GRE Quant Ep 19: Probability
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- Опубликовано: 16 июл 2024
- In a GRE probability question, do you know the difference between 'with replacement' and 'without replacement'? Do you know what the terms 'mutually exclusive' and 'independence' mean when it comes to probability? Are you confident using variables and algebra in a probability question?
If you answered 'no' to any of those questions then Dana -- a GMAT Ninja tutor -- will help you in this video. She'll show you how to tackle GRE probability problems efficiently and effectively.
This is video #19 in our complete GRE quant video series. For updates on upcoming videos, please subscribe!
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Chapters:
00:00 Introduction
02:38 Question 1 - Set theory
11:20 Question 2 - With or without replacement?
15:38 Question 3 - At least one
22:31 Question 4 - Formulas and mutual exclusivity
27:18 Question 5 - Formulas and independence
34:33 Question 6 - Probability with variables
40:32 Question 8 - Gratuitously hard final question
Nice playlist for GRE Quant prep. Dont stop making videos about GRE.THANKS
The last question is brilliant! Thanks for the video
Haha, I'm glad that you enjoyed it. Have fun studying!
awesome
Thank you!
I am still not sure about charles visits pub example. Why isnt the answer 1/256 ...as there is only one way to sit at each booth in 4 visits and total possible combinations is 4 to the power 4. Please explain.
In this question, we have to consider each visit separately as the number of places Charles can sit and still satisfy the restrictions in the question changes with each visit.
The first time Charles visits, it doesn't matter which booth he sits in. He can sit anywhere because he hasn't sat in any of the booths yet. This means the probability he is assigned a booth he hasn't yet sat in is 4/4.
The second time Charles visits, he can't sit in the booth he sat in during his first visit but he can sit in any of the other three booths. This means the probability he is assigned a booth he hasn't yet sat in is 3/4.
This pattern continues with each visit Charles makes to McSorleys. The third time Charles visits, the probability he is assigned a booth he hasn't yet sat in is 2/4. And the final time Charles visits, the probability he is assigned a booth he hasn't yet sat in is 1/4.
If we combine these visits, the total probability is (4/4)*(3/4)*(2/4)*(1/4) = 3/32.
I hope that helps!
@43:49 When you are selecting a group of two people, why are you doubling it? Could you explain why the order matters in this case?
In this question, there are four possible outcomes:
1. We could choose two Americans
2. We could choose two Europeans
3. We could choose an American and then a European
4. We could choose a European and then an American
Even though #3 and #4 leave us with the same final outcome (we chose one American and one European), they're counted as two different outcomes when we think about calculating the probability.
If the number of Americans in the group is x, then the number of Europeans is (10-x). This means the probability of outcome #3 occurring is (x/10)*((10-x)/9) and the probability of outcome #4 occurring is ((10-x)/10)*(x/9). Even though the order of the numerators is reversed in these two options, they give us the same final outcome (x(10-x))/90.
This means that to figure out the total probability of either one of outcomes #3 or #4 happening, we can calculate the probability of one of them and then double it.
I hope that helps!
@@GRENinjaTutoring Hi, thanks for replying back so quickly and with lots of details. I am starting to grasp it. I remember watching a previous quant video where a tutor mentioned that order doesn't matter when selecting people in a group or in a team, I am wondering how we are able to discern between these cases where order does and doesn't matter. I know in this example, there are two outcomes for the same group. But are we assuming or matters? Are there keywords? Thank you!
@@JennyLiang-oi9zc I'm sorry if I've misinterpreted your question here, but I think you're getting a little mixed up between finding the number of ways a thing can happen and finding the probability that a thing will happen.
In a probability question, you consider the selections as taking place one after the other even if the question just says something like "you choose two people". This means we need to consider all the possible outcomes:
- American and American
- American and European
- European and American
- European and European
However, in a combinations question, all we care about is the composition of the final group, so the order they were chosen in doesn't matter. The final option is that you're asked a permutations question which asks about the number of ways you can select a group, but the order of the selection does matter. For more on the distinction between combinations and permutations questions, check out our Combinatorics video.
For example, let's say the Americans are labeled A, B, C...and the Europeans are labeled Z, Y, X.... If we were looking for the number of ways you could choose one American and one European, then the order wouldn't matter as picking person A then person Z gives you the same group as if you were picking person Z then person A. However, if you need to find the probability of picking one American and one European, then we have to consider picking person A then person Z as a different outcome to picking person Z then person A, even though you end up with the same group at the end of it.
This is a really tricky thing to wrap your head around, but I hope that's helped a little. Please let me know if you have any other questions, and I'll do what I can to help!
in question 5 where does the 0.7 come from towards the end of the problem solving?
At that point, Dana had 0.3 + P(B) - 0.3P(B) = 0.58. If we do the subtraction P(B) - 0.3P(B), we'll get 0.7P(B).
I hope that helps!
in last question, we select E and A for first case , A and E for second case but what about if we choose A comes both 2 times or E comes both 2 times...What about these cases?????
In the solution to this question, we're examining the scenario provided that the probability of choosing 1 European and 1 American is greater than 8/15. Choosing two Americans or two Europeans wouldn't satisfy this scenario, so we can ignore those cases for the purposes of answering this question.
I hope that helps!
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