = 840 / 3360 = 1/4

Option E

METHOD-3

The total ways in which three balls can be Picked out of a total of 16 balls (without orders) = 16C3 = 560

The ways in which three balls of different colors can be taken out (with orders) = 7C1 * 5C1 *4C1 = 140

Probability = Favourable Outcome / Total Outcomes

i.e Probability = 140 / 560 = 1/4

Option E

d

step-by-step explanation:

i used the pitagoras theoreme