For Q6, I did it another way -- please let me know if my assumption was incorrect, but I got the same answer: If only 3 gets pie then the equations looks like the following. 3/4 * 3/4 * 3/4 * 1 = 27/64. Thats because if the first three get it, and then in order for you to see the probability that exactly 3 of the customers get pies, then you the know the probability of the last person which is 100% or 1. Maybe it was a bad assumption but it made sense in my head.
Hi Charles. For Q7, there would be 2 cases of picking 2 labradors (as all 9 dogs in their respective category is different from one another) so L x (L-1) < 18 that gives L 3.6, for NL = 5 it is valid but for NL=4,3 also it is valid hence statement 2 is also not sufficient. Both option 1 and 2 together are not sufficient as well so answer should be E. Please correct me if I am wrong
I'm not sure what you mean when you say there would be two cases of picking 2 labradors because we define a dog as either a labrador or not a labrador. The order of the choice doesn't matter here since we either pick two labradors or we don't. There are two cases when we pick one labrador and one non-labrador since we could choose a labrador then a non-labrador or we could choose a non-labrador then a labrador. However, if we want two labradors, there's only one way we can do that. If there are L labradors out of the 9 dogs in this scenario, the probability of picking two labradors is (L/9)[(L-1)/8]. Since we want to know whether this is less than 1/2, we can set this up as a question: is (L/9)[(L-1)/8] < 1/2. Multiplying by 72 on both sides, as Charles did in the video, gives L(L - 1) < 36. I hope that helps, but please let me know if I missed something in your question!
Thank you so much for the kind words! And you're 100% correct -- you can definitely use binomial expansion on Q6. I deliberately highlighted a different solution path in the video, just because most test-takers aren't familiar with binomial expansion, and it's not useful all that often on the GMAT. But if you happen to know how to use it, it's a nice tool for Q6. Thank you again, and have fun studying!
Thank you so much, Charles for all the videos! For Q2, I followed the below thought process, I dont seem to understand where I am going wrong- On visit one, he will sit on 1 of the 3 chairs On visit two, he will sit on 1 of the 2 chairs(since 1 chair he sat on visit 1 is out) On visit three, he will sit on 1 on the 1 remaining chair So - 1/3 X 1/2 X 1/1 = 1/6?
In this question, the denominators of the fractions will always be 3 because there are always 3 chairs available for Tim to sit on. The numerator of the fraction is where we list the number of chairs available for Tim to sit on that will satisfy the situation we want (sitting in a different chair each visit). This means the numerators of the fractions should decrease from 3 to 2 to 1 each time Tim visits the barber shop. This gives us a sum of 3/3 * 2/3 * 1/3 = 6/27 = 2/9. I hope that helps!
Hey Charles, in question 7 . How do you recognise that in statement 1, order matters and therefore you need to consider both ways of getting a non-lab and a lab. When setting up the question (before statement 1) you determined that order did not matter in this instance. Logically, order always matters, as when you pick one dog, the set of dogs remaining changes. So I would deal with both of these situations the same. Could you explain/define a rule for understanding when order matters? Thanks for the great content!
As you say, order always matters in questions like these for exactly the reason you suggest. When setting up the question, Charles looked at the scenarion in which the first choice is a labrador and the second choice is a labrador. We don't need to consider alternative paths to choosing two labradors because there is only one way of doing this: we choose a lab and then we choose a second lab. It's not that the order did not matter here, it's just that this is the only way of choose two labradors. When examining statment 1, we need to consider both routes to choosing exactly one lab: we could choose a lab and then choose a non-lab, or we could choose a non-lab and then choose a lab. Finally, when examining statement 2, we can see that there's only one way of choosing a non-lab and then a second non-lab, so we only need to consider one path here. I hope that helps!
When do we know that we need to take into consideration the order of selection, for example in question 7 , while checking the statement 1 , we multiplied by 2 and in question 8 we didn't take into consideration the order of cookies.??
Superlike to this video! 🤩I am amazed how you rephrased the Q7 DS ques! What is the level of that question? I was able to solve it but need to know the level.
Hi, for Question 7 for these types of questions the first thing I thought of if we think of L & NL as x & y the question then becomes can we solve for either x or y. Since similarly x = 9 - Y, for statement 1 you would get x^2 - 9x - 36. X ≠ 12 or 3 therefor statement 1 is invalid. Statement 2 you get y^2 = 7.2. Since this number is less than 9 we know that we can figure out x^2/72. Therefor B would be right. Can this method be successful in other questions like this?
