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13 Jan 2022, 09:30
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
78% (02:32) correct
22%
(02:19)
wrong
based on 105
sessions
History
Date
Time
Result
Not Attempted Yet
There are 8 marbles in a jar, 5 are red and 3 are green. If 4 marbles are selected at random form the jar, what is the probability that at least one red marble and one green marble will be selected?
(A) 4/5
(B) 10/11
(C) 11/12
(D) 12/13
(E) 13/14
(A) 4/5
(B) 10/11
(C) 11/12
(D) 12/13
(E) 13/14
22 Jan 2022, 06:58
Kudos
Bookmarks
Bunuel
From all cases if we subtract ALL RED cases and ALL GREEN cases we will get the required #of cases
SELECTING ALL RED CASES : \(5C4\)
TOTAL CASES : \(8C4\)
Probability \(= \frac{5C4}{8C4}\) = \(= \frac{5}{70}\)
SELECTING ALL GREEN CASES : \(0,\) Since there are only \(3 \) green marbles we cannot select \(4\) green marbles
Hence \(1 - ( \frac{5}{70} + 0 )= \frac{13}{14}\)
Ans E.
16 Jul 2025, 05:50
Kudos
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