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

 It is currently 17 Jun 2019, 07:59

### 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

# M07-12

Author Message
TAGS:

### Hide Tags

Manager
Joined: 09 Aug 2017
Posts: 61
Location: United States
Concentration: Technology
GMAT 1: 640 Q44 V33
GMAT 2: 630 Q47 V29
WE: Research (Investment Banking)

### Show Tags

13 Oct 2017, 07:23
Yeah, that looks like a negative sign there. I understand probability can't be negative, but it did take a few seconds of rereading that. Grammatically, you wouldn't put a dash there anyway. It should be a comma.
_________________
I'd love to hear any feedback or ways to improve my problem solving. I make a lot of silly mistakes. If you've had luck improving on stupid mistakes, I'd love to hear how you did it.

Also, I appreciate any kudos.
Intern
Joined: 11 Oct 2017
Posts: 1

### Show Tags

20 Dec 2017, 21:59
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 11 Apr 2016
Posts: 8
WE: Information Technology (Telecommunications)

### Show Tags

03 Apr 2018, 21:03
I think this the explanation isn't clear enough, please elaborate. As I thought that if of Ben looses the championship that means either Mike or Rob has won so I chose 6/7.
Math Expert
Joined: 02 Sep 2009
Posts: 55635

### Show Tags

03 Apr 2018, 21:57
himalayanmonk wrote:
I think this the explanation isn't clear enough, please elaborate. As I thought that if of Ben looses the championship that means either Mike or Rob has won so I chose 6/7.

You can check more solutions here: https://gmatclub.com/forum/if-ben-were- ... 11216.html

Hope it helps.
_________________
Manager
Joined: 26 Dec 2017
Posts: 156

### Show Tags

11 Jul 2018, 20:04
Bunuel wrote:
Official Solution:

If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$, and Rob - $$\frac{1}{3}$$ . If the probability of Ben being the winner is $$\frac{1}{7}$$, what is the probability that either Mike or Rob will win the championship? Assume that there can be only one winner.

A. $$\frac{1}{12}$$
B. $$\frac{1}{7}$$
C. $$\frac{1}{2}$$
D. $$\frac{7}{12}$$
E. $$\frac{6}{7}$$

This is a conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must lose: the probability of Ben losing is $$1-\frac{1}{7}=\frac{6}{7}$$.

Now out of these $$\frac{6}{7}$$ cases the probability of Mike winning is $$\frac{1}{4}$$ and the probability of Rob winning is $$\frac{1}{3}$$. So $$P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}$$.

Or consider the following:

Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).

Now, out of these 84 cases Ben will lose in $$\frac{6}{7}*84=72$$. Mike would be the winner in $$72*\frac{1}{4}=18$$ (1/4 th of the cases when Ben loses) and Rob would be the winner in $$72*\frac{1}{3}=24$$. Therefore $$P=\frac{18+24}{84}=\frac{1}{2}$$.

If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$, and Rob - $$\frac{1}{3}$$
I interpreted the red marked area as
Probability(B(not win))*P(M(win))=1/4
Probability(B(not win))*P(R(win))=1/3

Any one pls help me understand the above sentence the significance of "If Ben were to lose the championship"
_________________
--If you like my post pls give kudos
Math Expert
Joined: 02 Sep 2009
Posts: 55635

### Show Tags

11 Jul 2018, 20:53
tejyr wrote:
Bunuel wrote:
Official Solution:

If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$, and Rob - $$\frac{1}{3}$$ . If the probability of Ben being the winner is $$\frac{1}{7}$$, what is the probability that either Mike or Rob will win the championship? Assume that there can be only one winner.

A. $$\frac{1}{12}$$
B. $$\frac{1}{7}$$
C. $$\frac{1}{2}$$
D. $$\frac{7}{12}$$
E. $$\frac{6}{7}$$

This is a conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must lose: the probability of Ben losing is $$1-\frac{1}{7}=\frac{6}{7}$$.

