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805+ (Hard)|   Probability|      
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mikemcgarry
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On any given day, the probability that Bob will have breakfast is more 03 Mar 2017, 16:04
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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95% (hard)

Question Stats:

43% (02:40) correct 57% (03:03) wrong based on 238 sessions

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On any given day, the probability that Bob will have breakfast is more than 0.6. The probability that Bob will have breakfast and will have a sandwich for lunch is less than 0.5. The probability that Bob will have breakfast or will have a sandwich for lunch equals 0.7. Let P = the probability that, on any given day, Bob will have a sandwich for lunch. If all the statements are true, what possible range can be established for P?

(A) 0 < P < 0.6
(B) 0 ≤ P < 0.6
(C) 0 ≤ P ≤ 0.6
(D) 0 < P < 0.7
(E) 0 ≤ P < 0.7

This is a one of a set of 15 challenging GMAT Quant practice questions. For the whole collection, as well as the OE for this particular question, see:
Challenging GMAT Math Practice Questions

Mike :-)
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On any given day, the probability that Bob will have breakfast is more 04 Mar 2017, 23:53
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I am not the best at probability formulas but I chose A. Since Probability (Breakfast and Sandwich) is not = 0, probability of sandwich alone cannot be zero and only A depicts range > 0.
Show more

Hi,

P(A and B)<0.5, so ofcourse it can be 0 and even P(B) can be 0..
In this case, P(A) will be 0.7..

P(A and B)<0.5...Least 0 and max 4.9999999, just below 5
P(A)>0.6..(I)
P(A or B)=0.7(ll)..
From l and ll, P(A or B)=P(A)+P(B)-p(A and B)..
P(B)=(or)+(and)-P(A)
1) least value when (and) is 0 and P(a) is 0.7..
So P(B)= 0.7+0-0.7=0..
2) max value
When P(A) is just above 0.6 and P(A and B) is just below 0.5
So P(B)= 0.7+0.499999-0.6000001=1.1999999-0.6000001, something<0.6..

So range becomes 0 ≤ P < 0.6

B
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On any given day, the probability that Bob will have breakfast is more 04 Mar 2017, 13:06
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chetan2u
Hi mike,

You may have to relook into the answers ...

P(A and B)<0.5
P(A)>0.6..(I)
P(A or B)=0.7(ll)..
From l and ll, P(B)<0.1
Also if P(A) is 0.7, P(B)=0..
So range becomes 0 ≤ P < 0.1
This is not in choices
Show more
Dear chetan2u,

My intelligent colleague, remember that the probability OR formula is:
P(A or B) = P(A) + P(B) - P(A and B)
Say that P(A) = 0.65 and that P(A and B) = 0.35, both legal values. Then, this equation becomes
0.7 = 0.65 + P(B) - 0.35
0.7 = P(B) + 0.3
0.4 = P(B)
Thus, P(B) certainly can be more than 0.1

Does this make sense?
Mike :-)
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On any given day, the probability that Bob will have breakfast is more 04 Mar 2017, 22:37
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I am not the best at probability formulas but I chose A. Since Probability (Breakfast and Sandwich) is not = 0, probability of sandwich alone cannot be zero and only A depicts range > 0.
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On any given day, the probability that Bob will have breakfast is more 03 Jul 2017, 10:02
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chetan2u
Hi mike,

You may have to relook into the answers ...

P(A and B)<0.5
P(A)>0.6..(I)
P(A or B)=0.7(ll)..
From l and ll, P(B)<0.1
Also if P(A) is 0.7, P(B)=0..
So range becomes 0 ≤ P < 0.1
This is not in choices
Dear chetan2u,

My intelligent colleague, remember that the probability OR formula is:
P(A or B) = P(A) + P(B) - P(A and B)
Say that P(A) = 0.65 and that P(A and B) = 0.35, both legal values. Then, this equation becomes
0.7 = 0.65 + P(B) - 0.35
0.7 = P(B) + 0.3
0.4 = P(B)
Thus, P(B) certainly can be more than 0.1

Does this make sense?
Mike :-)
Show more

I'm ignoring the sandwich for lunch part and breaking this to only breakfast and lunch.

Hi, why is this not considered a mutually exclusive case? What is the condition of having breakfast to having lunch?

So assuming P(a) is breakfast and P(B) is lunch..

are we talking about cases where P(B|A). Meaning, Probablity of somebody having lunch given he already had breakfast ?

