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Bunuel
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If a bowl contains numbered tiles labeled with integers from 1 – 100, 31 Jan 2017, 09:58
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C
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35% (medium)

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67% (01:12) correct 33% (01:03) wrong based on 251 sessions

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If a bowl contains numbered tiles labeled with integers from 1 – 100, inclusive, what is the probability of selecting a tile with a three on it?

A. 1/5
B. 19/101
C. 19/100
D. 1/10
E. 1/100
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C
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If a bowl contains numbered tiles labeled with integers from 1 – 100, 31 Jan 2017, 10:59
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Bunuel
If a bowl contains numbered tiles labeled with integers from 1 – 100, inclusive, what is the probability of selecting a tile with a three on it?

A. 1/5
B. 19/101
C. 19/100
D. 1/10
E. 1/100
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Total # of tiles = 100

1 - 9 - one (3)
10 - 19 - one (13)
20 - 29 - 1
30 - 39 - all 10
40 - 49 - 1
...
90 - 100 - 1

All clusters of 10 (with exception of 30's) has one number with 3 (3 or X3). In the case of 30's we have all 10 numbers with 3 (9 with one 3 and one with two 3's - 33).

Our probability = (9 + 10/ 100 = 19/100

Answer C.
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If a bowl contains numbered tiles labeled with integers from 1 – 100, Updated on: 19 Feb 2019, 07:05
Originally posted by energetics on 18 Feb 2019, 11:10.
Last edited by energetics on 19 Feb 2019, 07:05, edited 1 time in total.
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There's 2 Cases for getting a 3:
1) 3 in units digit {03, 13, 23, 33, 43 ... 93} = 10 ways
2) 3 in tens digit: {31, 32, 33, 34 ... 39} 9 ways, remove 33 as it repeats

10+9 = 19/100
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If a bowl contains numbered tiles labeled with integers from 1 – 100, 18 Feb 2019, 19:24
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Solution



Given:
    • A bowl contains numbered tiles labelled with integers from 1 to 100, inclusive.

To find:
    • The probability of selecting a tile with a three on it

Approach and Working:
In the range between 1 to 100, inclusive, we have either 1-digit numbers or 2-digit numbers of 3-digit number.

Considering all the 1-digit numbers,
    • Number of numbers with the digit 3 = 1

Considering all the 2-digit numbers,
    • Number of numbers with digit 3 only in the units place = 8 (13, 23, 43, 53, 63, 73, 83, 93)
    • Number of numbers with digit 3 only in the tens place = 9 (30, 31, 32, 34, 35, 36, 37, 38, 39)
    • Number of numbers with digit 3 in both units and tens place = 1 (33)

Considering all the 3-digit numbers, there is no such number in the given range with the digit 3.

So, total number of instances = 1 + 8 + 9 + 1 = 19
Therefore, the required probability = \(\frac{19}{100}\)

Hence, the correct answer is option C.

Answer: C

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If a bowl contains numbered tiles labeled with integers from 1 – 100, 05 Jan 2026, 22:57
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Total number of outcomes = 100
Count number of times 3 appears between 1 to 100 = 3,13,23,43,53,63,73,83,93,30,31,32,33,34,35,36,37,38,39
=19
Therefore Prob = 19/100
Answer C
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