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# Raffle tickets ered consecutively from 201 through 450

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Intern
Joined: 05 May 2016
Posts: 39
Raffle tickets ered consecutively from 201 through 450 [#permalink]

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24 Aug 2016, 12:35
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Difficulty:

45% (medium)

Question Stats:

68% (02:15) correct 32% (02:28) wrong based on 69 sessions

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Raffle tickets ered consecutively from 201 through 450 are placed in a certain box. What is the probability of a ticket picked from the box at random having a number with at least one of its digits as 3?

A) 52/ 125
B) 58/125
C) 133/250
D) 71/125
E) 77/ 125
[Reveal] Spoiler: OA

Last edited by yosita18 on 28 Aug 2016, 08:06, edited 2 times in total.
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Re: Raffle tickets ered consecutively from 201 through 450 [#permalink]

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24 Aug 2016, 22:48
yosita18 wrote:
Raffle tickets ered consecutively from 201 through 450 are placed in a certain box. What is the probability of a ticket picked from the box at random having a number with at least one of its digits as 3?

If possible, please provide a solution using the "Complement" Method.

Ans: 133/250

Total number of outcomes = 250 (450 - 201 +1)

Favourable outcome is to pick at least one 3 :
between 201 - 299, there are 19 numbers with 3
between 300 - 399, there are 100 with 3
between 400 - 450 there are 14 with 3

So P = $$\frac{(19 + 100 + 14)}{250} = \frac{133}{250}$$

By "Complement" , you mean count the numbers with no 3 in them and then subtract from 250 ?
Intern
Joined: 05 May 2016
Posts: 39
Re: Raffle tickets ered consecutively from 201 through 450 [#permalink]

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25 Aug 2016, 04:44
Senthil1981 wrote:
Quote:
Raffle tickets ered consecutively from 201 through 450 are placed in a certain box. What is the probability of a ticket picked from the box at random having a number with at least one of its digits as 3?

If possible, please provide a solution using the "Complement" Method.

Ans: 133/250

Total number of outcomes = 250 (450 - 201 +1)

Favourable outcome is to pick at least one 3 :
between 201 - 299, there are 19 numbers with 3
between 300 - 399, there are 100 with 3
between 400 - 450 there are 14 with 3

So P = $$\frac{(19 + 100 + 14)}{250} = \frac{133}{250}$$

By "Complement" , you mean count the numbers with no 3 in them and then subtract from 250 ?

Quote:
Yes, sir. I want to use the complement, as it rules out the need to calculate 3 cases of units, tens, and hundreds digit.
Thanks again!

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Joined: 02 Aug 2009
Posts: 5663
Re: Raffle tickets ered consecutively from 201 through 450 [#permalink]

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27 Aug 2016, 04:47
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Expert's post
Hi yosita,
A solution if this is what you are looking for...
Let divide in three lots
1)201-299..
1st digit- only 1 that is 2..
2nd digit and 3rd digit any of 10 except 3, so 10-1=9
So total number with out 3=1*9*9=81..
This includes 200 also, so subtract 1... 81-1=80

2)300-399..... none

3)400-450..
1st digit... only 1 that is 4..
3rd..any of 10 except 3, so 9
2nd digit... 0,1,2,4, so 4..
Total 1*4*9=36
Add one more, 450, so overall 37

Total from 201-450= 80+37=117

So number with 3=250-117=133..
Prob=133/250..

Hope this is what you wanted..

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Re: Raffle tickets ered consecutively from 201 through 450 [#permalink]

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28 Aug 2016, 03:07
Hello,

Can you pls explain your response to the number category from 300-399. Actually, there are digits with at least one of its digits as 3 from 300-399. So, why are we ignoring them.

Thanks,
Pallavi
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Joined: 02 Aug 2009
Posts: 5663
Re: Raffle tickets ered consecutively from 201 through 450 [#permalink]

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28 Aug 2016, 03:58
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pallaviisinha wrote:
Hello,

Can you pls explain your response to the number category from 300-399. Actually, there are digits with at least one of its digits as 3 from 300-399. So, why are we ignoring them.

Thanks,
Pallavi

Hi pallavi,
The member posting the Q wanted a complement method that is total number 3 is not there and subtract from total....
That is why we calculated in each set of 100s, the number without any digit as 3 ..
So 300-399 has none as all have at least one digit as 3..

If I wanted to find numbers containing 3...
1) 201-299...
Units digit 3- 1*10*1=10...
Tens digit as3- 1*1*10...
Total 10+10-1=19 one is common in two, 233

2) 300-399
All 100

3)400-450...
Units digit...1*5*1=5, these 5 are 0,1,2,3,4
Tens digit...1*1*10=10..
Again subtract one from total, number being 433..
Total 10+5-1=14..

Overall - 19+100+14=133..
Prob 133/250
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

Intern
Joined: 17 Jan 2015
Posts: 38
Location: India
GMAT 1: 620 Q42 V34
GMAT 2: 710 Q49 V38
Re: Raffle tickets ered consecutively from 201 through 450 [#permalink]

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28 Aug 2016, 04:47
Thanks for your prompt and detailed response, chetan2u.
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Re: Raffle tickets ered consecutively from 201 through 450 [#permalink]

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31 Jan 2018, 16:05
yosita18 wrote:
Raffle tickets ered consecutively from 201 through 450 are placed in a certain box. What is the probability of a ticket picked from the box at random having a number with at least one of its digits as 3?

A) 52/ 125
B) 58/125
C) 133/250
D) 71/125
E) 77/ 125

We can use the formula:

P(at least one 3) = 1 - P(no 3’s)

From 201 to 229, there are three possible 3’s: 203, 213, and 223.

From 230 to 239 there are 10 possible 3’s.

From 240 to 299 there are 6 possible 3’s.

Since there are 99 numbers from 201 to 299, inclusive, there are 99 - 3 - 10 - 6 = 80 numbers with no 3’s.

Numbers 300 to 399:

From 300 to 399 inclusive, there are zero numbers with no 3’ss.

Numbers 400 to 450:

From 400 to 429, there are three possible 3’s: 403, 413, and 423.

From 430 to 439 there are 10 possible 3’s.

From 240 to 249 there is 1 possible 3.

Since there are 51 numbers from 400 to 450, inclusive, there are 51 - 3 - 10 - 1 = 37 numbers with no 3’s.

So, from 201 to 350, inclusive, there are 80 + 37 = 117 numbers with no threes.

Therefore, the probability of not getting a 3 is 117/250

Thus, the probability of getting at least one 3 is 1 - 117/250 = 133/250

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Re: Raffle tickets ered consecutively from 201 through 450   [#permalink] 31 Jan 2018, 16:05
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