Probability Help

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

fangtray: Master | Next Rank: 500 Posts; Posts: 273; Joined: Thu Sep 08, 2011 6:50 am; Thanked: 5 times; Followed by:3 members

Probability Help

by fangtray » Tue Feb 28, 2012 6:10 am

Hello, Could someone please help me solve the problems below without using the 1-x probability strategy?

In order to understand how probability works, I want to see how both ways can solve the problem. I'm having a tough time grasping the explanations.

1) A salesperson makes 5 sales calls. The probability of a successful sales call is 10%. What is the likelyhood the salesperson makes at least 1 successful sales call?

2) What is the probability that, on three rolls of a single die, AT LEAST ONE of the rolls with be a six?

thanks so much

ray

Quote

Source: Beat The GMAT — Problem Solving | Tue Feb 28, 2012 6:10 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Tue Feb 28, 2012 9:20 am

fangtray wrote: 2) What is the probability that, on three rolls of a single die, AT LEAST ONE of the rolls with be a six?

thanks so much

ray

P(at least one 6) = P(exactly one 6) + P(exactly two 6's) + P(exactly three 6's).

P(exactly one 6):
P(6 on the first roll) = 1/6.
P(not 6 on the second roll) = 5/6.
P(not 6 on the third roll) = 5/6.
Since we want these events to happen together, we multiply the fractions:
1/6 * 5/6 * 5/6.
Since the one 6 could happen on the first, second or third roll, we multiply by 3:
1/6 * 5/6 * 5/6 * 3 = 75/216.

P(exactly two 6's):
P(6 on the first roll) = 1/6.
P(6 on the second roll) = 1/6.
P(not 6 on the third roll) = 5/6.
Since we want these events to happen together, we multiply the fractions:
1/6 * 1/6 * 5/6.
Since the one "not 6" could happen on the first, second or third roll, we multiply by 3:
1/6 * 1/6 * 5/6 * 3 = 15/216.

P(exactly three 6's):
P(6 on the first roll) = 1/6.
P(6 on the second roll) = 1/6.
P(6 on the third roll) = 1/6.
Since we want these events to happen together, we multiply the fractions:
1/6 * 1/6 * 1/6 = 1/216.

P(at least one 6) = 75/216 + 15/216 + 1/216 = 91/216.

Quicker method:
P(at least one 6) = 1 - P(no 6's).

P(no 6's):
P(not 6 on the first roll) = 5/6.
P(not 6 on the second roll) = 5/6.
P(not 6 on the third roll) = 5/6.
Since we want these events to happen together, we multiply the fractions:
5/6 * 5/6 * 5/6 = 125/216.

P(at least one 6) = 1 - 125/216 = 91/216.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Quote

fangtray: Master | Next Rank: 500 Posts; Posts: 273; Joined: Thu Sep 08, 2011 6:50 am; Thanked: 5 times; Followed by:3 members

by fangtray » Tue Feb 28, 2012 3:26 pm

how does that plug into this formula P(A or B) = P(A) + P(B) - P(A and B)

thanks so much

Quote

krusta80: Master | Next Rank: 500 Posts; Posts: 143; Joined: Mon Mar 14, 2011 3:13 am; Thanked: 34 times; Followed by:5 members

by krusta80 » Tue Feb 28, 2012 3:46 pm

fangtray wrote:how does that plug into this formula P(A or B) = P(A) + P(B) - P(A and B)

thanks so much

Using the above formula will work for the die problem as follows:

Let A denote the probability of rolling a 6 on the first roll.
Let B denote the probability of rolling a 6 on the second roll.
Let C denote the probability of rolling a 6 on the third roll.

P(A or B or C) therefore denotes the probability or rolling at least one six. So, using the three-event version of your formula we get:

P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C)

P(A) = P(B) = P(C) = 1/6
P(A and B) = P(A and C) = P(B and C) = 1/36
P(A and B and C) = 1/216

Now we can plug it all in to get...
P(A or B or C ) = 1/6 + 1/6 + 1/6 - 1/36 - 1/36 - 1/36 + 1/216 = 91/216

-----------------------------------------------------
Hopefully, you can see how this can quickly get out of hand for more events (ie. your other example with five sales calls). When this happens, there is always going to be a shortcut way of doing it. Always keep the 1-P(no sixes) method ready to go when that happens.

