Raffle tickets ered consecutively from 201 through 450

Hi yosita,
Refer to your PM.
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..

Posted from my mobile device

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

Thanks for your prompt and detailed response, chetan2u.

We can use the formula:

P(at least one 3) = 1 - P(no 3’s)

Let’s start with the numbers from 201 to 299, inclusive.

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

Answer: C

tickets without 3 from 201 to 299 ; 1*9*9 ; 81
tickets without 3 from 400 to 450 ; 136
total digits ; 450-201+1 ; 250
300 to 399 ; 100 digits
with 3s' 250-81-136 ; 33 ;
total with 3s ; 133
P ; 133/250 ;
IMO C

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