Hey Charles! For the last question, you've got desired ways = 90 but I was additionally considering order of pick and multiplying desired outcomes with 4!/2!*2! . Why is this incorrect? Please help
Whenever you encounter a question like this one, you'll want to ask yourself: would changing the order give us a meaningfully different outcome? Do we care about rank or sequencing in some way? On Q8, our "desired" outcome is two cookies of each type. We don't care about their order, though -- we're not necessarily going to eat them in a specific sequence or anything. (Maybe we'll just stack them up, and eat a quadruple-decker cookie sandwich! Maybe with some gelato smeared between the cookies!! 😋😋😋) The bottom line: in this scenario, there's no need to adjust for the order of selection, since we don't care about the sequence at all. I hope that helps!
@@GMATNinjaTutoring But isn't the ordering taken care of when you choose? Let's say we have 9 dogs and 4 of them are labradors. If I'm picking 2 dogs, what is the probability of picking a lab and a non-lab? The probability of picking a lab in the first pick is 4/9, and picking a non-lab on second is 5/8. So we get (4/9)*(5/8). But since the ordering matters, the actual probability is (4/9)*(5/8)*2, which is 5/9. Alternatively, if I use combinatorics, favourable cases will be (4C1)*(5C1). Total cases will be (9C2). So the probability is (4C1)*(5C1)/(9C2), which is again 5/9.
For Q7 on the Labs, when calculating the number of non-labs to make statement p(no labs) > 1/10 true, should you multiply by 2 since it does not matter which order you pick the non-labs? This would make the right side 3.6 instead of 7.2 and thus the minimum number of non-labs could be 3. The correct answer is still B either way.
@LesterOaks No. If you will use PnC logic here, it would be much clearer why we didn't multiply this by 2. Let Non Labs = N We did Nc2 / 9c2 > 1/10 => N(N-1)/9x8 > 1/10 (basically 2 will cancel out 2)
Charles started with 4! / (2!2!), so we could rewrite that as (4*3*2*1)/(2*1*2*1). If we cancel the 2*1 in the numerator with one of the 2*1s in the denominator, we'll have (4*3)/(2*1), which is what Charles wrote at 52:59. You're absolutely right that it WAS 2!2! = 4 in the denominator, but there was a bit of canceling that happened after that to leave us with (4*3)/(2*1) = 2*3 = 6. I hope that helps!
We don't need to think of the four scenarios as representing the order the customers visit the restaurant. Instead, we could think of each entry in a scenario representing what a customer does. We could say the first entry in each scenario represents what happens if customer A does not buy a pie, the second entry represents what happens if customer B does not buy a pie, and so on. In this way of looking at the solution, it doesn't matter which order the customers visit the restaurant. All we need to know is whether a customer did or did not buy a pie. Since there are four different customers who might be the person who does not buy a pie, we need four different scenarios to represent every possible 'successful' outcome. I hope that helps!
hey guys great video, is there anyway of recognising question that might need to be answered using permutations combinations ? like any specific wordings or something of sort ?
Hey Charles, great video like all your lectures. For the lab question, gmat club says D is the answer. I am not entirely sure with the solutions there or here, it definitely went over my head but would appreciate your comment on why they said D and not B?
Are you referring to Q7 -- the DS question with the labradors? If so, we changed that question quite a bit for the current version of the video. What you're seeing on GMAT Club is probably the old version of the question. Since this video just published a few hours ago, I have no idea how GMAT Club would have already gotten its hands on the newer version of the question. :) I hope that helps a bit, but let me know if this doesn't explain the discrepancy!
Thank you so so much for this! I gave my first GMAT FE attempt yesterday and got a score of 82 (19 / 21 questions correct) in Quants. But I still feel I definitely need to work on Probability. I’m gonna reappear as my overall score is pretty low because of Verbal and DI. Apart from watching these videos, for quants - would you recommend practicing questions from the OG before I move on to the mocks ? I’m just not sure what material should I use to practice Quants questions from. Would love your inputs, thanks!
While we can't tell how many male chefs there are from the information provided, we can tell that the probability of selecting a male chef is less than 1/3 from the information provided in statement 2. If fewer than 20 of the attendees are chefs then we can definitely say that fewer than 25 attendees are male chefs. Since we know that fewer than 25 attendees are male chefs, the probability of choosing a male chef when selecting one attendee at random is less than 25/75 = 1/3. This is why the answer to this question is (B). I hope that helps!
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For Q6, I did it another way -- please let me know if my assumption was incorrect, but I got the same answer:
If only 3 gets pie then the equations looks like the following.
3/4 * 3/4 * 3/4 * 1 = 27/64.
Thats because if the first three get it, and then in order for you to see the probability that exactly 3 of the customers get pies, then you the know the probability of the last person which is 100% or 1.
Maybe it was a bad assumption but it made sense in my head.
Great Video! Thank you very much!