Quote

GMAT/MBA Expert

Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pm; Location: Milpitas, CA; Thanked: 1854 times; Followed by:523 members; GMAT Score:770

by Anurag@Gurome » Tue Feb 28, 2012 10:05 pm

What is the probability that, on three rolls of a single die, AT LEAST ONE of the rolls with be a six?

Probability of getting at least one six on three rolls = 1 - probability of NOT getting a six on three rolls
Probability of NOT getting a six on three rolls = 5/6 * 5/6 * 5/6 = 125/216
Therefore, probability of getting at least one six on three rolls = 1 - 125/216 = [spoiler]91/216[/spoiler]

Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/

Quote

GMAT/MBA Expert

Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pm; Location: Milpitas, CA; Thanked: 1854 times; Followed by:523 members; GMAT Score:770

by Anurag@Gurome » Wed Feb 29, 2012 12:52 am

1) A salesperson makes 5 sales calls. The probability of a successful sales call is 10%. What is the likelyhood the salesperson makes at least 1 successful sales call?

Probability of a successful sales call = 10% = 0.1
So, probability of an unsuccessful sales call = 90% = 0.09
Therefore, probability that the salesperson makes at least 1 successful sales call = 1 - (0.09)^5, which can be solved further.

Quote

fangtray: Master | Next Rank: 500 Posts; Posts: 273; Joined: Thu Sep 08, 2011 6:50 am; Thanked: 5 times; Followed by:3 members

by fangtray » Wed Feb 29, 2012 5:32 am

krusta80 wrote:
fangtray wrote:how does that plug into this formula P(A or B) = P(A) + P(B) - P(A and B)

thanks so much
Using the above formula will work for the die problem as follows:

Let A denote the probability of rolling a 6 on the first roll.
Let B denote the probability of rolling a 6 on the second roll.
Let C denote the probability of rolling a 6 on the third roll.

P(A or B or C) therefore denotes the probability or rolling at least one six. So, using the three-event version of your formula we get:

P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C)

P(A) = P(B) = P(C) = 1/6
P(A and B) = P(A and C) = P(B and C) = 1/36
P(A and B and C) = 1/216

Now we can plug it all in to get...
P(A or B or C ) = 1/6 + 1/6 + 1/6 - 1/36 - 1/36 - 1/36 + 1/216 = 91/216

-----------------------------------------------------
Hopefully, you can see how this can quickly get out of hand for more events (ie. your other example with five sales calls). When this happens, there is always going to be a shortcut way of doing it. Always keep the 1-P(no sixes) method ready to go when that happens.

Krusta,

Could you help me clarify? How come you do it in such a way that is different from the original formula?

Quote

krusta80: Master | Next Rank: 500 Posts; Posts: 143; Joined: Mon Mar 14, 2011 3:13 am; Thanked: 34 times; Followed by:5 members

by krusta80 » Wed Feb 29, 2012 1:22 pm

fangtray wrote:
krusta80 wrote:
fangtray wrote:how does that plug into this formula P(A or B) = P(A) + P(B) - P(A and B)

thanks so much
Using the above formula will work for the die problem as follows:

Let A denote the probability of rolling a 6 on the first roll.
Let B denote the probability of rolling a 6 on the second roll.
Let C denote the probability of rolling a 6 on the third roll.

P(A or B or C) therefore denotes the probability or rolling at least one six. So, using the three-event version of your formula we get:

P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C)

P(A) = P(B) = P(C) = 1/6
P(A and B) = P(A and C) = P(B and C) = 1/36
P(A and B and C) = 1/216

Now we can plug it all in to get...
P(A or B or C ) = 1/6 + 1/6 + 1/6 - 1/36 - 1/36 - 1/36 + 1/216 = 91/216

-----------------------------------------------------
Hopefully, you can see how this can quickly get out of hand for more events (ie. your other example with five sales calls). When this happens, there is always going to be a shortcut way of doing it. Always keep the 1-P(no sixes) method ready to go when that happens.
Krusta,

Could you help me clarify? How come you do it in such a way that is different from the original formula?

The original formula is for finding the "or" of two events: A,B
My formula is for three events: A,B,C

As I said in my previous post, it really is a waste of time to go beyond the formula for three events, so I won't go into any more detail. Needless to say, it is posible to extend your formula to as many events as you want.

So, assuming you take the formula I gave you as fact, do you understand how I applied it?