Now out of these $$\frac{6}{7}$$ cases the probability of Mike winning is $$\frac{1}{4}$$ and the probability of Rob winning is $$\frac{1}{3}$$. So $$P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}$$.

Or consider the following:

Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).

Now, out of these 84 cases Ben will lose in $$\frac{6}{7}*84=72$$. Mike would be the winner in $$72*\frac{1}{4}=18$$ (1/4 th of the cases when Ben loses) and Rob would be the winner in $$72*\frac{1}{3}=24$$. Therefore $$P=\frac{18+24}{84}=\frac{1}{2}$$.

If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$, and Rob - $$\frac{1}{3}$$
I interpreted the red marked area as
Probability(B(not win))*P(M(win))=1/4
Probability(B(not win))*P(R(win))=1/3

Any one pls help me understand the above sentence the significance of "If Ben were to lose the championship"

That sentence means that IF Ben loses, then the probability of Mike winning is 1/4 and IF Ben loses, then the probability of Ron winning is 1/3:
Mike wining = 6/7*1/4
Rob winning = 6/7*1/3

Mike or Rob wining = 6/7*1/4 + 6/7*1/3
_________________
Manager
Joined: 26 Dec 2017
Posts: 156

### Show Tags

15 Jul 2018, 19:53
1
Bunuel wrote:
tejyr wrote:
Bunuel wrote:
Official Solution:

If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$, and Rob - $$\frac{1}{3}$$ . If the probability of Ben being the winner is $$\frac{1}{7}$$, what is the probability that either Mike or Rob will win the championship? Assume that there can be only one winner.

A. $$\frac{1}{12}$$
B. $$\frac{1}{7}$$
C. $$\frac{1}{2}$$
D. $$\frac{7}{12}$$
E. $$\frac{6}{7}$$

This is a conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must lose: the probability of Ben losing is $$1-\frac{1}{7}=\frac{6}{7}$$.

Now out of these $$\frac{6}{7}$$ cases the probability of Mike winning is $$\frac{1}{4}$$ and the probability of Rob winning is $$\frac{1}{3}$$. So $$P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}$$.

Or consider the following:

Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7).

Now, out of these 84 cases Ben will lose in $$\frac{6}{7}*84=72$$. Mike would be the winner in $$72*\frac{1}{4}=18$$ (1/4 th of the cases when Ben loses) and Rob would be the winner in $$72*\frac{1}{3}=24$$. Therefore $$P=\frac{18+24}{84}=\frac{1}{2}$$.

If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$, and Rob - $$\frac{1}{3}$$
I interpreted the red marked area as
Probability(B(not win))*P(M(win))=1/4
Probability(B(not win))*P(R(win))=1/3

Any one pls help me understand the above sentence the significance of "If Ben were to lose the championship"

That sentence means that IF Ben loses, then the probability of Mike winning is 1/4 and IF Ben loses, then the probability of Ron winning is 1/3:
Mike wining = 7/6*1/4
Rob winning = 7/6*1/3

Mike or Rob wining = 7/6*1/4 + 7/6*1/3

Hi,
I did not get why Mike=7/6*1/4
and why not Mike=6/7*1/4 (Probability(mike win)*Prob(Ben lose))
Can you pls explain.
_________________
--If you like my post pls give kudos
Math Expert
Joined: 02 Sep 2009
Posts: 55635

### Show Tags

15 Jul 2018, 20:28
tejyr wrote:
Bunuel wrote:
tejyr wrote:
If Ben were to lose the championship, Mike would be the winner with a probability of $$\frac{1}{4}$$, and Rob - $$\frac{1}{3}$$
I interpreted the red marked area as
Probability(B(not win))*P(M(win))=1/4
Probability(B(not win))*P(R(win))=1/3

Any one pls help me understand the above sentence the significance of "If Ben were to lose the championship"

That sentence means that IF Ben loses, then the probability of Mike winning is 1/4 and IF Ben loses, then the probability of Ron winning is 1/3:
Mike wining = 7/6*1/4
Rob winning = 7/6*1/3

Mike or Rob wining = 7/6*1/4 + 7/6*1/3

Hi,
I did not get why Mike=7/6*1/4
and why not Mike=6/7*1/4 (Probability(mike win)*Prob(Ben lose))
Can you pls explain.

