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04 Mar 2021, 09:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
79% (02:22) correct
21%
(02:25)
wrong
based on 134
sessions
History
Date
Time
Result
Not Attempted Yet
An urn contains 3 white balls, 4 red balls and 5 black balls. Three balls are drawn at random and without replacement. What is the probability that they are the same color?
A. 41/44
B. 35/44
C. 3/11
D. 3/44
E. 1/22
A. 41/44
B. 35/44
C. 3/11
D. 3/44
E. 1/22
rvgmat12
Joined: 19 Oct 2014
Last visit: 27 Mar 2026
Posts: 352
Given Kudos: 189
Location: United Arab Emirates
Schools: Wharton '23 Booth '23 INSEAD Jan'21 Intake
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04 Mar 2021, 09:16
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3w + 3r + 3b
3/12*2/11*1/10 + 4/12*3/11*2/10 + 5/12*4/11*3/10 = 90/(12*11*10) = 3/44
Posted from my mobile device
3/12*2/11*1/10 + 4/12*3/11*2/10 + 5/12*4/11*3/10 = 90/(12*11*10) = 3/44
Posted from my mobile device
04 Mar 2021, 09:36
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Hello,
I'm not sure I found the correct answer but here's my reasoning:
1st draw:
- P(W) = 3/12
- P(R) = 4/12
- P(B) = 5/12
2nd draw:
- P(W) = 2/11
- P(R) = 3/11
- P(B) = 4/11
3rd draw:
-P(W) = 1/10
-P(R) = 2/10
-P(B) = 3/10
So the probability to have 3 of the same color is the sum of the probability of drawing 3 W, 3 R or 3 B
P(3 equal color) = [(3/12)*(2/11)*(1/10)]+[(4/12)*(3/11)*(2/10)]+[(5/12)*(4/11)*(3/10)] = [1/220 + 1/55 + 1/22] = 15/220 = 3/44 (D)
Thank you,
Waiting for the OA
I'm not sure I found the correct answer but here's my reasoning:
1st draw:
- P(W) = 3/12
- P(R) = 4/12
- P(B) = 5/12
2nd draw:
- P(W) = 2/11
- P(R) = 3/11
- P(B) = 4/11
3rd draw:
-P(W) = 1/10
-P(R) = 2/10
-P(B) = 3/10
So the probability to have 3 of the same color is the sum of the probability of drawing 3 W, 3 R or 3 B
P(3 equal color) = [(3/12)*(2/11)*(1/10)]+[(4/12)*(3/11)*(2/10)]+[(5/12)*(4/11)*(3/10)] = [1/220 + 1/55 + 1/22] = 15/220 = 3/44 (D)
Thank you,
Waiting for the OA
