GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Dec 2018, 12:23

### 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

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

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Typical Day of a UCLA MBA Student - Recording of Webinar with UCLA Adcom and Student

December 14, 2018

December 14, 2018

10:00 PM PST

11:00 PM PST

Carolyn and Brett - nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session.
• ### Free GMAT Strategy Webinar

December 15, 2018

December 15, 2018

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

# Two dice are tossed once. The probability of getting an even

Author Message
TAGS:

### Hide Tags

Intern
Joined: 15 Mar 2010
Posts: 11
Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

23 Jan 2012, 20:20
7
37
00:00

Difficulty:

95% (hard)

Question Stats:

44% (02:07) correct 56% (02:18) wrong based on 668 sessions

### HideShow timer Statistics

Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is

A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36
Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: Probability - tossing two dice  [#permalink]

### Show Tags

24 Jan 2012, 01:47
6
14
Macsen wrote:
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36

OR probability:
If Events A and B are independent, the probability that Event A OR Event B occurs is equal to the probability that Event A occurs plus the probability that Event B occurs minus the probability that both Events A and B occur: $$P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B)$$.

This is basically the same as 2 overlapping sets formula:
{total # of items in groups A or B} = {# of items in group A} + {# of items in group B} - {# of items in A and B}.

Note that if event are mutually exclusive then $$P(A \ and \ B)=0$$ and the formula simplifies to: $$P(A \ or \ B) = P(A) + P(B)$$.

Also note that when we say "A or B occurs" we include three possibilities:
A occurs and B does not occur;
B occurs and A does not occur;
Both A and B occur.

AND probability:
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events: $$P(A \ and \ B) = P(A)*P(B)$$.

This is basically the same as Principle of Multiplication: if one event can occur in $$m$$ ways and a second can occur independently of the first in $$n$$ ways, then the two events can occur in $$mn$$ ways.

BACK TO THE ORIGINAL QUESTION:
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36

1. The probability of getting an even number at the first die is 1/2 (as the probability of even = the probability of odd = 1/2);
2. The probability of getting a total of 8 is 5/6^2, as there are 5 different favorable scenarios: (2,6), (6,2), (3,5), (5,3) and (4,4);
3. The probability of getting an even number at the first die AND a total of 8 is 3/6^2 (from above case);

Hence, The probability of getting an even number at the first die OR a total of 8 is 1/2+5/36-3/36=20/36.

_________________
Intern
Joined: 26 Dec 2011
Posts: 6
Re: Probability - tossing two dice  [#permalink]

### Show Tags

23 Jan 2012, 20:58
6
4
total possible outcomes for two dice: 36

case 1
prob of getting an even number on the first die
first die can have 2,4,6 second die can have any of 1,2,3,4,5,6
or favorable outcomes 3x6 = 18
probability= 18/36

case 2
getting a sum of 8 (2,6)(6,2)(3,5)(5,3)(4,4)
probability=5/36

now both of the above cases have some cases common to them
i.e. when the first die has an even number and the sum is also 8
there are 3 cases of this kind (2,6) (6,2) (4,4)
prob=3/36

also P(A or B)=P(A) + P(B) - P(A & B)
so we have P(even or sum of 8) = 18/36 + 5/36 - 3/36
20/36
##### General Discussion
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1171
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

19 Feb 2012, 07:18
+1 D

Great question.
Remember that the "OR formula" is:

P(A) OR P(B) = P(A) + P(B) - P(A and B)

In this case, A and B can take place together, so the value of P(A and B) is greater than 0.
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings

Director
Joined: 23 Apr 2010
Posts: 547
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

Updated on: 21 Feb 2012, 01:24
Bunuel, could you please explain why the probability of getting 'four' on both dices is 1/36 and not 1/18? Why don't we make a difference between 4 on dice 1 and four on dice ?

I mean why the combination 4'4'' is the same as 4''4' (apostrophe symbolizes the dice: ' = dice number one, '' = dice number two).

I have encountered many probability problems, where events that are identical are differentiated. For example:

Quote:
A 5 meter long wire is cut into two pieces. The longer piece is then used to form a perimeter of a square. What is the probability that the area of the square will be more than 1 if the original wire is cut at an arbitrary point?

1/6
1/5
3/10
1/3
2/5

OA:
(E)

In the example above the probability is 2/5.

Could you please explain the difference?

Originally posted by nonameee on 21 Feb 2012, 01:10.
Last edited by nonameee on 21 Feb 2012, 01:24, edited 1 time in total.
Director
Joined: 23 Apr 2010
Posts: 547
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

21 Feb 2012, 01:18
One more question: When calculating probability, how should we know when to count the same instances and when not?
Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

21 Feb 2012, 01:38
nonameee wrote:
Bunuel, could you please explain why the probability of getting 'four' on both dices is 1/36 and not 1/18? Why don't we make a difference between 4 on dice 1 and four on dice ?