It's a typo there. Of course it's 6/7 in both cases. The probability cannot be more than 1.
_________________
Intern
Joined: 23 Jan 2018
Posts: 18

### Show Tags

03 Aug 2018, 10:10
Bunuel
Are we not considering the probability(1-1/3) of Rob losing the game when we are taking the winning probability of Mike since Mike's winning probability already include all the possibilities, even that of Rob losing the game?
Is my thinking correct?
_________________
"Winning is all about right strategy"

Set S.M.A.R.T Objectives : Specific | Measurable | Attainable | Relevant | Time-Bound
Math Expert
Joined: 02 Sep 2009
Posts: 55635

### Show Tags

03 Aug 2018, 10:28
Setback wrote:
Bunuel
Are we not considering the probability(1-1/3) of Rob losing the game when we are taking the winning probability of Mike since Mike's winning probability already include all the possibilities, even that of Rob losing the game?
Is my thinking correct?

We are told that there can be only one winner. So, if Mike wins, Rob looses and if Rob wins, Mike looses (automatically).
_________________
Director
Joined: 04 Jun 2018
Posts: 536
Location: Germany
Concentration: General Management, Finance
GPA: 3.4
WE: Analyst (Transportation)

### Show Tags

09 Aug 2018, 13:42
Bunuel wrote:
dyg wrote:
Hello,

I think question should be edited bcs it seems like Rob's probability of winning is -1/3. Minus should be eliminated from the question.

Probability cannot be negative, so it's clearly a dash a not a minus sign.

While I am aware that probability cannot be negative, I was briefly thrown off by the dash as well.
_________________
A couple of things that helped me in verbal:
https://gmatclub.com/forum/verbal-strategies-268700.html#p2082192

Gmat Prep CAT #1: V42, Q34, 630
Gmat Prep CAT #2: V46, Q35, 660
Gmat Prep CAT #3: V41, Q42, 680

On the mission to improve my quant score, all help is appreciated!
Intern
Joined: 14 Aug 2017
Posts: 13

### Show Tags

04 Sep 2018, 22:50
Hi Bunuel,

I understood the official solution. But I have a doubt in the wording of the question, although I landed up with the correct answer , but I wanted to get it confirmed from you.

According to the question " If Ben were to lose the championship, Mike would be the winner with a probability of 1/4, and Rob - 1/3 ", I thought that since only one person can win at at time , P(Mike winning) = 1/4 and P(Rob Losing) = 1/3 are given, so then
P(Rob Winning) = 1- P(Rob Losing)= 1-(1/3) =2/3
and P(Ben Winning) = 1/7, which is already given . So P(Ben Losing) = 1 - P(Ben Winning) = 1-(1/7)= 6/7
Now we have to calculate the probability of winning of either Mike or Rob ,

P(Mike Win) = P(Ben Losing)*P(Mike Winning)*P(Rob Losing) = (6/7)*(1/4)*(1/3)=1/14
P(Rob Win) = P(Ben Losing)*P(Mike Losing)*P(Rob Winning) = (6/7)*(3/4)*(2/3)=3/7

P(Mike Win or Rob Win) = P(Mike Win) + P(Rob Win) = (1/14)+(3/7) = 1/2

The final answer is same i.,e 1/2 . But my interpretation was different as compared to official explanation.
So I wanted to get it confirmed from you. Whether my approach/understanding is correct?

Thanks
Intern
Joined: 15 Nov 2018
Posts: 3

### Show Tags

11 Apr 2019, 19:55
Please change the value for rob from - 1/3 to 1/3. It throws off the whole question.
Re: M07-12   [#permalink] 11 Apr 2019, 19:55

Go to page   Previous    1   2   [ 33 posts ]

Display posts from previous: Sort by

# M07-12

Moderators: chetan2u, Bunuel