Last visit was: 27 Apr 2026, 08:28 It is currently 27 Apr 2026, 08:28
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 Apr 2026
Posts: 109,928
Own Kudos:
811,563
 [1]
Given Kudos: 105,914
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,928
Kudos: 811,563
 [1]
M37-64 30 Nov 2021, 03:54
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Nine cards numbered from 1 to 9 are placed in an empty bowl. Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number (first card gives the thousands digit, second card gives the hundreds digit, third card gives the tens digit and fourth card gives the units digit). What it the probability that the four-digit number is even?

A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{3}{9}\)
D. \(\frac{4}{9}\)
E. \(\frac{5}{9}\)
ShowHide Answer
Official Answer
D
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 Apr 2026
Posts: 109,928
Own Kudos:
811,563
Given Kudos: 105,914
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,928
Kudos: 811,563
M37-64 30 Nov 2021, 03:54
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Official Solution:

Nine cards numbered from 1 to 9 are placed in an empty bowl. Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number (first card gives the thousands digit, second card gives the hundreds digit, third card gives the tens digit and fourth card gives the units digit). What it the probability that the four-digit number is even?

A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{3}{9}\)
D. \(\frac{4}{9}\)
E. \(\frac{5}{9}\)


The four-digit number will be even if the units digit is even. So, the question basically asks about the probability that 4th card picked will be even. There are 4 even cards from 9, so the probability that 4th card will be even is \(\frac{4}{9}\).

To elaborate more: the initial probability of drawing an even card is \(\frac{4}{9}\). Without knowing the other results, the probability of drawing an even card will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).


Answer: D
User avatar
Shivang29
User avatar
Joined: 16 Sep 2024
Last visit: 05 Nov 2025
Posts: 23
Own Kudos:
13
 [1]
Given Kudos: 30
Posts: 23
Kudos: 13
 [1]
M37-64 24 Feb 2025, 11:03
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Nine cards numbered from 1 to 9 are placed in an empty bowl. Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number (first card gives the thousands digit, second card gives the hundreds digit, third card gives the tens digit and fourth card gives the units digit). What it the probability that the four-digit number is even?

A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{3}{9}\)
D. \(\frac{4}{9}\)
E. \(\frac{5}{9}\)
Show more
I am not sure I understand the reasoning.

if we do it the long way and determine all possible numbers as 9*8*7*6 then try to find all possible combinations that would allow for a number to still be even.

Example:
OddOddOddEven
OddOddEvenEven
OddEvenEvenEven
EvenEvenEvenEven

we would get 1344 possible ways and when divided by all the possible outcomes we get 2/3.

Am I missing something obvious?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 Apr 2026
Posts: 109,928
Own Kudos:
811,563
 [1]
Given Kudos: 105,914
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,928
Kudos: 811,563
 [1]
M37-64 24 Feb 2025, 11:27
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Shivang29
Bunuel
Nine cards numbered from 1 to 9 are placed in an empty bowl. Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number (first card gives the thousands digit, second card gives the hundreds digit, third card gives the tens digit and fourth card gives the units digit). What it the probability that the four-digit number is even?

A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{3}{9}\)
D. \(\frac{4}{9}\)
E. \(\frac{5}{9}\)
I am not sure I understand the reasoning.

if we do it the long way and determine all possible numbers as 9*8*7*6 then try to find all possible combinations that would allow for a number to still be even.

Example:
OddOddOddEven
OddOddEvenEven
OddEvenEvenEven
EvenEvenEvenEven

we would get 1344 possible ways and when divided by all the possible outcomes we get 2/3.

Am I missing something obvious?
Show more

You should get 4/9 with any correct method. You are missing couple of cases there:

Odd - Odd - Odd - Even = 5/9 * 4/8 * 3/7 * 4/6 = 5/63

Even - Odd - Odd - Even = 4/9 * 5/8 * 4/7 * 3/6 = 5/63
Odd - Even - Odd - Even = 5/9 * 4/8 * 4/7 * 3/6 = 5/63
Odd - Odd - Even - Even = 5/9 * 4/8 * 4/7 * 3/6 = 5/63

Even - Even - Odd - Even = 4/9 * 3/8 * 5/7 * 2/6 = 5/126
Even - Odd - Even - Even = 5/126
Odd - Even - Even - Even = 5/126

