Answer:
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
Answer: 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
Answer: 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
Answer: Option E
