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15 Feb 2023, 21:41
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B
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:
77% (02:03) correct
23%
(02:00)
wrong
based on 189
sessions
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A set has ten positive integers six even and four odd. Two integers are picked randomly and added. What is the probability that the sum is even?
A. \(\frac{1}{2}\)
B. \(\frac{7}{15}\)
C. \(\frac{8}{15}\)
d. \(\frac{2}{3}\)
E. \(\frac{3}{5}\)
A. \(\frac{1}{2}\)
B. \(\frac{7}{15}\)
C. \(\frac{8}{15}\)
d. \(\frac{2}{3}\)
E. \(\frac{3}{5}\)
15 Feb 2023, 23:52
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TBT
A set has ten positive integers six even and four odd. Two integers are picked randomly and added. What is the probability that the sum is even?
a. \(\frac{1}{2}\)
b. \(\frac{7}{15}\)
c. \(\frac{8}{15}\)
d. \(\frac{2}{3}\)
e. \(\frac{3}{5}\)
a. \(\frac{1}{2}\)
b. \(\frac{7}{15}\)
c. \(\frac{8}{15}\)
d. \(\frac{2}{3}\)
e. \(\frac{3}{5}\)
For the sum to be even, either both the terms chosen must be even or both the terms chosen must be odd.
Probability that both the terms are even = \(\frac{^6C_2 }{ ^{10}C_2}\)
Probability that both the terms are odd= \(\frac{^4C_2 }{ ^{10}C_2}\)
Net Proabiblity = \(\frac{^6C_2 }{ ^{10}C_2}\) + \(\frac{^4C_2 }{ ^{10}C_2}\)
Solving we get \(\frac{7}{15}\)
Option B
Iserhard
Joined: 07 Mar 2021
Last visit: 01 Jul 2025
Posts: 8
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Location: Brazil
Schools: IESE '27 (A) Judge '26 (A) IMD'26 (II)
GMAT Focus 1: 605 Q79 V82 DI79 
GPA: 8.4
27 Dec 2024, 11:07
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Hello, one question
Can we do 1 - Prob (EO)?
Idk what Im doing wrong here.
1 - (4/10 * 6/9) =
(1 - 4/15) = 3/5
Cheers
Can we do 1 - Prob (EO)?
Idk what Im doing wrong here.
1 - (4/10 * 6/9) =
(1 - 4/15) = 3/5
Cheers
