Raffle tickets numbered consecutively from 101 through 350 are placed

Question

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

Bunuel · Accepted Answer

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.

Answer: A.

ScottTargetTestPrep · Answer

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.

Answer A.

navkaran · Answer

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}

Ergenekon · Answer

Bunuel, why do we think that the question means 'inclusive' 101 and 350?

Bunuel · Answer

"Raffle tickets numbered consecutively   from  101  through  350 ", so both 101 and 350 are inclusive.

Ergenekon · Answer

Then if it didn't mean inclusive, how else it could present the problem? Could you answer, pls?

Bunuel · Answer

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".

Ergenekon · Answer

Thank you very much Bunuel for your help.

Lucky2783 · Answer

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

option A.

GGrunthal · Answer

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!

GMATSkilled · Answer

(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

Richlove · Answer

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

generis · Answer

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}

Answer C. Hope that helps.

*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."

EMPOWERgmatRichC · Answer

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

Final Answer:  A

GMAT assassins aren't born, they're made, 
Rich

CLIMBTHELADDER · Answer

The question asks for hundreds digit of 2. Range of possibilities are  from  101  through 350. 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

Answer choice A

HouseStark · Answer

Total no.= 350-101+1=250

total no. in which hundreds digit of 2= 299-200+1=100

probability= \frac{100}{250}=\frac{2}{5}

Answer A

CEdward · Answer

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

Denominator: 
# of integers: (350 - 101) + 1 = 250

Numerator: 
# of integers starting with 2: (299 - 200) + 1 = 100

P(starting with 2) = 100 / 250 = 2/5

A.

ArnauG · Answer

To solve this problem, we need to determine the number of tickets with a hundreds digit of 2 and divide it by the total number of tickets in the box.

The range of ticket numbers is from 101 to 350, inclusive. To find the number of tickets with a hundreds digit of 2, we need to count the numbers between 200 and 299.

The number of tickets with a hundreds digit of 2 is 100, as there are 100 numbers between 200 and 299 (inclusive).

The total number of tickets is the difference between the highest and lowest ticket numbers, plus 1:

Total number of tickets = 350 - 101 + 1 = 250.

Therefore, the probability is given by the number of tickets with a hundreds digit of 2 divided by the total number of tickets:

Probability = (number of tickets with a hundreds digit of 2) / (total number of tickets) = 100 / 250 = 2 / 5.

Among the given answer choices, the fraction 2/5 is represented by option (A).

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Raffle tickets numbered consecutively from 101 through 350 are placed

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