Quote

fangtray: Master | Next Rank: 500 Posts; Posts: 273; Joined: Thu Sep 08, 2011 6:50 am; Thanked: 5 times; Followed by:3 members

by fangtray » Fri Mar 02, 2012 6:34 am

krusta80 wrote:
fangtray wrote:
krusta80 wrote:
fangtray wrote:how does that plug into this formula P(A or B) = P(A) + P(B) - P(A and B)

thanks so much
Using the above formula will work for the die problem as follows:

Let A denote the probability of rolling a 6 on the first roll.
Let B denote the probability of rolling a 6 on the second roll.
Let C denote the probability of rolling a 6 on the third roll.

P(A or B or C) therefore denotes the probability or rolling at least one six. So, using the three-event version of your formula we get:

P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(A and B and C)

P(A) = P(B) = P(C) = 1/6
P(A and B) = P(A and C) = P(B and C) = 1/36
P(A and B and C) = 1/216

Now we can plug it all in to get...
P(A or B or C ) = 1/6 + 1/6 + 1/6 - 1/36 - 1/36 - 1/36 + 1/216 = 91/216

-----------------------------------------------------
Hopefully, you can see how this can quickly get out of hand for more events (ie. your other example with five sales calls). When this happens, there is always going to be a shortcut way of doing it. Always keep the 1-P(no sixes) method ready to go when that happens.
Krusta,

Could you help me clarify? How come you do it in such a way that is different from the original formula?
The original formula is for finding the "or" of two events: A,B
My formula is for three events: A,B,C

As I said in my previous post, it really is a waste of time to go beyond the formula for three events, so I won't go into any more detail. Needless to say, it is posible to extend your formula to as many events as you want.

So, assuming you take the formula I gave you as fact, do you understand how I applied it?

no i'm not questioning whether you are right or wrong. I just don't understand how you applied it and why you applied it that way.

From the formula i thought it might work like P(A or B or C) = P(A) + P(B) + P(C) - P(A and B and C)

at least from the formula it looks as if it should work that way

Quote

krusta80: Master | Next Rank: 500 Posts; Posts: 143; Joined: Mon Mar 14, 2011 3:13 am; Thanked: 34 times; Followed by:5 members

by krusta80 » Fri Mar 02, 2012 7:06 am

I'm glad that you are asking these questions; this is actually at the core of all questions involving sets. I'm not sure whether you're familiar with Venn Diagrams, but they are a visual representation of groups of things (or when probabilities are concerned: all possibilities for a given set of events). They are a really big help when trying to conceptualize formulas like the one I gave you.

Let's keep with the dice example. First let's assume there are two dice: A and B. Each die has six possible values: 1 through 6. Below is a Venn diagram representing the event of rolling two dice:

Imagine that the red and blue circles are slightly transparent and laying on top of one another. This is why the middle is purple. To find the area of the top of this figure, we simply find the area of each circle and subtract the overlapped region. This is because the area of each circle takes the overlapped region into account, but we only want to count it once.

This is at the heart of the P(A or B) formula. If we let each circle represent the entire number of possibilities for A and for B, then the overlapped region represents (A and B). Further, in order to figure out P(A or B), we can simply add P(A) to P(B) and then subtract P(A and B), which represents the extra overlapped region.

Now, taking this concept to three events, the Venn diagram changes to:

Again, our goal for finding P(A or B or C) is to count each overlapped region exaclty once...

P(A)+P(B)+P(C) counts the red, yellow, and blue regions once. But what about the purple, orange, and green regions? These were each counted twice (as in the last example), so let's subtract the intersection of each pair.

P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(A and C)

OK, so now the purple, green, and orange sections are each counted once. That leaves us with the brown section, which is P(A and B and C). We counted this section three times when counting each circle once but then removed this section three times when removing each two-circle overlapped region once. Therefore, we need to add it back and then we are done!

P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(A and C) + P(A and B and C)

Quote

fangtray: Master | Next Rank: 500 Posts; Posts: 273; Joined: Thu Sep 08, 2011 6:50 am; Thanked: 5 times; Followed by:3 members

by fangtray » Sat Mar 03, 2012 4:00 am

krusta80 wrote:I'm glad that you are asking these questions; this is actually at the core of all questions involving sets. I'm not sure whether you're familiar with Venn Diagrams, but they are a visual representation of groups of things (or when probabilities are concerned: all possibilities for a given set of events). They are a really big help when trying to conceptualize formulas like the one I gave you.