Even - Even - Even - Even = 4/9 * 3/8 * 2/7 * 1/6 = 1/126

The sum = 4/9.
User avatar
ragnarok13
User avatar
Joined: 15 Oct 2019
Last visit: 16 Feb 2026
Posts: 28
Own Kudos:
8
Given Kudos: 297
Location: India
GMAT Focus 1: 615 Q80 V84 DI77
GMAT Focus 2: 655 Q88 V84 DI76
GMAT Focus 2: 655 Q88 V84 DI76
Posts: 28
Kudos: 8
M37-64 14 Mar 2025, 11:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel, I am pretty sure there is something really really wrong with my understanding, but can you please help. I approached it as,
Last digit should be even. For the first three it is irrespective whether they are odd or even.
Case 1 : Last digit is 2. Then for the first digit there are 8 possible options.
Probability = (8/9) * (7/8) * (6/7)
And then same when the last digit is either 4 or 6 or 8
So total probability = 4 * ( (8/9) * (7/8) * (6/7) )
I know this is giving me a value >1, and of course this is wrong, so could you please explain the flaw in my understanding. Thank you
Bunuel
Shivang29
Bunuel
Nine cards numbered from 1 to 9 are placed in an empty bowl. Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number (first card gives the thousands digit, second card gives the hundreds digit, third card gives the tens digit and fourth card gives the units digit). What it the probability that the four-digit number is even?

A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{3}{9}\)
D. \(\frac{4}{9}\)
E. \(\frac{5}{9}\)
I am not sure I understand the reasoning.

if we do it the long way and determine all possible numbers as 9*8*7*6 then try to find all possible combinations that would allow for a number to still be even.

Example:
OddOddOddEven
OddOddEvenEven
OddEvenEvenEven
EvenEvenEvenEven

we would get 1344 possible ways and when divided by all the possible outcomes we get 2/3.

Am I missing something obvious?

You should get 4/9 with any correct method. You are missing couple of cases there:

Odd - Odd - Odd - Even = 5/9 * 4/8 * 3/7 * 4/6 = 5/63

Even - Odd - Odd - Even = 4/9 * 5/8 * 4/7 * 3/6 = 5/63
Odd - Even - Odd - Even = 5/9 * 4/8 * 4/7 * 3/6 = 5/63
Odd - Odd - Even - Even = 5/9 * 4/8 * 4/7 * 3/6 = 5/63

Even - Even - Odd - Even = 4/9 * 3/8 * 5/7 * 2/6 = 5/126
Even - Odd - Even - Even = 5/126
Odd - Even - Even - Even = 5/126

Even - Even - Even - Even = 4/9 * 3/8 * 2/7 * 1/6 = 1/126

The sum = 4/9.
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 Apr 2026
Posts: 109,928
Own Kudos:
811,563
Given Kudos: 105,914
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,928
Kudos: 811,563
M37-64 14 Mar 2025, 11:52
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ragnarok13
Bunuel, I am pretty sure there is something really really wrong with my understanding, but can you please help. I approached it as,
Last digit should be even. For the first three it is irrespective whether they are odd or even.
Case 1 : Last digit is 2. Then for the first digit there are 8 possible options.
Probability = (8/9) * (7/8) * (6/7)
And then same when the last digit is either 4 or 6 or 8
So total probability = 4 * ( (8/9) * (7/8) * (6/7) )
I know this is giving me a value >1, and of course this is wrong, so could you please explain the flaw in my understanding. Thank you
Bunuel
Shivang29
Nine cards numbered from 1 to 9 are placed in an empty bowl. Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number (first card gives the thousands digit, second card gives the hundreds digit, third card gives the tens digit and fourth card gives the units digit). What it the probability that the four-digit number is even?

A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{3}{9}\)
D. \(\frac{4}{9}\)
E. \(\frac{5}{9}\)

I am not sure I understand the reasoning.

if we do it the long way and determine all possible numbers as 9*8*7*6 then try to find all possible combinations that would allow for a number to still be even.

Example:
OddOddOddEven
OddOddEvenEven
OddEvenEvenEven
EvenEvenEvenEven

we would get 1344 possible ways and when divided by all the possible outcomes we get 2/3.

Am I missing something obvious?