Hi Charles, In question 1, how did you decide he takes shoes one after the other or simultaneously picks up both shoes at random
Hi Charles. For Q7, there would be 2 cases of picking 2 labradors (as all 9 dogs in their respective category is different from one another) so L x (L-1) < 18 that gives L 3.6, for NL = 5 it is valid but for NL=4,3 also it is valid hence statement 2 is also not sufficient.
Both option 1 and 2 together are not sufficient as well so answer should be E.
Please correct me if I am wrong
I'm not sure what you mean when you say there would be two cases of picking 2 labradors because we define a dog as either a labrador or not a labrador. The order of the choice doesn't matter here since we either pick two labradors or we don't. There are two cases when we pick one labrador and one non-labrador since we could choose a labrador then a non-labrador or we could choose a non-labrador then a labrador. However, if we want two labradors, there's only one way we can do that.
If there are L labradors out of the 9 dogs in this scenario, the probability of picking two labradors is (L/9)[(L-1)/8]. Since we want to know whether this is less than 1/2, we can set this up as a question: is (L/9)[(L-1)/8] < 1/2. Multiplying by 72 on both sides, as Charles did in the video, gives L(L - 1) < 36.
I hope that helps, but please let me know if I missed something in your question!
For Q6, one can use Binomial Probability Formula to solve it. Many thanks for the video, I learned hell a lot.
Thank you so much for the kind words!
And you're 100% correct -- you can definitely use binomial expansion on Q6. I deliberately highlighted a different solution path in the video, just because most test-takers aren't familiar with binomial expansion, and it's not useful all that often on the GMAT. But if you happen to know how to use it, it's a nice tool for Q6.
Thank you again, and have fun studying!
Thank you so much, Charles for all the videos! For Q2, I followed the below thought process, I dont seem to understand where I am going wrong-
On visit one, he will sit on 1 of the 3 chairs
On visit two, he will sit on 1 of the 2 chairs(since 1 chair he sat on visit 1 is out)
On visit three, he will sit on 1 on the 1 remaining chair
So - 1/3 X 1/2 X 1/1 = 1/6?
In this question, the denominators of the fractions will always be 3 because there are always 3 chairs available for Tim to sit on. The numerator of the fraction is where we list the number of chairs available for Tim to sit on that will satisfy the situation we want (sitting in a different chair each visit). This means the numerators of the fractions should decrease from 3 to 2 to 1 each time Tim visits the barber shop. This gives us a sum of 3/3 * 2/3 * 1/3 = 6/27 = 2/9.
I hope that helps!
@@GMATNinjaTutoring thank you so much! You guys are the best :)
I literally danced on my seat for solving the last one...HAA!!!
Hi, can you share free resources for GMAT ?
Hi contact me
Hey Charles, in question 7 . How do you recognise that in statement 1, order matters and therefore you need to consider both ways of getting a non-lab and a lab.
When setting up the question (before statement 1) you determined that order did not matter in this instance.
Logically, order always matters, as when you pick one dog, the set of dogs remaining changes. So I would deal with both of these situations the same. Could you explain/define a rule for understanding when order matters? Thanks for the great content!
As you say, order always matters in questions like these for exactly the reason you suggest. When setting up the question, Charles looked at the scenarion in which the first choice is a labrador and the second choice is a labrador. We don't need to consider alternative paths to choosing two labradors because there is only one way of doing this: we choose a lab and then we choose a second lab. It's not that the order did not matter here, it's just that this is the only way of choose two labradors.
When examining statment 1, we need to consider both routes to choosing exactly one lab: we could choose a lab and then choose a non-lab, or we could choose a non-lab and then choose a lab. Finally, when examining statement 2, we can see that there's only one way of choosing a non-lab and then a second non-lab, so we only need to consider one path here.
I hope that helps!
@@GMATNinjaTutoring Yes ! Thanks for taking the time to explain further, really appreciate it
When do we know that we need to take into consideration the order of selection, for example in question 7 , while checking the statement 1 , we multiplied by 2 and in question 8 we didn't take into consideration the order of cookies.??
Superlike to this video! 🤩I am amazed how you rephrased the Q7 DS ques! What is the level of that question? I was able to solve it but need to know the level.
hey, how u up with the gmat progress, have any tips for the distance problems, when time is different?
Hi, for Question 7 for these types of questions the first thing I thought of if we think of L & NL as x & y the question then becomes can we solve for either x or y. Since similarly x = 9 - Y, for statement 1 you would get x^2 - 9x - 36. X ≠ 12 or 3 therefor statement 1 is invalid. Statement 2 you get y^2 = 7.2. Since this number is less than 9 we know that we can figure out x^2/72. Therefor B would be right. Can this method be successful in other questions like this?
Hey Charles! For the last question, you've got desired ways = 90 but I was additionally considering order of pick and multiplying desired outcomes with 4!/2!*2! . Why is this incorrect? Please help
Whenever you encounter a question like this one, you'll want to ask yourself: would changing the order give us a meaningfully different outcome? Do we care about rank or sequencing in some way?