Let's keep with the dice example. First let's assume there are two dice: A and B. Each die has six possible values: 1 through 6. Below is a Venn diagram representing the event of rolling two dice:

Imagine that the red and blue circles are slightly transparent and laying on top of one another. This is why the middle is purple. To find the area of the top of this figure, we simply find the area of each circle and subtract the overlapped region. This is because the area of each circle takes the overlapped region into account, but we only want to count it once.

This is at the heart of the P(A or B) formula. If we let each circle represent the entire number of possibilities for A and for B, then the overlapped region represents (A and B). Further, in order to figure out P(A or B), we can simply add P(A) to P(B) and then subtract P(A and B), which represents the extra overlapped region.

Now, taking this concept to three events, the Venn diagram changes to:

Again, our goal for finding P(A or B or C) is to count each overlapped region exaclty once...

P(A)+P(B)+P(C) counts the red, yellow, and blue regions once. But what about the purple, orange, and green regions? These were each counted twice (as in the last example), so let's subtract the intersection of each pair.

P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(A and C)

OK, so now the purple, green, and orange sections are each counted once. That leaves us with the brown section, which is P(A and B and C). We counted this section three times when counting each circle once but then removed this section three times when removing each two-circle overlapped region once. Therefore, we need to add it back and then we are done!

P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(A and C) + P(A and B and C)

thanks for the detailed explanation krusta. i think i almost got it. but why do you need to add the brown one last time? is it because with 3 circles, you've never counted for it when you add the brown 3 times, and then subtract it 3 times?

whereas in 2 circles , you add the overlap twice, so you subtract it once?

Quote

krusta80: Master | Next Rank: 500 Posts; Posts: 143; Joined: Mon Mar 14, 2011 3:13 am; Thanked: 34 times; Followed by:5 members

by krusta80 » Sat Mar 03, 2012 8:57 am

OK, I'm going to take the visual a bit further now to really drive home the point. Let's say that you have three coins (or poker chips or any other disc-shaped item).

First for the two-part formula, take two of the coins and overlap them as per my diagram. Now, take a permanent marker and outline where the top circle overlaps the bottom. The bottom coin will now have the purple shape outlined on it after separating the two coins.

Now let's say you have a laser or really powerful knife that lets you cut the coins however you want. We want to cut the bottom coin so that there is no overlap when you recreate the shape (aka the coins lay flat on the surface)...like puzzle pieces. How do you do that?

You would cut along the bottom coin's marker line, right? Then, like a puzzle, the full coin and the cut coin will fit together to make the shape. Now we have a nice, concrete representation of the formula: everything that is part of the puzzle is added to the formula, and each part you cut out is subtracted from the formula.

In this case, the first coin, P(A), and second coin, P(B), are both part of the puzzle, whereas the overlap that you cut out, P(A and B), is not: P(A or B) = P(A) + P(B) - P(A and B)

I'm going to stop here to make sure that you follow this example before moving on to the three-coin example.

Quote

fangtray: Master | Next Rank: 500 Posts; Posts: 273; Joined: Thu Sep 08, 2011 6:50 am; Thanked: 5 times; Followed by:3 members

by fangtray » Sat Mar 03, 2012 11:20 pm

krusta80 wrote:OK, I'm going to take the visual a bit further now to really drive home the point. Let's say that you have three coins (or poker chips or any other disc-shaped item).

First for the two-part formula, take two of the coins and overlap them as per my diagram. Now, take a permanent marker and outline where the top circle overlaps the bottom. The bottom coin will now have the purple shape outlined on it after separating the two coins.

Now let's say you have a laser or really powerful knife that lets you cut the coins however you want. We want to cut the bottom coin so that there is no overlap when you recreate the shape (aka the coins lay flat on the surface)...like puzzle pieces. How do you do that?

You would cut along the bottom coin's marker line, right? Then, like a puzzle, the full coin and the cut coin will fit together to make the shape. Now we have a nice, concrete representation of the formula: everything that is part of the puzzle is added to the formula, and each part you cut out is subtracted from the formula.

In this case, the first coin, P(A), and second coin, P(B), are both part of the puzzle, whereas the overlap that you cut out, P(A and B), is not: P(A or B) = P(A) + P(B) - P(A and B)

I'm going to stop here to make sure that you follow this example before moving on to the three-coin example.

yea i got it. So if you were flipping coins, and you wanted at least 1 heads out of 3 flips, you would take 1 -1/8 = 7/8.

or if you wanted to do it conceptually..

