Raffle tickets numbered consecutively from 101 through 350

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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

Answer: A

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by [email protected] » Mon Apr 25, 2016 3:49 pm
Hi boomgoesthegmat,

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

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by OptimusPrep » Mon Apr 25, 2016 8:26 pm
boomgoesthegmat 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

Answer: A
This is a straight forward probability question.

Always remember, the total numbers between a and b = b - a + 1

Total number of tickets = 350 - 101 + 1 = 250
Tickets with 2 in the hundreds digit = 299 - 200 + 1 = 100

Probability of picking a ticket with 2 in the hundredth digit = 100/250 = 2/5

Correct Option: A

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by Brent@GMATPrepNow » Tue Apr 26, 2016 9:35 am
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
Rich and Ankur have provided nice solutions, so I won't solve the question again.
However, I do want to add a pro tip to remember when answering questions using calculating probabilities using the formula:

P(event A occurs) = (# of outcomes where event A occurs)/(total # of outcomes)

In these cases, calculate the denominator first
There are two reasons for this:
1) The denominator is usually the easier value to calculate
2) If you can't calculate the numerator, you can probably use the denominator to eliminate answer choices.

Here's what I mean:

In this question, P(number has a hundreds digit of 2) = (# of integers with hundreds digit of 2)/(# of integers to choose from)

# of integers to choose from = 350 - 101 + 1 = 250

So, P(number has a hundreds digit of 2) = ??)/(250)

This tells us that the correct answer EITHER has 250 in its denominator OR, when the probability is simplified, the new denominator is a factor of 250

At this point, if we can't calculate the numerator, we eliminate some answer choices
A) 2/5 [5 is a factor of 250, so this answer COULD be correct]

B) 2/7 [7 is NOT a factor of 250. ELIMINATE B]

C) 33/83 [83 is NOT a factor of 250. ELIMINATE C]

D) 99/250 [250 is a factor of 250, so this answer COULD be correct]

E) 100/249 [249 is NOT a factor of 250. ELIMINATE E]

So, we were able to deduce that the correct answer is either A or D.

For more on this concept, see the following videos:
- Introduction to Probability: https://www.gmatprepnow.com/module/gmat ... /video/742
- General Probability Strategies: https://www.gmatprepnow.com/module/gmat ... /video/757

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by OptinumGmat » Tue Apr 26, 2016 12:44 pm
OptimusPrep wrote:
boomgoesthegmat 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

Answer: A
This is a straight forward probability question.

Always remember, the total numbers between a and b = b - a + 1

Total number of tickets = 350 - 101 + 1 = 250
Tickets with 2 in the hundreds digit = 299 - 200 + 1 = 100

Probability of picking a ticket with 2 in the hundredth digit = 100/250 = 2/5

Correct Option: A
How do you simply 100/250 to 2/5?

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by [email protected] » Tue Apr 26, 2016 6:33 pm
Hi boomgoesthegmat,

When simplifying a fraction, you can do the math in 'stages' or you can do it all at once (depending on your comfort level with the work involved). When dealing with a fraction, you can multiply BOTH the numerator and the denominator by the same number or you can divide BOTH the numerator and the denominator by the same number.

With 100/250, if you recognize that both numbers are multiples of 50, then you can divide 100 by 50 and 250 by 50... you end up with 2/5.

You could also divide in 'stages' though. Notice how both 100 and 250 both end in a 0... That means that you can divide both by 10:

100/250 = 10/25

Now you can divide both by 5... and you'll still end up with 2/5.

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by Matt@VeritasPrep » Wed Apr 27, 2016 2:39 pm
In this problem,

Probability = # of target events / # of possible events

The target is "hundreds digit of 2". We have 200 -> 299, or 00 -> 99. That's 00 + (01 -> 99), for a total of 100 numbers.

The total is "anything from 101 to 350". We want everything from 1 to 350 MINUS everything from 1 to 100, or 350 - 100, or 250 numbers.

That gives us

100 / 250 => 2/5

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by Scott@TargetTestPrep » Thu Mar 15, 2018 9:26 am
boomgoesthegmat 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
There are a total of 350 - 101 + 1 = 250 raffle tickets in the box, and there are 299 - 200 + 1 = 100 raffle tickets with a hundreds digit of 2; thus, the probability is

100/250 = 2/5

Answer: A

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