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26 May 2025, 00:30
Kudos
Bookmarks
C
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:
72% (03:16) correct
28%
(03:16)
wrong
based on 114
sessions
History
Date
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Not Attempted Yet
A gumball machine contains 21 gumballs, of which 8 are red, 7 are blue, and 6 are green. If a child buys 4 gumballs from the machine, what is the probability that the gumballs will all be of the same color?
A. 2/399
B. 4/399
C. 8/399
D. 11/399
E. 13/399
Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
A. 2/399
B. 4/399
C. 8/399
D. 11/399
E. 13/399
Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
26 May 2025, 01:05
Kudos
Bookmarks
P(all gumballs of the same color) = P(all red) + P(all blue) + P(all green)
Total = 21C4
P(all red) = 8C4/21C4
P(all blue) = 7C4/21C4
P(all green) = 6C4/21C4
P(all gumballs of the same color) = 8C4/21C4 + 7C4/21C4 + 6C4/21C4
P(all gumballs of the same color) = (8C4 + 7C4 + 6C4)/21C4
P(all gumballs of the same color) = 8/399
Total = 21C4
P(all red) = 8C4/21C4
P(all blue) = 7C4/21C4
P(all green) = 6C4/21C4
P(all gumballs of the same color) = 8C4/21C4 + 7C4/21C4 + 6C4/21C4
P(all gumballs of the same color) = (8C4 + 7C4 + 6C4)/21C4
P(all gumballs of the same color) = 8/399
Bunuel