1/2+1/2+1/2-1/4-1/4-1/4+1/8 = 7/8

I guess what I'm not understanding is how the venn diagram can be used represent probability and flips a coin or rolls of a dice..perhaps it is because it is a little too abstract.

So when i apply that concept to 4 flips of a coin or 4 rolls of a die, i'm not even sure how to draw the venn diagrams.

Quote

krusta80: Master | Next Rank: 500 Posts; Posts: 143; Joined: Mon Mar 14, 2011 3:13 am; Thanked: 34 times; Followed by:5 members

by krusta80 » Sat Mar 03, 2012 11:46 pm

fangtray wrote:
krusta80 wrote:OK, I'm going to take the visual a bit further now to really drive home the point. Let's say that you have three coins (or poker chips or any other disc-shaped item).

First for the two-part formula, take two of the coins and overlap them as per my diagram. Now, take a permanent marker and outline where the top circle overlaps the bottom. The bottom coin will now have the purple shape outlined on it after separating the two coins.

Now let's say you have a laser or really powerful knife that lets you cut the coins however you want. We want to cut the bottom coin so that there is no overlap when you recreate the shape (aka the coins lay flat on the surface)...like puzzle pieces. How do you do that?

You would cut along the bottom coin's marker line, right? Then, like a puzzle, the full coin and the cut coin will fit together to make the shape. Now we have a nice, concrete representation of the formula: everything that is part of the puzzle is added to the formula, and each part you cut out is subtracted from the formula.

In this case, the first coin, P(A), and second coin, P(B), are both part of the puzzle, whereas the overlap that you cut out, P(A and B), is not: P(A or B) = P(A) + P(B) - P(A and B)

I'm going to stop here to make sure that you follow this example before moving on to the three-coin example.
yea i got it. So if you were flipping coins, and you wanted at least 1 heads out of 3 flips, you would take 1 -1/8 = 7/8.

or if you wanted to do it conceptually..

1/2+1/2+1/2-1/4-1/4-1/4+1/8 = 7/8

I guess what I'm not understanding is how the venn diagram can be used represent probability and flips a coin or rolls of a dice..perhaps it is because it is a little too abstract.

So when i apply that concept to 4 flips of a coin or 4 rolls of a die, i'm not even sure how to draw the venn diagrams.

As I alluded to in one of my earlier replies to your post, going beyond three events with this method will probably NOT be the most efficient way to answer a GMAT question. The point of the diagrams was to explain why the formula for three events was the one I used. I agree that it's a little hard to visualize how probabilities exactly fit in a Venn Diagram, but the ideas of union ("or"ing) and intersection ("and"ing) are nicely represented by them I think.

Moving on to the more complex examples you mentioned, however, there is a set formula that is perfect for these sorts of questions! It's called the Binomial Distribution, and it's pretty easy to setup and apply:

Let p represent the probability of a success for one event (ie. flipping heads or rolling a 6)
Let n represent the total number of observed events (ie. coin flips or dice rolls)
Let k represent the desired number of successes (ie. number of heads or number of rolled 6's)

Using the binomial, we can quickly calculate the probability of k successes out of n events...

P(k) = nCk * p^k * (1-p)^(n-k)

Example 1: What is the probability of flipping exactly 6 heads out of 10 coin flips?

P(6 heads) = 10C6 * (1/2)^6 * (1/2)^4 = 210/1024 = 105/512

Example 2: What is the probability of rolling exactly 2 sixes with three dice?

P(2 sixes) = 3C2 * (1/6)^2 * (5/6)^1 = 15/216 = 5/72

The numerator of each answer (before simplification) will always represent the number of WAYS to perform the k successes. Obviously you can perform the binomial for any number of successes and then add them up as needed. Also, notice that for k = 0, the binomial simplifies to (1-p)^n, which means that for k >= 1, we use 1 - (1-p)^n

If you'd like to go any deeper into this and see exactly how the binomial relates to Venn Diagrams and the other formulas...as well as Pascal Triangles...I'd be happy to go as deeply into it as you'd like. It may best be done via PM though; I feel that we're drifting a bit from the scope of the GMAT. That said, I'm hoping to help you gain a deeper understanding of these concepts.