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. 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.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 01 Jan 2007
Posts: 23

A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
Updated on: 14 Aug 2017, 03:01
Question Stats:
41% (01:17) correct 59% (00:59) wrong based on 199 sessions
HideShow timer Statistics
A fair die is rolled once and a fair coin is flipped once. What is the probability that either the die will land on 3 or that the coin will land on heads? A. 1/2 B. 5/12 C. 7/12 D. 2/3 E. 3/4
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Regards,
Rocky
Originally posted by tekno9000 on 30 Dec 2007, 08:29.
Last edited by Bunuel on 14 Aug 2017, 03:01, edited 1 time in total.
Added the options.




CEO
Joined: 17 Nov 2007
Posts: 3440
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
30 Dec 2007, 11:19
I totally agree with eschn3am
and would like to add some important note that either X or Y means X alone, Y alone, and both XY (!)
P(X or Y)=P(X)+P(Y)P(both X,Y)
where, P(X) means probability of X regardless of Y
where, P(Y) means probability of Y regardless of X
in our case: 1/2+1/61/12=(6+21)/12=7/12
there is a little bit of theory: http://richardbowles.tripod.com/maths/p ... y/prob.htm




Director
Joined: 12 Jul 2007
Posts: 843

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
30 Dec 2007, 08:54
There are 12 possibilities for outcomes:
Coin: H or T
Dice: 1, 2, 3, 4, 5 or 6
Odds of the coin landing on heads: 1/2
Odds of the dice landing on three when the coin lands on tails: 1/12
1/2+1/12 = 7/12 probability
Make sure you don't try to add 1/2 and 1/6. We've already counted the dice landing on 3 when the coin lands on heads (included in the 1/2), so we're looking for the chance of the coin landing on tails and the dice landing on 3, which is 1/12
H1, H2, H3, H4, H5, H6, T3 are the 7 possibilities that work
T1, T2, T4, T5, T6 are the 5 possibilities that don't work




SVP
Joined: 07 Nov 2007
Posts: 1664
Location: New York

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
25 Aug 2008, 10:02
tekno9000 wrote: A fair die is rolled once and a fair coin is flipped once. What is the probaility that either the die will land on 3 or that the coin will land on heads?
Please solve and explain.
Thanks,
Tekno9000 P = 1/2 +1/6  (1/2)*(1/6) = 7/12
_________________
Your attitude determines your altitude Smiling wins more friends than frowning



Manager
Joined: 27 Oct 2008
Posts: 177

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
27 Sep 2009, 10:22
A fair die is rolled once and a fair coin is flipped once. What is the probaility that either the die will land on 3 or that the coin will land on heads?
Soln: The probability that either the die will land on 3 or coin will land heads is = Prob(die lands 3) + Prob(coin land heads)  Prob(that both happen) = 1/6 + 1/2  1/2*1/6 = 7/12



Intern
Joined: 16 Jan 2010
Posts: 9

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
01 Feb 2010, 13:47
what about this way =
P(X or Y)= 1P(NOT X)x P(NOTY)
1  5/6*1/2=7/12



Intern
Joined: 24 Feb 2012
Posts: 31

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
29 Feb 2012, 10:55
I understand that the formula here is: p(a or b) = p(a) + p(b)  p(a and b) = 1/6 + 3/6  (1/12) = 4/6  1/12 = 7/12
But, I'm curious as to what's wrong with this logic: p(die landing on 3) = 1/6 p(heads) = 1/2 Therefore p(die landing on 3 or coin landing on heads) = 1/6 + 1/2 = 2/3. Why do I have to account for the time when both events occur, and subtract it from 2/3?



Math Expert
Joined: 02 Sep 2009
Posts: 51215

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
29 Feb 2012, 11:48
fortsill wrote: I understand that the formula here is: p(a or b) = p(a) + p(b)  p(a and b) = 1/6 + 3/6  (1/12) = 4/6  1/12 = 7/12
But, I'm curious as to what's wrong with this logic: p(die landing on 3) = 1/6 p(heads) = 1/2 Therefore p(die landing on 3 or coin landing on heads) = 1/6 + 1/2 = 2/3. Why do I have to account for the time when both events occur, and subtract it from 2/3? Because 1/6 is basically 1/6*1: the die landing on 3 and the coin landing on any side. The same way 1/2 is basically 1/2*1: the coin landing on heads and the die landing on any side. So, both 1/6 and 1/2 count (include) the probability of the case when the die lands on 3 and the coin lands on heads, hence we should subtract P(3 on the die and heads on the coin)=1/6*1/2 once, to avoid double counting. P=1/6+1/21/6*1/2=7/12. A fair die is rolled once and a fair coin is flipped once. What is the probaility that either the die will land on 3 or that the coin will land on heads?This question can be solved with an easier approach: P(3 on a die OR heads on a coin)=1P(neither 3 on a die nor heads on a coin)=15/6*1/2=7/12. 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. Hope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 14 Apr 2012
Posts: 2