On Q8, our "desired" outcome is two cookies of each type. We don't care about their order, though -- we're not necessarily going to eat them in a specific sequence or anything. (Maybe we'll just stack them up, and eat a quadruple-decker cookie sandwich! Maybe with some gelato smeared between the cookies!! 😋😋😋)
The bottom line: in this scenario, there's no need to adjust for the order of selection, since we don't care about the sequence at all.
I hope that helps!
@@GMATNinjaTutoring
But isn't the ordering taken care of when you choose?
Let's say we have 9 dogs and 4 of them are labradors. If I'm picking 2 dogs, what is the probability of picking a lab and a non-lab?
The probability of picking a lab in the first pick is 4/9, and picking a non-lab on second is 5/8. So we get (4/9)*(5/8). But since the ordering matters, the actual probability is (4/9)*(5/8)*2, which is 5/9.
Alternatively, if I use combinatorics, favourable cases will be (4C1)*(5C1). Total cases will be (9C2). So the probability is (4C1)*(5C1)/(9C2), which is again 5/9.
For Q7 on the Labs, when calculating the number of non-labs to make statement p(no labs) > 1/10 true, should you multiply by 2 since it does not matter which order you pick the non-labs? This would make the right side 3.6 instead of 7.2 and thus the minimum number of non-labs could be 3. The correct answer is still B either way.
@LesterOaks No. If you will use PnC logic here, it would be much clearer why we didn't multiply this by 2.
Let Non Labs = N
We did Nc2 / 9c2 > 1/10 => N(N-1)/9x8 > 1/10 (basically 2 will cancel out 2)
Hi Charles, 52:59 isn't it 2!2!= 4 in denominator??
Charles started with 4! / (2!2!), so we could rewrite that as (4*3*2*1)/(2*1*2*1). If we cancel the 2*1 in the numerator with one of the 2*1s in the denominator, we'll have (4*3)/(2*1), which is what Charles wrote at 52:59.
You're absolutely right that it WAS 2!2! = 4 in the denominator, but there was a bit of canceling that happened after that to leave us with (4*3)/(2*1) = 2*3 = 6.
I hope that helps!
q6 - why does there have to be 4 different scenarios? the question's language doesn't indicate that the order of the customers matters right?
We don't need to think of the four scenarios as representing the order the customers visit the restaurant. Instead, we could think of each entry in a scenario representing what a customer does. We could say the first entry in each scenario represents what happens if customer A does not buy a pie, the second entry represents what happens if customer B does not buy a pie, and so on.
In this way of looking at the solution, it doesn't matter which order the customers visit the restaurant. All we need to know is whether a customer did or did not buy a pie. Since there are four different customers who might be the person who does not buy a pie, we need four different scenarios to represent every possible 'successful' outcome.
I hope that helps!
hey guys great video, is there anyway of recognising question that might need to be answered using permutations combinations ? like any specific wordings or something of sort ?
It's the word ' at least ' in the question
I got Q5 right lol I was so happy
Q7: wont P(L,L)+ P(L,NL)+ P(NL,L)+ P(NL) = 1
Hey Charles, great video like all your lectures. For the lab question, gmat club says D is the answer. I am not entirely sure with the solutions there or here, it definitely went over my head but would appreciate your comment on why they said D and not B?
Are you referring to Q7 -- the DS question with the labradors? If so, we changed that question quite a bit for the current version of the video.
What you're seeing on GMAT Club is probably the old version of the question. Since this video just published a few hours ago, I have no idea how GMAT Club would have already gotten its hands on the newer version of the question. :)
I hope that helps a bit, but let me know if this doesn't explain the discrepancy!
@@GMATNinjaTutoring Thanks for clarifying, I see statement one differs in the two questions.
Thank you so so much for this! I gave my first GMAT FE attempt yesterday and got a score of 82 (19 / 21 questions correct) in Quants. But I still feel I definitely need to work on Probability. I’m gonna reappear as my overall score is pretty low because of Verbal and DI. Apart from watching these videos, for quants - would you recommend practicing questions from the OG before I move on to the mocks ? I’m just not sure what material should I use to practice Quants questions from. Would love your inputs, thanks!
For Q4 won’t the answer would be E
While we can't tell how many male chefs there are from the information provided, we can tell that the probability of selecting a male chef is less than 1/3 from the information provided in statement 2.
If fewer than 20 of the attendees are chefs then we can definitely say that fewer than 25 attendees are male chefs. Since we know that fewer than 25 attendees are male chefs, the probability of choosing a male chef when selecting one attendee at random is less than 25/75 = 1/3. This is why the answer to this question is (B).
I hope that helps!