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GRE Whiz Representative
Joined: 25 May 2022
Posts: 25
Given Kudos: 1
Set S consists of all the two-digit numbers that do not have 0 in them
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Updated on: 06 Sep 2022, 04:04
Updated on: 06 Sep 2022, 04:04
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Expert Reply
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Question Stats:
93% (02:48) correct
6% (01:39) wrong based on 16 sessions
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Set S consists of all the two-digit numbers that do not have 0 in them. If a number is picked at random from the set S, what is the probability that its unit digit is greater than the tens digit?
A. \(\frac{1}{9}\)
B. \(\frac{4}{9}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{9}\)
E. \(\frac{8}{9}\)
Source: GREWhiz
A. \(\frac{1}{9}\)
B. \(\frac{4}{9}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{9}\)
E. \(\frac{8}{9}\)
Source: GREWhiz
Originally posted by SaquibHGREWhiz on 06 Sep 2022, 02:39.
Last edited by SaquibHGREWhiz on 06 Sep 2022, 04:04, edited 1 time in total.
Last edited by SaquibHGREWhiz on 06 Sep 2022, 04:04, edited 1 time in total.
GRE Whiz Representative
Joined: 25 May 2022
Posts: 25
Given Kudos: 1
Set S consists of all the two-digit numbers that do not have 0 in them
[#permalink]
Updated on: 13 Sep 2022, 17:26
Updated on: 13 Sep 2022, 17:26
2
Expert Reply
Solution:
Thus, the required probability \(=\frac{36}{81}=\frac{4}{9}\)
Hence the right answer is Option B
- Total number of 2-digit numbers \(=99-10+1=90\)
- Number of 2-digit numbers with 0 in them = 10, 20, 30...90 \(= 9\)
- Therefore, total number of 2-digit numbers that do not have 0 in them \(=90-9=81\)
- An alternate way to count this would be by multiplying \(9 \times 9\)
- From 11 to 19
- Number of digit with unit digit greater than the tens digit \(=8\) (12, 13, 14...19)
- From 21 to 29
- Number of digit with unit digit greater than the tens digit \(=7\) (23, 24, 25...29)
- .
- .
- .
- From 81 to 89
- Number of digit with unit digit greater than the tens digit \(=1\) (89)
- From 91 to 99
- Number of digit with unit digit greater than the tens digit \(=0\)
- Therefore, total number of 2-digit numbers with unit digit greater than the tens digit \(=8+7+6+5+4+3+2+1=36\)
Thus, the required probability \(=\frac{36}{81}=\frac{4}{9}\)
Hence the right answer is Option B
Originally posted by SaquibHGREWhiz on 06 Sep 2022, 02:40.
Last edited by SaquibHGREWhiz on 13 Sep 2022, 17:26, edited 2 times in total.
Last edited by SaquibHGREWhiz on 13 Sep 2022, 17:26, edited 2 times in total.
Re: Set S consists of all the two-digit numbers that do not have 0 in them
[#permalink]
06 Sep 2022, 09:21
06 Sep 2022, 09:21
2
Set of Integers with no zeros = 9*9 = 81
Total Number of digits with ones place greater than tens place -->
For 11-19 there will be 8 such numbers
For 21-29 there will be 7 such numbers
…
So on
…
Up to 1 such number from 81-89
Hence the total will be = 36 {8+7+6+5+4+3+2+1}
The total probability would then be = Number of Favourable Digits / Total
i.e. =39/81 = 4/9
Total Number of digits with ones place greater than tens place -->
For 11-19 there will be 8 such numbers
For 21-29 there will be 7 such numbers
…
So on
…
Up to 1 such number from 81-89
Hence the total will be = 36 {8+7+6+5+4+3+2+1}
The total probability would then be = Number of Favourable Digits / Total
i.e. =39/81 = 4/9
