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# Raffle tickets numbered consecutively from 101 through 350

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Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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30 Jul 2012, 01:05
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Difficulty:

25% (medium)

Question Stats:

74% (01:15) correct 26% (01:31) wrong based on 2597 sessions

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The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

Raffle tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random will have a number with a hundreds digit of 2 ?

(A) 2/5
(B) 2/7
(C) 33/83
(D) 99/250
(E) 100/249

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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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30 Jul 2012, 01:06
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SOLUTION

Raffle tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random will have a number with a hundreds digit of 2 ?

(A) 2/5
(B) 2/7
(C) 33/83
(D) 99/250
(E) 100/249

The number of integers from 101 to 350, inclusive is 250, out of which 100 (from 200 to 299) will have a hundreds digit of 2. Thus the probability is 100/250=2/5.

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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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30 Jul 2012, 01:47
1
numbers with hundreds digit 2 are 200-299
100 such numbers

total numbers are 101-350
250 such numbers

probability would be $$\frac{100}{250}$$
$$\frac{2}{5}$$
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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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15 Mar 2015, 07:45
Bunuel, why do we think that the question means 'inclusive' 101 and 350?
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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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15 Mar 2015, 07:50
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Ergenekon wrote:
Bunuel, why do we think that the question means 'inclusive' 101 and 350?

"Raffle tickets numbered consecutively from 101 through 350", so both 101 and 350 are inclusive.
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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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15 Mar 2015, 07:53
Then if it didn't mean inclusive, how else it could present the problem? Could you answer, pls?
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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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15 Mar 2015, 07:57
Ergenekon wrote:
Then if it didn't mean inclusive, how else it could present the problem? Could you answer, pls?

Tickets numbered consecutively from 101, for me, naturally means including 101. There are other ways to write that if 101 were not included, for example "tickets numbered consecutively from 102".
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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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15 Mar 2015, 08:00
1
Thank you very much Bunuel for your help.
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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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15 Mar 2015, 08:15
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Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

Raffle tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random will have a number with a hundreds digit of 2 ?

(A) 2/5
(B) 2/7
(C) 33/83
(D) 99/250
(E) 100/249

Practice Questions
Question: 10
Page: 153
Difficulty: 600

GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide for GMAT® Review, 13th Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
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Thank you!

Total numbers = 350 -101 +1 = 250
favorable cases = 299-200 +1 = 100
Probability = 100/250 = 2/5

option A.
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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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03 May 2016, 10:29
2
1
Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

Raffle tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random will have a number with a hundreds digit of 2 ?

(A) 2/5
(B) 2/7
(C) 33/83
(D) 99/250
(E) 100/249

The probability of an event is equal to: favorable outcomes/total outcomes

In this particular problem we have:

favorable outcomes = numbers with a hundreds digit of 2

total outcomes = consecutive integers from 101 through 350

Let’s start with the total outcomes. Although the word “inclusive” is not actually used, it is implied because we are told the raffle tickets start at 101 and end at 350. Thus, the number of tickets from 101 to 350, inclusive, is 350 – 101 + 1 = 250. (Note that we had to add 1 because we counted BOTH tickets 101 and 350.)

The favorable outcomes are the number of tickets with a hundreds digit of 2. Since all the numbers from 200 to 299 are included, there are 299 – 200 +1 = 100 numbers with a hundreds digit of 2.

Therefore, the probability that a randomly selected ticket with have a hundreds digit of 2 is 100/250 = 2/5.

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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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25 Oct 2017, 05:25
1
Can someone please explain to me when we have to add 1?

If it includes both 101 and 350, we have to + 1.

When it does not include neither 101 nor 350, we do not add.

What hapens when we include just 101, or just 350?

Thank you!
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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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02 Nov 2017, 08:52
1
GGrunthal wrote:
Can someone please explain to me when we have to add 1?

If it includes both 101 and 350, we have to + 1.

When it does not include neither 101 nor 350, we do not add.