I mean why the combination 4'4'' is the same as 4''4' (apostrophe symbolizes the dice: ' = dice number one, '' = dice number two).

We can get combination of (2, 6) in two ways: 2 on die A and 6 on die B OR 2 on die B and 6 on die A;

Whereas (4, 4) has only one combination: 4 on on die A and 4 on die B, it has no second combination, since 4 on die B and 4 on die A is exact same combination.

nonameee wrote:
I have encountered many probability problems, where events that are identical are differentiated. For example:

Quote:
A 5 meter long wire is cut into two pieces. The longer piece is then used to form a perimeter of a square. What is the probability that the area of the square will be more than 1 if the original wire is cut at an arbitrary point?

1/6
1/5
3/10
1/3
2/5

OA:
(E)

In the example above the probability is 2/5.

Could you please explain the difference?

This is completely different problem. In order the area of a square to be more than 1 the side of it must be more than 1, or the perimeter more than 4. So the longer piece must be more than 4. Look at the diagram.

-----

If the wire will be cut anywhere at the red region then the rest of the wire (longer piece) will be more than 4 meter long. The probability of that is 2/5 (2 red pieces out of 5).

Also discussed here: a-5-meter-long-wire-is-cut-into-two-pieces-if-the-longer-106448.html

Hope it helps.
_________________
Director
Joined: 23 Apr 2010
Posts: 547
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

21 Feb 2012, 01:46
I have re-read my posts and I think it's clear now.

Quote:
why the probability of getting 'four' on both dices is 1/36 and not 1/18? Why don't we make a difference between 4 on die1 and four on die?

This is because we fix dices one and two.

Quote:
A 5 meter long wire is cut into two pieces. The longer piece is then used to form a perimeter of a square. What is the probability that the area of the square will be more than 1 if the original wire is cut at an arbitrary point?

This is because we can cut the wire from both ends. When we fix the ends, we can cut 1 m from the left, or we can cut 1 m from the right end. Both options are different since we have fixed the ends.

Quote:
When calculating probability, how should we know when to count the same instances and when not?

We don't count twice the same instances of the fixed "property" (don't know how to call that). But if two fixed properties (e.g., two dices: die_1 = property_one, die_2 = property_two) yield the same result by taking various values, we need to count those options (as in the above example with dices producing the sum of 8).

Bunuel, does the above make sense?
Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

21 Feb 2012, 03:20
nonameee wrote:
I have re-read my posts and I think it's clear now.

Quote:
why the probability of getting 'four' on both dices is 1/36 and not 1/18? Why don't we make a difference between 4 on die1 and four on die?

This is because we fix dices one and two.

Quote:
A 5 meter long wire is cut into two pieces. The longer piece is then used to form a perimeter of a square. What is the probability that the area of the square will be more than 1 if the original wire is cut at an arbitrary point?

This is because we can cut the wire from both ends. When we fix the ends, we can cut 1 m from the left, or we can cut 1 m from the right end. Both options are different since we have fixed the ends.

Quote:
When calculating probability, how should we know when to count the same instances and when not?

We don't count twice the same instances of the fixed "property" (don't know how to call that). But if two fixed properties (e.g., two dices: die_1 = property_one, die_2 = property_two) yield the same result by taking various values, we need to count those options (as in the above example with dices producing the sum of 8).

Bunuel, does the above make sense?

I'm not sure that I understand completely the last part of your post, though it seems that you got the main point.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

07 Jun 2013, 06:03
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html

_________________
Intern
Status: Waiting
Joined: 11 Dec 2012
Posts: 49
Location: Bahrain
Concentration: Healthcare, General Management
GMAT 1: 640 Q49 V24
GMAT 2: 720 Q49 V40
WE: Sales (Other)
Re: Probability - tossing two dice  [#permalink]

### Show Tags

21 Jun 2013, 01:38
Bunuel wrote:
2. The probability of getting a total of 8 is 5/6^2, as there are 5 different favorable scenarios: (2,6), (6,2), (3,5), (5,3) and (4,4);

Dear Bunuel, my question pertains to the quoted text.

My reasoning to get sum 8 is:
Select any 5 numbers from first dice ie. probability: 5/6
To make the sum 8, only a unique number is to be chosen from 2nd dice. Hence the probability of selecting 1 number in second dice is : 1/6
Probability for sum 8 is 5/6^2. However I multiplied this number by 2 since you can perform the same task by selecting the second dice first and choosing a corresponding unique value in 1st dice.
Net probability is 5*2/6^2, but i believe this is wrong. Please let me know under what situations do we employ my understanding. Sorry if it is a silly question. Thank you for your time.
SVP
Joined: 06 Sep 2013
Posts: 1721
Concentration: Finance
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

23 Apr 2014, 05:15
I've done other problems in which the answer would simply come out from

1/2 + (1/2)(5/36)

That is first probability that we get an even number = 1/2
Second probability that we DON'T get an even number * Probability that we get a sum of 8 = (1/2)*(5/36)

Answer should be sum of both

Could someone please clarify why this approach is NOT valid?
Thanks!
Cheers
J
Manager
Joined: 30 Jul 2014
Posts: 131
GPA: 3.72
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

07 Sep 2017, 04:42
It is a very good question - I forgot to subtract the 3 cases in which the second dice will also show an even number - hence marked E, when the answer should be D...
_________________

A lot needs to be learned from all of you.