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
14 Apr 2012, 13:55
The above formula is incorrect. Let me explain and then prove it. When you add the probabilities for the dice hitting 3 (1/6) to the coin toss resulting in a heads (1/2) and then subtract the probability of BOTH a heads and a 3 (1/12), you aren't factoring that that probability was counted TWICE in your original sum. Once during your probability for 3 and once during your probability for a heads result.
Here is the proof.. there are 12 possibilities. I've put an asterisk next to the ones which satisfy the scenario (#s signify dice and t is tails, h is heads):
1t 1h* 2t 2h* 3t* 3h 4t 4h* 5t 5h* 6t 6h*
3h is obviously excluded because it includes both heads and 3. Now if you see, there are only 12 possible scenarios, and only 6 satisfy our condition (heads OR 3). Therefore, the probability is 6/12 = 1/2.
Now, I've thought of a simpler way of approaching these type of problems... Probability that it will be a HEADS and NOT A 3... 1/2 x 5/6 = 5/12
Probability that it will be a 3 and NOT HEADS... 1/6 x 1/2 = 1/12
ADD probabilities... 1/12 + 5/12 = 6/12 = 1/2



Math Expert
Joined: 02 Sep 2009
Posts: 51215

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
14 Apr 2012, 14:05
Germainte wrote: The above formula is incorrect. Let me explain and then prove it. When you add the probabilities for the dice hitting 3 (1/6) to the coin toss resulting in a heads (1/2) and then subtract the probability of BOTH a heads and a 3 (1/12), you aren't factoring that that probability was counted TWICE in your original sum. Once during your probability for 3 and once during your probability for a heads result.
Here is the proof.. there are 12 possibilities. I've put an asterisk next to the ones which satisfy the scenario (#s signify dice and t is tails, h is heads):
1t 1h* 2t 2h* 3t* 3h 4t 4h* 5t 5h* 6t 6h*
3h is obviously excluded because it includes both heads and 3. Now if you see, there are only 12 possible scenarios, and only 6 satisfy our condition (heads OR 3). Therefore, the probability is 6/12 = 1/2.
Now, I've thought of a simpler way of approaching these type of problems... Probability that it will be a HEADS and NOT A 3... 1/2 x 5/6 = 5/12
Probability that it will be a 3 and NOT HEADS... 1/6 x 1/2 = 1/12
ADD probabilities... 1/12 + 5/12 = 6/12 = 1/2 This is MGMAT question (from Manhattan GMAT Strategy Guide) and the OA is 7/12, not 1/2. I guess your reading of the question in different from that of the autors of the question.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 12 Oct 2011
Posts: 112
GMAT 1: 700 Q48 V37 GMAT 2: 720 Q48 V40

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
15 Apr 2012, 10:16
Rolling a dice and throwing a coin are independent events, because they can occur at the same time, right? In the probability part of the GMAT Club mathbook it says that two events are independent if the occurence of one event does not influence the occurence of the other event. Then it says that "tossing a coin and rolling a die are independent events". So what's correct? And if they were independent, we wouldn't have to subtract the probability of both occuring at same time, would we?



Math Expert
Joined: 02 Sep 2009
Posts: 51215

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
15 Apr 2012, 10:18



Manager
Joined: 30 Jun 2011
Posts: 199

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
17 Apr 2012, 05:45
BN1989 wrote: Rolling a dice and throwing a coin are independent events, because they can occur at the same time, right? In the probability part of the GMAT Club mathbook it says that two events are independent if the occurence of one event does not influence the occurence of the other event. Then it says that "tossing a coin and rolling a die are independent events". So what's correct? And if they were independent, we wouldn't have to subtract the probability of both occuring at same time, would we? See it this way, both the events can happen together. and that prob would be 1/2*1/6 . What you mentioned is the case when both cannot occur together . e.g. prob of winning a race for a or b = prob(a) + prob(b) Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring. BUT they can occur together and i.e. prob of occurrence together is prob(a)*prob(b) ( product of their individual prob)



Director
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 785
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
31 Jul 2013, 04:34
Thumb rule : 1. Events which can occur in one attempt subtract their double counting. E.g : In above case dice and coin can be flipped together and we can get three possible outcomes. 1 head no 3 1 tail and 3 1 head and 3 (thus both are overlapping and its possible in one attempt) E.g : Choose Spade or a king, overlapping is at King of spade. 2. Events which can not occur in one attempt, simply add their probabilities. E.g : Draw at least 2 red balls in 4 draws : case 1 : 2 red ball 2 black ball. case 2 : 3 red ball 1 black ball. case 4 : 4 red ball 0 black ball. These all cases can not occur in one attempt, thus simply add their probability, no overlapping. I tried to understand this concept in this way.
_________________
Piyush K
 Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press> Kudos My Articles: 1. WOULD: when to use?  2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".



