Raffle tickets ered consecutively from 201 through 450

Question

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

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

Senthil1981 · Answer

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 ?

yosita18 · Answer

chetan2u · Answer

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

pallaviisinha · Answer

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

chetan2u · Answer

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

ScottTargetTestPrep · Answer

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

Archit3110 · Answer

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

kreel11 · Answer

scott  can you please clarify how you got 10 3's between 230 to 239 or 430 to 439 as there are two 3's in 233 and 433 which makes the count to 11 in each case

Bunuel · Answer

From 230 to 239 ten numbes have at least one 3 in them:
230
231
232
233
234
235
236
237
238
239

Hope it's clear.

djangobackend · Answer

I did by - 
answer = 1- (no 3s)

so from 201-299 - 
we have 1*9*9 = 81 ways
1 = 2
9 = all values except 3
9 = all values except 3 
BUT this will also include 200 but our range starts from 201 so remove that 81-1 = 80 ways

now 300 to 399 - eliminated

for 400-450 
we can do 2 approaches- 
1st approach (we will count till 449 only) - 1*4*9 = 36 ways
1 = 4
4 = (0,1,2,4)
9 = all values except 3 
now since we calculated till 449 only but range includes 450 so include that so it becomes 36+1 = 37 ways

2nd approach (include all 450s as well) - 1*5*9 = 45 ways 
1 = 4
5 = (0,1,2,4,5)
9 = all values except 3 
now since we have values 451, 452, 454, 455, 456, 457, 458, 459 - we remove these 8 values so overall becomes 45-8 = 37 ways

now, 
total ways = 80+37 = 117 
so without 3's probability becomes = 117/250 
but we require at least one 3 so it answer becomes = 1- (no 3) = 1 - 117/250 = 133/250

kk07950796 · Answer

Can someone correct me please, 
for 3 in Hundred's place there are 100 Values, for 3 in tens place there are 20 Values(hundred's place 2 values*tens place 1 value * Units place 10 Values) and 3 in Units place 18 Values (2*9*1) so my numerator adds up to 138 . which 5 values am i missing?

Bunuel · Answer

Units place - 15 numbers: 
203
213
223
 233 
243
253
263
273
283
293

403
413
423
 433 
443

Tens place - 20 numbers: 
230
231
232
 233 
234
235
236
237
238
239

430
431
432
 433 
434
435
436
437
438
439

However, there is an overlap of two numbers, 233 and 433, so the total = 15 + 20 - 2 = 33.

kk07950796 · Answer

Thanks A Lot Bunuel

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Raffle tickets ered consecutively from 201 through 450

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