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## Comment on

General GMAT Probability Strategies## Hi Brent, Can you help me

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

"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

## 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

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

## Here are a couple of examples

Here are a couple of examples:

http://www.beatthegmat.com/og-on-saturday-morning-malachi-will-begin-cam...

http://www.beatthegmat.com/probablity-t206270.html

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

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 ?

Please let me know

## 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 !

## 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

I assumed P(Getting job from A & Getting Job from B) = 1-P(Not Getting job from A & not getting job from B)

Isn't P(Doing a task X) = 1 - P(Not doing a task X)? Please help me understand the flaw in my reasoning.

## 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

## That's correct.

That's correct.

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

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

## Question link: https:/

Question link: https://gmatclub.com/forum/if-two-points-a-and-b-are-randomly-placed-on-...

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

## 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,

needed help with this-

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

thank you

tanya

## Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/each-of-the-45-books-on-a-shelf-is-written-21...

## 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

## Question link: https:/

Question link: https://gmatclub.com/forum/a-box-contains-100-balls-numbered-from-1-to-1...

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.

## https://gmatclub.com/forum

Help on this question please ? :)

## Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/triplets-adam-bruce-and-charlie-enter-a-triat...

## https://gmatclub.com/forum

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

## Question link: https:/

Question link: https://gmatclub.com/forum/mary-and-joe-are-to-throw-three-dice-each-the...

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

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