What hapens when we include just 101, or just 350?

Thank you!

(i) Lets say you have consecutive numbers from a to b, then the total numbers would be b - a + 1.

(ii) But if we are asked for consecutive numbers between a & b (in this case a and b are not included), then we would have the total number as b - a - 1

However, if we are asked for consecutive numbers between a & b inclusive then we use the same as (i), which is b - a + 1.

Hope you got it
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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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11 Jan 2018, 02:32
1
Please explain to me how the favorable cases inclusive are from 200 - 299 . Most especially the hundred digit of 2.
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Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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12 Jan 2018, 12:51
Bunuel wrote:
Raffle tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random will have a number with a hundreds digit of 2 ?

(A) 2/5
(B) 2/7
(C) 33/83
(D) 99/250
(E) 100/249

Richlove wrote:
Please explain to me how the favorable cases inclusive are from 200 - 299 . Most especially the hundred digit of 2.

Richlove , I am not quite sure what you are asking.
Please be more specific next time?
I think your question involves inclusive counting.*

Favorable cases
Any integer of the form 2 _ _ is a "favorable case"

How many of those 2 _ _ integers / terms are there?
• The first term in form "2 _ _" is 200
• The last term in the form "2 _ _" is 299
• Number of favorable cases? All the numbers from 200 to 299
• To find the number of favorable cases, use inclusive counting formula:
(Last term - first term) PLUS ONE

Subtraction only, (Last - First), yields the difference between integers
Subtraction does not yield the number of integers.
Take a small sample to see why

Small sample
How many integers are there from 2 to 5?
2, 3, 4, 5: FOUR integers
But if we subtract? 5 - 2 = 3. Not correct
Add 1 to (5 - 2) = 3. Then (3 + 1) = FOUR integers

Counting: probability
For the problem, we need $$\frac{FavorableCases}{PossibleOutcomes}$$

Total possible outcomes?
All integers from 101 to 350
(350 - 101) = 249 + 1 = 250 all possible outcomes
(The box contains ALL the numbers: the 100s group, the 200s group, and the 300s group)

Favorable cases?
All the integers in the group with form 2 _ _
First term is 200, last term is 299
All integers in the 200s group?
(299 - 200) = 99 + 1 = 100 integers from 200 to 299

Probability?
Probability that you will pick a number from the 200s group?

$$\frac{Favorable}{Possible} = \frac{200s}{AllTickets'Numbers} = \frac{100}{250} = \frac{2}{5}$$

*Other than "numbers that have hundreds digit of 2," what could be the "favorable cases" here? If you are asking why integers with 2 in the hundreds place are "favorable"? Because the prompt defines "numbers with a hundreds digit of 2" as "success."
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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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04 Feb 2018, 11:29
Hi All,

This is an example of a 'fence post' problem (meaning that you have to remember to count the tickets at the 'beginning' and 'end' of each sub-list.

We're asked for the probability of selecting a ticket with a "2" in the hundreds digit from a group of tickets numbered 101 through 350, inclusive.

The number of tickets is 350 - 101 + 1 = 250 total tickets
The number that have a 2 in the hundreds spot = 100 (200 through 299, inclusive).

So the probability is 100/250 = 2/5

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Re: Raffle tickets numbered consecutively from 101 through 350  [#permalink]

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22 Apr 2018, 02:01
The question asks for hundreds digit of 2. Range of possibilities are from 101 through350. This includes all the integers in that range.

Hence, that this is a probability question: thus

Number of favorable outcomes
---------------------------------------
Number of total outcomes

Favorable outcome are the ones with a hundreds digit of 2. So 200-299. These are 100 numbers, as we have to include 200 and 299.
Number of total outcomes we can find by using the inclusive counting formula: last-first+1 = 350-101+1 = 250.

Let's set up the equation:

100/250
10/25
2/5

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Re: Raffle tickets numbered consecutively from 101 through 350 &nbs [#permalink] 22 Apr 2018, 02:01
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