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There are 20 balls in an urn, of which x are red balls and rest are bl
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21 Aug 2023, 03:38
21 Aug 2023, 03:38
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Question Stats:
70% (01:17) correct
30% (02:55) wrong based on 10 sessions
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There are 20 balls in an urn, of which x are red balls and rest are black balls. If y red balls and z black balls are added to the urn. If a ball is selected at random, what is the probability that the selected ball is a black ball?
A. \(\dfrac{x+y}{20+y+z}\)
B. \(\dfrac{20-x+y}{20+y+z}\)
C. \(\dfrac{20-(x+z)}{20+y+z}\)
D. \(\dfrac{x+z}{20+y+z}\)
E. \(\dfrac{20-x+z}{20+y+z}\)
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A. \(\dfrac{x+y}{20+y+z}\)
B. \(\dfrac{20-x+y}{20+y+z}\)
C. \(\dfrac{20-(x+z)}{20+y+z}\)
D. \(\dfrac{x+z}{20+y+z}\)
E. \(\dfrac{20-x+z}{20+y+z}\)
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE Math Essentials - A most comprehensive handout!!
\(\Longrightarrow\) ALL the GRE Quant and Verbal Practice Directories in ONE place (2023)
\(\Longrightarrow\) Best GRE Test Books of 2023
Re: There are 20 balls in an urn, of which x are red balls and rest are bl
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11 Sep 2023, 04:59
11 Sep 2023, 04:59
1
Before the addition of y and z,
X + (20-X) = 20.
After additional y and z,
RED(X + Y) and Black(20-X + Y). The total = 20+Y+Z
20-X+Y/20+Y+Z
Posted from my mobile device
X + (20-X) = 20.
After additional y and z,
RED(X + Y) and Black(20-X + Y). The total = 20+Y+Z
20-X+Y/20+Y+Z
Posted from my mobile device
gmatclubot
Re: There are 20 balls in an urn, of which x are red balls and rest are bl [#permalink]
11 Sep 2023, 04:59
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