Let C be the collection of all integers between 100,000 and 999,999 in

Question

Let C be the collection of all integers between 100,000 and 999,999 inclusive. What is the probability that the number randomly selected from C has exactly 5 occurrences of the digit 3 in the decimal representation?

A) 1/150,000
B) 53/899,999
C) 54/899,999
D) 53/900,000
E) 3/50,000

A) 1/150,000
B) 53/899,999
C) 54/899,999
D) 53/900,000
E) 3/50,000

gmatophobia · Accepted Answer

Total number of digits between 999,999 and 100,000 , inclusive = (999,999 - 100,000) + 1 We can infer that each of the numbers between the range is a six-digit number. In that, we want to identify numbers in which the digit 3 occurs 5 times. Hence, the sixth digit can be any digit between 0 and 9 , inclusive, except for 3 . Two cases are possible 1) The number does not start with 3 2) The number starts with 3 Case 1 : The number does not start with 3 i.e. the format of the number is X33333 If the number doesn't start with three, the first place, X , can be filled by 1, 2, 4, 5, 6, 7, 8, and 9 ⇒ So 8 ways The next five digits of the number will be filled with 3s Total = 8 ways Case 2 : The number starts with 3 If the number starts with three, four places of the remaining five places will contain a 3. The fifth place can be placed in 9 ways ( 0, 1, 2, 4, 5, 6, 7, 8, and 9) We can multiple arrangements of the five digits. This can be done in \frac{5!}{4!} ways (all possible arrangments of 3333X , where X is a value between 0 and 9, except 3) Total number of ways = 9 * \frac{5!}{4!} = 45 Favourable cases = 45 + 8 = 53 Total cases = 900000 Required Probability = \frac{53}{900,000} What am I missing here?

KarishmaB · Answer

We have all 900,000 six digit numbers here. If we have five 3s and we have one occurence of an additional digit, then we need to consider 2 cases: Any additional digit except 0 and 3: A digit can be chosen in 8 ways from 1/2/4/5/.../9. Five 3s and another digit can be arranged in 6 distinct ways (333334, 333343, 333433... 433333) This gives us 8*6 = 48 numbers The additional digit is 0: 0 cannot be the first digit from left but it can take any other 5 positions such as 333330, 333303 etc. Hence 5 more numbers. Total we have 53 such numbers. Required Probability = Favorable Cases/ Total number of cases = 53/900,000 Answer (D) Video on Probability: https://youtu.be/0BCqnD2r-kY

NoviceBoy · Answer

My bad  gmatophobia  !! Your answer is correct. I have marked the wrong answer in the question. The correct option is D.

sujoykrdatta · Answer

Assuming no digit is zero, we have: 
Number of ways of choosing the positions of 3 = 6C5 = 6 ways
Number of ways of choosing the 6th digit = 8 ways (ignoring 0 and 3) 
Thus, total number of such numbers = 6 * 8 = 48

Assuming zero is present, the first digit must be 3. Zero can occur in the 2nd, 3rd, 4th, 5th and 6th positions i.e. 5 ways

Thus, total favorable cases = 48 + 5 = 53
Total cases = 999999 - 100000 + 1 = 900000
Probability = 53/900000

Answer D

javierjohnson · Answer

Case 1:   Dont starts with 3

In this case we have (considering that the first number cant be 0)

8/9 x 1/10 x 1/10 x 1/10 x 1/10 x 1/10 = 8/9 x 1/100.000

Case 2:   Starts with 3

In this case we have (considering that the first number cant be 0)

1/9 x 9/10 x 1/10 x 1/10 x 1/10 x 1/10 Notice that after the first 3, the rest can be ordered un 5 possible ways, 5!/4!, so:

5 x 1/100.000

Finally, adding both cases:

8/9 x 1/100.000 + 5 x 1/100.000

(1/100.000)(5 + 8/9)

(1/100.000)(53/9)

53/900.000

MonishBhawale · Answer

smart guess can be helpful to solve this question < 30sec
scan the options
B C D are very close so answer can be either of them 
the total number of outcomes is the total number of integers which in the case is 900,000
so only option d has the said value

answer - D

einstein801 · Answer

decimal representation??? Doesn't that mean 0.33333?

KarishmaB · Answer

We use the decimal system of numbers which means we use digits 0 - 9. 
Other number systems are binary (digits 0 and 1) or hexadecimal (0 - 9, A - F) etc.

Here decimal representation just means that the number is represented using digits 0 to 9 and not in binary or another form.

emont · Answer

Bunuel where can i see more questions like this please? struggling with this counting integers in large sets. thank you!

Bunuel · Answer

Search in Probability section: https://gmatclub.com/forum/search.php?s ... &tag_id=54 Check Constructing Numbers, Codes and Passwords Questions in our Special Questions Directory . Hope it helps.

Adit_ · Answer

Total numbers=900,000
Favourable outcomes:
9c1(9 digits to choose from apart from 3)*6!/5!(arranging them) = 54 
Now we have one case with 033333 and not possible 
Hence 54-1=53
Total Probability=53/900,000

sujoykrdatta · Answer

3 needs to occur 5 times out of 6. We need to select the places in 6c5 = 6 ways. The remaining digit can be chosen in 9 ways (since we cannot use 3 again)
So total cases = 9*6 = 54

However, when the last 5 digits are 3, the first digit cannot be 0. So the number 033333 has to be removed.

Thus, number of favorable cases = 54-1=53
Total = 999999-100000+1 = 900000

Probability = 53/900000

Answer D

Rickooreoisb · Answer

Since you said in the end what are you missing

It is fully correct just to make it air tight

Writing 5!/4! is fine mathematically (it equals 5),
 but combinatorially it is clearer to say:
(5C4) 
 Your reasoning structure is solid.

Adit_ · Answer

You shouldn't really use such strategies here as you are excluding the possiblity of simplification.
Option E is very close too.

Usernamevisible · Answer

Total numbers in C:

100,000 to 999,999

These are all 6-digit numbers.

Count:

999,999 - 100,000 + 1 = 900,000

So the denominator is 900,000.

---

We want exactly 5 occurrences of the digit 3.

A 6-digit number with exactly five 3s looks like:

33333x

where one position contains a digit other than 3.

We count according to the position of the non-3 digit.

Case 1: The non-3 digit is in the first position.

x33333

The first digit cannot be 0 (otherwise it would not be a 6-digit number), and it cannot be 3.

Choices:

1, 2, 4, 5, 6, 7, 8, 9

So there are 8 possibilities.

---

Case 2: The non-3 digit is in one of the remaining five positions.

3x3333, 33x333, etc.

Now x can be any digit except 3, including 0.

So each position gives 9 choices.

Total:

5 * 9 = 45

---

Total favorable numbers:

8 + 45 = 53

Therefore,

Probability = 53 / 900,000

Answer: D

---

A common mistake:

You might think:

Choose 1 of the 6 positions for the non-3 digit, and give it 9 choices:

6 * 9 = 54

But this incorrectly counts 033333, which is not a 6-digit number and is not in the set C.

So we subtract that one invalid case:

54 - 1 = 53

giving the correct probability:

53 / 900,000.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=88939 --- Harsha GMAT-Club-Forum-fueg81j8.png

BongBideshini · Answer

The 6th digit could be X X = 0, 1, 2, 4, 5, 6, 7, 8, or 9.

Total arrangement of 33333X=6!/5! =6
Total arrangement for different X = 9*6 - 1 = 53 (-1 because we are discarding 033333)
P=53/900,000

Ans D

Let C be the collection of all integers between 100,000 and 999,999 in

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Let C be the collection of all integers between 100,000 and 999,999 in

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards