# Lesson: General GMAT Probability Strategies

## Comment on General GMAT Probability Strategies

### Hi Brent, Can you help me

Hi Brent, Can you help me understand the following question:

The bear alarm at Grizzly’s Peak ski resort sounds an average of once every thirty days, but the alarm is so sensitively calibrated that it sounds an average of ten false alarms for every undetected bear. Despite this, the alarm only sounds for three out of four bears that actually appear at the resort.

1. If the alarm sounds, what is the probability that a bear has actually been sighted?
2. In any given day at the resort, what is the approx. probability that there is neither an alarm nor an undetected bear?
3. If the alarm were to sound an average of ten false alarms for every detected bear, the probability that a sounded alarm would indicate an actual bear would be?
4. Approx. how many bears appear in the resort each year?

### The bear alarm at Grizzly’s

NOTE: I'm not crazy about the wording of any of these question, but here goes....

QUESTION 1

The bear alarm at Grizzly’s Peak ski resort sounds an average of once every thirty days, but the alarm is so sensitively calibrated that it sounds an average of ten false alarms for every undetected bear. Despite this, the alarm only sounds for three out of four bears that actually appear at the resort.

1. If the alarm sounds, what is the probability that a bear has actually been sighted?

The alarm only sounds for three out of four bears that actually appear at the resort.

So, for every 4 bears, 3 are detected by the alarm system and 1 is not.

Total number of alarms for the 3 detected bears = 3.

There are 10 false alarms for every undetected bear. So, the number of alarms from the 1 undetected bear = 10.

Total number of alarms = 13

So, for every 13 alarms, 3 are real and 10 are false.

P(an alarm = an actual bear) = 3/13

### QUESTION 2

QUESTION 2

This isn't a very good question. I'd skip it.
More here: https://gmatclub.com/forum/the-bear-alarm-at-grizzly-s-peak-ski-resort-s...

### Where did we get the 14 here

Where did we get the 14 here from?

"We know the alarm goes off once every 30 days, so the alarm would go off 13 times in 390 days. We just found out that there will be 1 undetected bear if the alarm goes off 13 times. So in a 390 day period, the alarm will go off 13 times, and there will be 1 undetected bear, so on 14 days there will be an alarm or an undetected bear, and thus on 376 days there will be neither. So the answer is 376/390."

### I don't think there's much

I don't think there's much benefit in examining question #2 any further. It's not a great representation of what you'll see on test day (which is probably why Veritas pulled the question from circulation).

### QUESTION 3

QUESTION 3
The bear alarm at Grizzly’s Peak ski resort sounds an average of once every thirty days, but the alarm is so sensitively calibrated that it sounds an average of ten false alarms for every undetected bear. Despite this, the alarm only sounds for three out of four bears that actually appear at the resort.

3. If the alarm were to sound an average of ten false alarms for every detected bear, the probability that a sounded alarm would indicate an actual bear would be?

Go back to the information from question 1

For every 4 bears, 3 are detected by the alarm system and 1 is not.

Total number of alarms for the 3 detected bears = 3.

There are 10 false alarms for every DETECTED bear. So, the number of alarms from the 3 undetected bears = 30.

Total number of alarms = 33

So, for every 33 alarms, 3 are real and 30 are false.

P(an alarm = an actual bear) = 3/33 = 1/11

### But the question says that

But the question says that for every UNDETECTED bear there are 10 false alarms.

### Yes, that's what the original

Yes, that's what the original question states. However, question 3 asks: "If the alarm were to sound an average of ten false alarms for every DETECTED bear, the probability that a sounded alarm would indicate an actual bear would be?"

### QUESTION 4

QUESTION 4

The bear alarm at Grizzly’s Peak ski resort sounds an average of once every thirty days, but the alarm is so sensitively calibrated that it sounds an average of ten false alarms for every undetected bear. Despite this, the alarm only sounds for three out of four bears that actually appear at the resort.

4. Approx. how many bears appear in the resort each year?

From question #1, we see that, for every 13 alarms, 3 are real and 10 are false.

The question tells us that there is 1 alarm every 30 days.

So, we'll get 13 alarms in 390 days.

In those 390 days (with 13 alarms), there are 3 detected bears and 1 undetected bear.

So, in 390 days, a total of 4 bears visit the resort.

So, in 365 days, there will be a slightly fewer than 4 bear visits.

### Hi Brent-can you give example

Hi Brent-can you give example where outcomes are not equally likely?

### https://gmatclub.com/forum/if

https://gmatclub.com/forum/if-two-points-a-and-b-are-randomly-placed-on-the-circumference-234196.html

Confused. My approach -
Chord = line thru the circle and longest chord = diameter.