P(A and B) = P(A) * P(B|A)

What I am trying to understand is how is this different from lets say, we roll a 2 in a dice and a head in a coin? (This one feels like mutual exclusive), but then how, is having a lunch dependent on having breakfast? Or does it depend upon the question? So, I can turn the "dice and coin" statement into dependent also right? the probablity of getting a head and getting a 2(Assuming getting a 2 is success and anything else a failure).
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On any given day, the probability that Bob will have breakfast is more 03 Jul 2017, 11:01
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SOUMYAJIT_
I'm ignoring the sandwich for lunch part and breaking this to only breakfast and lunch.

Hi, why is this not considered a mutually exclusive case? What is the condition of having breakfast to having lunch?

So assuming P(a) is breakfast and P(B) is lunch..

are we talking about cases where P(B|A). Meaning, Probablity of somebody having lunch given he already had breakfast ?

P(A and B) = P(A) * P(B|A)

What I am trying to understand is how is this different from lets say, we roll a 2 in a dice and a head in a coin? (This one feels like mutual exclusive), but then how, is having a lunch dependent on having breakfast? Or does it depend upon the question? So, I can turn the "dice and coin" statement into dependent also right? the probablity of getting a head and getting a 2(Assuming getting a 2 is success and anything else a failure).
Show more
Dear SOUMYAJIT_,

I'm happy to respond. :-)

It's clear to me that you understand a great deal of math. That's great. I think you need a little clarification on the English terminology.

I. Mutually Exclusive (a.k.a. Disjoint)
If J and K are mutually exclusive, then they never can happen together: the fact that one happens guarantees that the other is impossible. In this case,
P(J and K) = 0
P(J or K) = P(J) + P(K)
Examples of mutually exclusive events would be
(a) getting H and T on a single coin flip
(b) rolling a 3 and 6 in a single die role
(c) a single day is both Tuesday and Friday

II. Independent
If J and K are independent, then whether J happens has absolutely no influence on whether K happens. If I have information about J, that tells me absolutely nothing about K, and vice versa. If J and K are independent quantitative variables, then they have zero correlation. In this case,
P(J and K) = P(J)P(K)
Examples of independent events would be
(a) two separate coins
(b) two separate dice
(c) the day of the week and inches of rain that day
(d) inches of rain on a given day and the net gain/loss of the Dow Jone Industrial Average
Those are events that are independent.

For more on the terminology, see:
GMAT Math: Probability Rules

When two events are not mutually exclusive and not independent, then we have to use conditional probabilities for the "and" rule.
P(B) = P(A)*P(B|A)
This is a little more advanced than the GMAT expects you to know, but if this helps you decipher a problem, so much the better.

Now, in this problem, breakfast and lunch certainly aren't mutually exclusive. I think what you meant to ask whether we could regard breakfast and lunch as independent.

As a general rule, the ideas of "mutually exclusive" and "independent" are clean, tidy, mathematical ideas. They describe very simple systems, such as coins & dice. Human beings are messy. Everything is interconnected with everything else in ways that defy neat mathematical conventions. As a general rule, unless you are told something quite specifically, you never should assume one of these neat categories applies to human behavior.

In this particular case, think about it. Suppose somebody tells us that today Bob did not have breakfast. Of course, that means he probably will be a little hungrier by midday than when he does have breakfast. In other words, we have at least some information about (a) whether he will have lunch, and (b) what kind of lunch he might have. As long as one variable gives any kind of information, any kind of preferential guess, about the other variable, they are not independent.

Does all this make sense?
Mike :-)
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On any given day, the probability that Bob will have breakfast is more 03 Jul 2017, 11:21
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What's the answer for this one?


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On any given day, the probability that Bob will have breakfast is more 03 Jul 2017, 11:25
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What's the answer for this one?


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You can check Official Answers (OAs) under the spoiler in the first post of a topic. For this question OA is B.
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On any given day, the probability that Bob will have breakfast is more 03 Jul 2017, 22:02
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SOUMYAJIT_
I'm ignoring the sandwich for lunch part and breaking this to only breakfast and lunch.

Hi, why is this not considered a mutually exclusive case? What is the condition of having breakfast to having lunch?

So assuming P(a) is breakfast and P(B) is lunch..

are we talking about cases where P(B|A). Meaning, Probablity of somebody having lunch given he already had breakfast ?

P(A and B) = P(A) * P(B|A)

What I am trying to understand is how is this different from lets say, we roll a 2 in a dice and a head in a coin? (This one feels like mutual exclusive), but then how, is having a lunch dependent on having breakfast? Or does it depend upon the question? So, I can turn the "dice and coin" statement into dependent also right? the probablity of getting a head and getting a 2(Assuming getting a 2 is success and anything else a failure).
Dear SOUMYAJIT_,

I'm happy to respond. :-)

It's clear to me that you understand a great deal of math. That's great. I think you need a little clarification on the English terminology.