You should get 4/9 with any correct method. You are missing couple of cases there:

Odd - Odd - Odd - Even = 5/9 * 4/8 * 3/7 * 4/6 = 5/63

Even - Odd - Odd - Even = 4/9 * 5/8 * 4/7 * 3/6 = 5/63
Odd - Even - Odd - Even = 5/9 * 4/8 * 4/7 * 3/6 = 5/63
Odd - Odd - Even - Even = 5/9 * 4/8 * 4/7 * 3/6 = 5/63

Even - Even - Odd - Even = 4/9 * 3/8 * 5/7 * 2/6 = 5/126
Even - Odd - Even - Even = 5/126
Odd - Even - Even - Even = 5/126

Even - Even - Even - Even = 4/9 * 3/8 * 2/7 * 1/6 = 1/126

The sum = 4/9.
Show more
You're right—the first three digits don’t matter. Only the last digit determines whether the number is even. The probability of that happening is 4/9. That’s it!

You can check more on this question here: https://gmatclub.com/forum/nine-cards-n ... 47082.html

Hope it helps/
User avatar
Rebaz
User avatar
Joined: 14 Feb 2014
Last visit: 25 Apr 2026
Posts: 145
Own Kudos:
38
Given Kudos: 4,772
Posts: 145
Kudos: 38
M37-64 23 Jun 2025, 07:17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

Quote:
Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number
Show more
--------This basically means that we cannot put back the card in the empty bowl once they are drawn out of the empty bowl, right?

If yes, what if the thousands, hundreds and tens digits are even (even though the first three digits donot matter for the four digit number to be even), then we are left with one even digit from a total of 6 digits left (5 odd digits and 1 even digit)and in that case the probability that the fourth digit would be even is 1/6?

Where do i go wrong in my understanding/approach?

Thanks in advance!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 Apr 2026
Posts: 109,928
Own Kudos:
811,563
 [1]
Given Kudos: 105,914
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,928
Kudos: 811,563
 [1]
M37-64 23 Jun 2025, 07:50
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Rebaz

Bunuel
Official Solution:

Nine cards numbered from 1 to 9 are placed in an empty bowl. Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number (first card gives the thousands digit, second card gives the hundreds digit, third card gives the tens digit and fourth card gives the units digit). What it the probability that the four-digit number is even?

A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{3}{9}\)
D. \(\frac{4}{9}\)
E. \(\frac{5}{9}\)


The four-digit number will be even if the units digit is even. So, the question basically asks about the probability that 4th card picked will be even. There are 4 even cards from 9, so the probability that 4th card will be even is \(\frac{4}{9}\).

To elaborate more: the initial probability of drawing an even card is \(\frac{4}{9}\). Without knowing the other results, the probability of drawing an even card will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).


Answer: D
Bunuel

Quote:
Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number

This basically means that we cannot put back the card in the empty bowl once they are drawn out of the empty bowl, right?

If yes, what if the thousands, hundreds and tens digits are even (even though the first three digits donot matter for the four digit number to be even), then we are left with one even digit from a total of 6 digits left (5 odd digits and 1 even digit)and in that case the probability that the fourth digit would be even is 1/6?

Where do i go wrong in my understanding/approach?

Thanks in advance!
Show more

The point is that we don't know what the first three cards were, so without that information, nothing changes when calculating the probability for the fourth card.

Consider this: there are 8 cards, 5 spades, and 3 hearts. What is the probability that the first card you pick will be a spade? Clearly, it’s 5/8.

Now, suppose I randomly remove three cards without telling you which cards were thrown away. What is the probability now that you pick a spade out of the 5 remaining cards? Even though the sample has been reduced, the probability still remains 5/8. Why should it change?

Another scenario: using the same 8 cards (5 spades and 3 hearts), imagine you are allowed to pick one card, but only from a randomly selected subset of 5 cards. What is the probability of picking a spade? Again, it remains 5/8.

In all these cases, the probability does not change. If it were to increase or decrease, it would imply that the probability of picking a heart would inversely change. But there is no reason for this to happen, as the process remains random, and we lack information about the removed or previously picked cards.

Check all cases here: https://gmatclub.com/forum/m37-376460.html#p3532064

Similar questions to practice:
https://gmatclub.com/forum/m28-184525.html
https://gmatclub.com/forum/m14-183995.html
https://gmatclub.com/forum/a-certain-te ... 31321.html
https://gmatclub.com/forum/a-certain-te ... 27423.html
https://gmatclub.com/forum/a-medical-re ... 27396.html
https://gmatclub.com/forum/a-certain-cl ... 27730.html
https://gmatclub.com/forum/a-certain-cl ... 34891.html
https://gmatclub.com/forum/a-bag-contai ... 00023.html
https://gmatclub.com/forum/a-box-contai ... 90272.html
https://gmatclub.com/forum/if-40-people ... 97015.html
Moderators:
Bunuel
Math Expert
109928 posts
bb
Founder
43163 posts