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23 Apr 2020, 10:17
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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:
5%
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Question Stats:
90% (01:30) correct
10%
(01:52)
wrong
based on 83
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A jar consists of 21 sweets. 12 are green and 9 are blue. William picked two sweets at random. What is the probability that one sweet is blue and one sweet is green.
A)18/36
B)11/23
C)12/33
D)14/27
E) 18/35
A)18/36
B)11/23
C)12/33
D)14/27
E) 18/35
firas92
Current Student
Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
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23 Apr 2020, 10:32
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The sweets can be picked in two ways - Blue, Green or Green, Blue
(12/21)(9/20)+(9/21)(12/20)
2(12/21)(9/20)=18/35
Answer is (E)
Posted from my mobile device
(12/21)(9/20)+(9/21)(12/20)
2(12/21)(9/20)=18/35
Answer is (E)
Posted from my mobile device
23 Apr 2020, 20:59
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Number of ways in which 2 sweets can be taken out of 21: 21C2= (21*20)/2
Number of favorable cases= (1 out of 12 green sweets)*(1 out of 9 blue sweets)
= 12*9
Probability= Favorable cases/ total cases = 18/35.
Number of favorable cases= (1 out of 12 green sweets)*(1 out of 9 blue sweets)
= 12*9
Probability= Favorable cases/ total cases = 18/35.