I. Mutually Exclusive (a.k.a. Disjoint)
If J and K are mutually exclusive, then they never can happen together: the fact that one happens guarantees that the other is impossible. In this case,
P(J and K) = 0
P(J or K) = P(J) + P(K)
Examples of mutually exclusive events would be
(a) getting H and T on a single coin flip
(b) rolling a 3 and 6 in a single die role
(c) a single day is both Tuesday and Friday

II. Independent
If J and K are independent, then whether J happens has absolutely no influence on whether K happens. If I have information about J, that tells me absolutely nothing about K, and vice versa. If J and K are independent quantitative variables, then they have zero correlation. In this case,
P(J and K) = P(J)P(K)
Examples of independent events would be
(a) two separate coins
(b) two separate dice
(c) the day of the week and inches of rain that day
(d) inches of rain on a given day and the net gain/loss of the Dow Jone Industrial Average
Those are events that are independent.

For more on the terminology, see:
GMAT Math: Probability Rules

When two events are not mutually exclusive and not independent, then we have to use conditional probabilities for the "and" rule.
P(B) = P(A)*P(B|A)
This is a little more advanced than the GMAT expects you to know, but if this helps you decipher a problem, so much the better.

Now, in this problem, breakfast and lunch certainly aren't mutually exclusive. I think what you meant to ask whether we could regard breakfast and lunch as independent.

As a general rule, the ideas of "mutually exclusive" and "independent" are clean, tidy, mathematical ideas. They describe very simple systems, such as coins & dice. Human beings are messy. Everything is interconnected with everything else in ways that defy neat mathematical conventions. As a general rule, unless you are told something quite specifically, you never should assume one of these neat categories applies to human behavior.

In this particular case, think about it. Suppose somebody tells us that today Bob did not have breakfast. Of course, that means he probably will be a little hungrier by midday than when he does have breakfast. In other words, we have at least some information about (a) whether he will have lunch, and (b) what kind of lunch he might have. As long as one variable gives any kind of information, any kind of preferential guess, about the other variable, they are not independent.

Does all this make sense?
Mike :-)
Show more


Very Nice explanation!! Thanks!! You should know that I bumped into this SUM while doing the quizzes in Magoosh! Thanks a ton!!
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On any given day, the probability that Bob will have breakfast is more 17 Dec 2019, 13:13
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chetan2u
jedit
I am not the best at probability formulas but I chose A. Since Probability (Breakfast and Sandwich) is not = 0, probability of sandwich alone cannot be zero and only A depicts range > 0.

Hi,

P(A and B)<0.5, so ofcourse it can be 0 and even P(B) can be 0..
In this case, P(A) will be 0.7..

P(A and B)<0.5...Least 0 and max 4.9999999, just below 5
P(A)>0.6..(I)
P(A or B)=0.7(ll)..
From l and ll, P(A or B)=P(A)+P(B)-p(A and B)..
P(B)=(or)+(and)-P(A)
1) least value when (and) is 0 and P(a) is 0.7..
So P(B)= 0.7+0-0.7=0..
2) max value
When P(A) is just above 0.6 and P(A and B) is just below 0.5
So P(B)= 0.7+0.499999-0.6000001=1.1999999-0.6000001, something<0.6..

So range becomes 0 ≤ P < 0.6

B
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Hey, in the second line of your solution, how did you derive the fact that P(A) will be 0.7?
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On any given day, the probability that Bob will have breakfast is more 28 Apr 2020, 04:45
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chetan2u
jedit
I am not the best at probability formulas but I chose A. Since Probability (Breakfast and Sandwich) is not = 0, probability of sandwich alone cannot be zero and only A depicts range > 0.

Hi,

P(A and B)<0.5, so ofcourse it can be 0 and even P(B) can be 0.
In this case, P(A) will be 0.7..

P(A and B)<0.5...Least 0 and max 4.9999999, just below 5
P(A)>0.6..(I)
P(A or B)=0.7(ll)..
From l and ll, P(A or B)=P(A)+P(B)-p(A and B)..
P(B)=(or)+(and)-P(A)
1) least value when (and) is 0 and P(a) is 0.7..
So P(B)= 0.7+0-0.7=0..
2) max value
When P(A) is just above 0.6 and P(A and B) is just below 0.5
So P(B)= 0.7+0.499999-0.6000001=1.1999999-0.6000001, something<0.6..

So range becomes 0 ≤ P < 0.6

B

Hey, in the second line of your solution, how did you derive the fact that P(A) will be 0.7?
Show more

Hi fireflyejd1,

General OR formula for two events A, B is given by:
P(A OR B)=P(A)+P(B)-P(A AND B)

if P(A AND B)=0 and P(B)=0, the above equation yields P(A)=0.7.
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