Intern
Joined: 09 Dec 2013
Posts: 1

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
07 Jan 2014, 15:27
How about this approach?
There are 3 desired scenarios:
scenario 1: Die lands on 3 AND coin DOESN'T land on tails. scenario 2: Die DOESN'T land on 3 AND coin lands on heads. scenario 3: Die lands on 3 AND coin lands on heads.
So I add the probability of the 3 scenarios:
(scenario 1) + (scenario 2) + (scenario 3)
(Die condition * coin condition) + (Die condition * coin condition) + (Die condition * coin condition)
(1/6 * 1/2) + (5/6 * 1/2) + (1/6 * 1/2) = 7/12
scenario 1= (1/6 * 1/2) Only 3 is the desired outcome AND only tails is the desired outcome (To meet the condition of NO heads). scenario 2= (5/6 * 1/2) All number are desired outcomes but 3 (To meet the condition of NO 3) AND only heads is the desired outcome. scenario 3= (1/6 * 1/2) Only 3 is the desired outcome AND only heads is the desired outcome.
At the end of the day is the same as applying the formula. But is easier for me to "prevent" the overlapping count and forget about the last step of subtracting this overlap.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8678
Location: Pune, India

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
07 Jan 2014, 19:25
Carmona wrote: How about this approach?
There are 3 desired scenarios:
scenario 1: Die lands on 3 AND coin DOESN'T land on tails. scenario 2: Die DOESN'T land on 3 AND coin lands on heads. scenario 3: Die lands on 3 AND coin lands on heads.
So I add the probability of the 3 scenarios:
(scenario 1) + (scenario 2) + (scenario 3)
(Die condition * coin condition) + (Die condition * coin condition) + (Die condition * coin condition)
(1/6 * 1/2) + (5/6 * 1/2) + (1/6 * 1/2) = 7/12
scenario 1= (1/6 * 1/2) Only 3 is the desired outcome AND only tails is the desired outcome (To meet the condition of NO heads). scenario 2= (5/6 * 1/2) All number are desired outcomes but 3 (To meet the condition of NO 3) AND only heads is the desired outcome. scenario 3= (1/6 * 1/2) Only 3 is the desired outcome AND only heads is the desired outcome.
At the end of the day is the same as applying the formula. But is easier for me to "prevent" the overlapping count and forget about the last step of subtracting this overlap. This approach is fine too but here is a problem with enumerating cases  we might forget a case or two under time pressure. This question is extremely simple and you could think of overlapping sets here. The question asks you the probability of A or B or both. This is your case of the number of elements which are in set A or set B or both. How do you calculate that? n(A) + n(B)  n(A and B) (because n(A and B) is counted twice  once in n(A) and another time in n(B) so you need to subtract it out once) This is the same concept. P(A) + P(B)  P(A and B) = 1/6 + 1/2  (1/6*1/2) = 7/12
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 24 Nov 2014
Posts: 18

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
24 Feb 2015, 05:00
I still quite don't understand this, my logic is this:
P that BOTH Events occur: 1/6 * 1/2 = 1/12 P that NONE of the Events occur: 5/6 * 1/2 = 5/12 ;that means no 3 and no heads
So 1  6/12 = 6/12 which is that either one or the other event must occur. 3/noH or no3/H



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8678
Location: Pune, India

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
25 Feb 2015, 19:04
madmax1000 wrote: I still quite don't understand this, my logic is this:
P that BOTH Events occur: 1/6 * 1/2 = 1/12 P that NONE of the Events occur: 5/6 * 1/2 = 5/12 ;that means no 3 and no heads
So 1  6/12 = 6/12 which is that either one or the other event must occur. 3/noH or no3/H "A occurs or B occurs" implies that either only A occurs or only B occurs or both occur. In logic, "or" includes "both". Hence you will not subtract 6/12 but only 5/12 from 1 to get 1  5/12 = 7/12.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



SVP
Joined: 06 Nov 2014
Posts: 1880

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
28 Feb 2015, 04:37
tekno9000 wrote: A fair die is rolled once and a fair coin is flipped once. What is the probability that either the die will land on 3 or that the coin will land on heads? Event A = Fair die is rolled once Event B = Fair coin is flipped once. P(A or B) = P(A) + P(B)  P(A and B) = 1/6 + 1/2  (1/6)(1/2) = 7/12  Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: http://www.optimusprep.com/gmatondemandcourse



Math Expert
Joined: 02 Aug 2009
Posts: 7106

Re: A fair die is rolled once and a fair coin is flipped once. What is the
[#permalink]
Show Tags
28 Feb 2015, 05:18
VeritasPrepKarishma wrote: madmax1000 wrote: I still quite don't understand this, my logic is this:
P that BOTH Events occur: 1/6 * 1/2 = 1/12 P that NONE of the Events occur: 5/6 * 1/2 = 5/12 ;that means no 3 and no heads
So 1  6/12 = 6/12 which is that either one or the other event must occur. 3/noH or no3/H "A occurs or B occurs" implies that either only A occurs or only B occurs or both occur. In logic, "or" includes "both". Hence you will not subtract 6/12 but only 5/12 from 1 to get 1  5/12 = 7/12. hi, either A or B should not involve a scenario where both occur.... incase both together is also a scenario, the question should have been... "ways in which atleast one of the events occur?"... is there any Q from OG which tests this concept so that one can read what the GMAC people expect?
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor




Re: A fair die is rolled once and a fair coin is flipped once. What is the &nbs
[#permalink]
28 Feb 2015, 05:18



Go to page
1 2
Next
[ 22 posts ]