Diameter = 12 hence chords can be 1 - 12.
So essentially the problem is - what is the probability to get at least 6 from a set of numbers from 1 - 12.
Which should be
P(6) ... P(12) = 7/12
Where did I go wrong ?

### I love this approach. The

I love this approach!! The only problem is that each possible chord length (from 0 to 12) is not equally likely.

For example, if we randomly choose a second point, we are twice as likely to get a chord that is greater than 6 than less than 6.

So, even though the chord length can range from 0 to 12, we cannot assume that each outcome (chord length) is equally likely.

Here's an analogous case: Harry, a teenager from Canada, wants to determine the probability that he will marry German Chancellor Angela Merkel some day.

There are two possible outcomes:
1) Harry will marry Angela Merkel some day
2) Harry will NOT marry Angela Merkel some day

Since there are 2 possible outcomes, P(Harry marries Angela Merkel) = 1/2

As we might expect, the two outcomes are not equally likely. So, we can't use this approach.

### Thanks that helps !

Thanks that helps !

### Hi Brent,

Hi Brent,

Please refer the Q : https://gmatclub.com/forum/jill-has-applied-for-a-job-with-each-of-two-different-companies-what-220335.html

Could you please explain a detailed solution using the techniques taught by you? I am getting really confused with "AT LEAST, "AT MOST" and "EXACTLY"

Thanks & Regards,
Abhirup

### Phew, that one took a while!

Phew, that one took a while!
Here's my full solution: https://gmatclub.com/forum/jill-has-applied-for-a-job-with-each-of-two-d...

Cheers,
Brent

### Thanks Brent for that

Thanks Brent for that wonderful insight.
I assumed P(Getting job from A & Getting Job from B) = 1-P(Not Getting job from A & not getting job from B)

### You're correct to say that P

You're correct to say that P(Doing a task X) = 1 - P(Not doing a task X)

However, the task in this case involves TWO events.

Let's say that a certain club will only admit people who are left-handed AND own a motorcycle.

So, if Joe is admitted to the club, we know that Joe is left-handed AND he owns a motorcycle.

What if Joe is NOT admitted to the club?
What can we conclude?
We know that Joe did not meet the admission criteria (which requires left-handedness AND motorcycle ownership)

So, ONE possibility is that Joe is RIGHT-handed, and he does NOT own a motorcycle.
However, there are 2 more possibilities:
- Joe is RIGHT-handed AND he owns a motorcycle.
- Joe is left-handed AND he does NOT own a motorcycle.

The same applies to the question we're talking about.

Does that help?

Cheers,
Brent

### So that means to get P(Job

So that means to get P(Job offers from both companies) we have to eliminate all the probabilities that have either company offering job or neither offering job i.e. 1 -[ P(job from A but not B) +P(job from B but not A) + P(no job from both) ]. This is how we achieve P(Not doing the task) in this case.

That's correct.

### https://gmatclub.com/forum/if

https://gmatclub.com/forum/if-two-points-a-and-b-are-randomly-placed-on-the-circumference-234196.html

I do not understand this solution. Why are we taking ratio of angles.

Great question!

KEY CONCEPT: the length of a circle's arc is proportional to its central angle.
For example, if we have a circle with radius 10, then an arc with a central angle of 180° (i.e., half the circle) will be TWICE the length of an arc with a central angle of 90° (i.e., one-quarter) of the circle.

So, rather than compare the lengths of the desired arcs, we can just compare the central angles.
That said, if we were to calculate the arc lengths instead, we'd arrive at the same probability.

Does that help?

Cheers,
Brent

### Hi Brent, would you mind

Hi Brent, would you mind explaining again the difference between mutually and not mutually exclusive events please? Thanks very much x

### Those two concepts are

Those two concepts are covered here: https://www.gmatprepnow.com/module/gmat-probability/video/746

Once you review those videos, please let me know whether you need an clarification.

Cheers,
Brent

### hello sir,

hello sir,
needed help with this-

https://gmatclub.com/forum/each-of-the-45-books-on-a-shelf-is-written-210669.html

thank you
tanya

### Hi Brent,

Hi Brent,

https://gmatclub.com/forum/a-box-contains-100-balls-numbered-from-1-to-100-if-three-b-109279.html

I am not able to understand why order would matter of 'ooo' or 'oee'. These type of question trick me, I am not able to understand when order matters and when it doesn't

Thanks

If we're going to use counting techniques to solve a probability question, we must ensure that all of the possible outcomes are EQUALLY LIKELY.

Here's an extreme example:
A box contains 49 block balls and 1 white ball. If one ball is randomly chosen from the box, what is the probability that the selected ball is white?

