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If a number is selected from 100 to 199. What is the probability that
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21 Aug 2023, 22:49
21 Aug 2023, 22:49
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If a number is selected from 100 to 199. What is the probability that the number will contain either 2 or 3 as at least one of the digits?
A. 34/100
B. 34/99
C. 35/100
D. 35/99
E. 36/100
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A. 34/100
B. 34/99
C. 35/100
D. 35/99
E. 36/100
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: If a number is selected from 100 to 199. What is the probability that
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11 Sep 2023, 05:57
11 Sep 2023, 05:57
2
For every group of 10digit, there is one 2. Total of 10group = 10.
For every group of 10 digits, there is one 3, total of 10group = 10.
For number beginning with twenties, there are 10 but to avoid double counting we remove 1 = 9.
For numbers beginning with 3, there are also 9
Numbers that have 2 and 3 are 32 and 23.
P(2) + P(3) - P(2 and 3)
19 + 19 - 2 = 36
Posted from my mobile device
For every group of 10 digits, there is one 3, total of 10group = 10.
For number beginning with twenties, there are 10 but to avoid double counting we remove 1 = 9.
For numbers beginning with 3, there are also 9
Numbers that have 2 and 3 are 32 and 23.
P(2) + P(3) - P(2 and 3)
19 + 19 - 2 = 36
Posted from my mobile device
Re: If a number is selected from 100 to 199. What is the probability that
[#permalink]
11 Sep 2023, 05:57
11 Sep 2023, 05:57
Joshuautazi2021 wrote:
For every group of 10digit, there is one 2. Total of 10group = 10.
For every group of 10 digits, there is one 3, total of 10group = 10.
For number beginning with twenties, there are 10 but to avoid double counting we remove 1 = 9.
For numbers beginning with 3, there are also 9
Numbers that have 2 and 3 are 32 and 23.
P(2) + P(3) - P(2 and 3)
19 + 19 - 2 = 36
Posted from my mobile device
For every group of 10 digits, there is one 3, total of 10group = 10.
For number beginning with twenties, there are 10 but to avoid double counting we remove 1 = 9.
For numbers beginning with 3, there are also 9
Numbers that have 2 and 3 are 32 and 23.
P(2) + P(3) - P(2 and 3)
19 + 19 - 2 = 36
Posted from my mobile device
I will appreciate 🙏 a faster way to tackle this problem.