Senior Manager
Joined: 07 Sep 2014
Posts: 357
Concentration: Finance, Marketing
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

08 Sep 2017, 03:20
no calc required

half of the time the first dice will have even no. so prob will be greater than 1/2 as some cases extra when sum is 8.
so A,B,C ruled out

E gives 5 cases for sum 8.
but we know that sum cases will be common with "even condition" so D
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4295
Location: United States (CA)
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

11 Sep 2017, 10:35
Macsen wrote:
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is

A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36

When we roll two dice, there are 6 x 6 = 36 equally likely outcomes. From these 36 outcomes, we can get a total of 8 in the following ways:

6,2

2,6

4,4

5,3

3,5

The probability of each of these outcomes is 1/36, so the probability of getting a total of 8 is 5/36.

The probability of getting an even number on the first die is 1/2.

Recall that the probability of A or B is P(A or B) = P(A) + P(B) - P(A and B).

Let’s concentrate on P(A and B), which includes outcomes that have an even first roll AND a total of 8. The probability that the first die is even and the total of the two dice is 8 is 3/36 since (6,2), (2,6), and (4,4) are 3 of the 36 outcomes in which the first die is even and the total is 8.

Thus, the total probability of getting an even number on the first die OR a total of 8 is:

5/36 + 1/2 - 3/36 = 5/36 + 18/36 - 3/36 = 20/36

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Manager
Joined: 02 Jan 2017
Posts: 73
Location: Pakistan
Concentration: Finance, Technology
GMAT 1: 650 Q47 V34
GPA: 3.41
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

03 Oct 2017, 15:58
Bunuel wrote:
Macsen wrote:
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36

OR probability:
If Events A and B are independent, the probability that Event A OR Event B occurs is equal to the probability that Event A occurs plus the probability that Event B occurs minus the probability that both Events A and B occur: $$P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B)$$.

This is basically the same as 2 overlapping sets formula:
{total # of items in groups A or B} = {# of items in group A} + {# of items in group B} - {# of items in A and B}.

Note that if event are mutually exclusive then $$P(A \ and \ B)=0$$ and the formula simplifies to: $$P(A \ or \ B) = P(A) + P(B)$$.

Also note that when we say "A or B occurs" we include three possibilities:
A occurs and B does not occur;
B occurs and A does not occur;
Both A and B occur.

AND probability:
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events: $$P(A \ and \ B) = P(A)*P(B)$$.

This is basically the same as Principle of Multiplication: if one event can occur in $$m$$ ways and a second can occur independently of the first in $$n$$ ways, then the two events can occur in $$mn$$ ways.

BACK TO THE ORIGINAL QUESTION:
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36

1. The probability of getting an even number at the first die is 1/2 (as the probability of even = the probability of odd = 1/2);
2. The probability of getting a total of 8 is 5/6^2, as there are 5 different favorable scenarios: (2,6), (6,2), (3,5), (5,3) and (4,4);
3. The probability of getting an even number at the first die AND a total of 8 is 3/6^2 (from above case);

Hence, The probability of getting an even number at the first die OR a total of 8 is 1/2+5/36-3/36=20/36.

As you have mentioned above

When two events are independent, the probability of both occurring is the product of the probabilities of the individual events: $$P(A \ and \ B) = P(A)*P(B)$$.

Could you explain why can we not get probability of ( even number & total of 8)= by multiplying the two probabilities ?

Like this = 1/2 * 5/36?

Or am i missing something quite simple here ( I feel like i am)
Intern
Joined: 08 Dec 2017
Posts: 16
Re: Two dice are tossed once. The probability of getting an even  [#permalink]

### Show Tags

27 Dec 2017, 15:51
We van just list out the possibilites.

2,1
2,2
2,3
2,4
2,5

2,6

3,5

4,1
4,2
4,3
4,4
4,5
4,6

5,3

6,1
6,2
6,3
6,4
6,5
6,6

So 20/36. 18 cases occur from the first dice being even. The other two cases are 3,5 and 5,3

Posted from my mobile device
Re: Two dice are tossed once. The probability of getting an even &nbs [#permalink] 27 Dec 2017, 15:51
Display posts from previous: Sort by