One (incorrect) approach would be to say that there are only two possible outcomes:
1) The selected ball is white
2) The selected ball is black
So, P(ball is white) = 1/2
The problem with this approach is that the two possible outcomes that we listed are not equally likely.

The same goes for this question.
If we say that order does not matter, then we have exactly four possible outcomes:
Outcome #1) All 3 numbers are even
Outcome #2) 2 of the 3 numbers are even
Outcome #3) 1 of the 3 numbers is even
Outcome #4) 0 of the 3 numbers are even

Notice that these 4 outcomes are not equally likely.
For example, there's only 1 way to achieve Outcome #1: EEE
But, there are 3 ways to achieve Outcome #2: EEO, EOE, and OEE

Does that help?

### Thanks, understood.

Thanks, understood.

### https://gmatclub.com/forum

Help on this question please ? :)

### https://gmatclub.com/forum

https://gmatclub.com/forum/mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-86407.html

Hi Brent,

I solved the above question using 1-p(event not happening) and was able to get the right answer. Could you maybe elaborate another method ? Or the same method, i feel it was a stroke of luck i got the correct answer.

Thanks

This is a very very tricky question. Good for you to solve the question using the complement! I'm sure that approach took a lot of time and effort.

The only way to answer this question in a reasonable amount of time is to recognize the symmetry involved with various sums (as Bunuel did here: https://gmatclub.com/forum/mary-and-joe-are-to-throw-three-dice-each-the...)

I'll take a second to elaborate on his approach.

First recognize that there is only 1 way to get a sum of 3: 1-1-1
Also recognize that there is only 1 way to get a sum of 18: 6-6-6

Now recognize that there are 3 ways to get a sum of 4: 1-1-2, 1-2-1 and 2-1-1
Also recognize that there are 3 ways to get a sum of 17: 6-6-5, 6-5-6 add 5-6-6

Similarly, there are 6 ways to get a sum of 5: 1-1-3, 1-3-1, 3-1-1, 2-2-1, 2-1-2 and 1-2-2
And there are 6 ways to get a sum of 16: 6-6-4, 6-4-6, 4-6-6, 5-5-6, 5-6-5 and 6-5-5

And so the symmetry goes.
More importantly, the "line" of symmetry is at 10.5.
So, the number of ways to get a sum of 10 is equal to the number of ways to get a sum of 11
And the number of ways to get a sum of 9 is equal to the number of ways to get a sum of 12
etc.

So, P(sum is greater than 10.5) = P(sum is less than 10.5) = 1/2

Cheers,
Brent

### Hi Brent,

Hi Brent,
Just to make sure that I understood everything correctly, can you explain again what do you mean by "equally likely outcomes?" or could you give a counterexample where its not the case?
Thank you!!

### Here are some examples:

Here are some examples:

A box contains 4 balls labelled A, B, C and D.
If we select one ball at random, the four possible outcomes are:
1) Ball A is selected
2) Ball B is selected
3) Ball C is selected
4) Ball D is selected
These 4 possible outcomes are equally likely
So, P(A is selected) = P(B is selected) = P(C is selected) = P(D is selected) = 1/4

A different example:
I'm thinking about going visit Deep Lake tomorrow.
The two possible outcomes are:
1) A giant meteorite strikes Deep Lake tomorrow and destroys everything.
2) A giant meteorite does NOT strike Deep Lake tomorrow.
So, P(A giant meteorite strikes Deep Lake and destroys everything) = 1/2
In this case, the probability is incorrect since the two possible outcomes are not equally likely.

Does that help?

### Hi Brent,

Hi Brent,
I found the following question interesting. https://gmatclub.com/forum/a-box-has-4-apples-and-3-oranges-robert-picks-fruits-randomly-one-aft-346378.html

In most of the cases, we always calculate the probability of occurrence of first event, second event and third event, the probability is P(1)*P(2)*P(3) etc.

In this question, P(selecting an apple) is given. P(selecting the third fruit as an apple is given). Since they don't really care about what is selected in the first and second pick, it seems that just the probability of picking the third apple alone is the probability after the first two are picked.

I was confused where i assumed that the probability of picking the third fruit as an apple was actually P(1)*P(2)*P(3) in four different ways and adding them.

Thanks
Rahul

### That's not a very good

That's not a very good question.
Ian (https://gmatclub.com/forum/a-box-has-4-apples-and-3-oranges-robert-picks...) has a nice response.

It sounds like Bunuel (the person who made up the question for GMAT Club) is talking about the probability that the third piece of fruit is an apple GIVEN THAT we've already chosen 2 pieces of fruit AND we know what those two pieces of fruit were. However that's not really explained anywhere.

I suggest you just ignore this question.

thank you Brent.