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08 Jul 2015, 03:10
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (01:47) correct
34%
(02:10)
wrong
based on 911
sessions
History
Date
Time
Result
Not Attempted Yet
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?
A. 35/1024
B. 1/24
C. 1/8
D. 105/512
E. 1/4
Kudos for a correct solution.
A. 35/1024
B. 1/24
C. 1/8
D. 105/512
E. 1/4
Kudos for a correct solution.
08 Jul 2015, 03:38
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Bunuel
Total balls in the Machine = 7+5+4 = 16
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
General Discussion
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,319
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Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
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Updated on: 13 Jul 2015, 12:31
Originally posted by ENGRTOMBA2018 on 08 Jul 2015, 04:17.
Last edited by ENGRTOMBA2018 on 13 Jul 2015, 12:31, edited 1 time in total.
Last edited by ENGRTOMBA2018 on 13 Jul 2015, 12:31, edited 1 time in total.
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Bunuel
Probability of selecting a red ball = P(R) = 4/16
Probability of selecting a green ball = P(G) = 5/15
Probability of selecting a blue ball = P(B) = 7/14
Total number of ways we can select the RBG combination =3!
Thus the total probability is = (4/16)*(5/15)*(7/14)*3! = 1/4. Thus E is the correct answer.
