Free GMAT Prep Hour: Probability

[@shreyasaxena5160]

please increase the number of free lectures..it really helps us a lot

[@manhattanprepgmat6791]

We always have Free Prep Hours on the calendar! You can see our upcoming events here: www.manhattanprep.com/gmat/free-gmat-prep-hour/

[@TvikrPritam]

Manhattan Prep Hours are really high quality stuff, I'm thinking of enrolling for your live classes.

[@manhattanprepgmat6791]

Thank you! You can also attend the first class of any of our courses for free if you'd like to try it out! You can sign up here: www.manhattanprep.com/gmat/classes/free/

[@footlongchen]

Thanks for this Jeff!

[@MrSubho11]

thank you for the session

[@manhattanprepgmat6791]

You're so welcome!

[@roshan1970]

Thanks Jeff. You are awesome!!

[@giovanniluigirosano5934]

The way I approached problem 4 it's a little different. I listed 1-6 in two columns. I realized for each odd # listed on the "first column" there were only 3 way to get an even result if you multiply it for a # on the other column, at the same time the even number in the first column will give you an even result for all 6 numbers in the second column. Meaning 3+3+3+6+6+6 = 27/36 = 3/4. This approach may be not ideal if you have 3 dice but I worked for problem 4

[@giovanniluigirosano5934]

*it worked

[@jeffreyvollmer8806]

@@giovanniluigirosano5934 Yes, this is totally viable too! And the logic underpinning your approach has a lot of similarity to the logic used in our discussion of Problem 4. Dare I say, they are the same method just dressed up a little differently? Let's put it this way: half of the first column's numbers are odd, and half of the second column's numbers are even. That's one way to get to an even product: (1/2 x 1/2 = 1/4). Then, the remaining half of the first column's numbers are even, and all of the second column's numbers will combine with those even values from the first column to give an even product: (1/2 x 1 = 1/2). The 1/4 we got earlier represents the 9 (3 + 3 + 3) odd-even combos out of a possible 36 total combos. The product of 1/2 represents the 18 (6 + 6 + 6) even-whatever combos out of a possible 36 total combos. Add those two options up: 1/4 + 1/2, = [(3 + 3 + 3) + (6 + 6 + 6)]/36 = 3/4. And we can totally incorporate the "1 minus" approach this way. There would only be 3 + 3 + 3 ways to roll an odd product, and your two-column chart would show that pretty clearly. So 9/36 ways will give us an odd product, meaning the remaining 27/36 ways would give us an even product. Depending on the context, I can be a big fan of just counting (adding) the ways to get to a certain result, rather than multiplying out a chain of probabilities. I certainly wouldn't want to solve Problem 3 by multiplying out and adding up the various roll-probabilities and possibilities of dice 1, 2, and 3 in order to get to a sum of 5!

[@3ashooorable]

I would solve it as follow: Find number of odd numbers for both columns, which will be 3 each = 9. Subtract 9 from 36 = 27 (number of times even product occurs) Last, divide 27 by 36 = 3/4.

[@sabinazus8371]

Really good! Thank you!

[@manhattanprepgmat6791]

You're very welcome!

[@kanishkpatel911]

Great video

[@manimehrotra7481]

Thanks once again, Jeff! I think I found Manhattan Prep Hour videos way too late. :(

[@manhattanprepgmat6791]

We're happy you're here now!

[@aytanhamidli9659]

Thanks a lot .I think I found Manhattan Prep Hour videos so late. ((((

[@ashishsinha9035]

Very good

[@yarashamas9712]

Thank you, sir, for this lecture, but can you please tell me how we chose E in problem 7, after we made a calculation and reached the part of 5*2/3^5?

[@jeffreyvollmer8806]

Sorry! The 5 x (2/ 3^5) represents the probability that 4 of the 5 marbles are red. There is also a 1/(3^5) chance that all five are red. If we add those up, that is a total of 11/(3^5) that 4 or 5 of the marbles are red. We are searching for the "opposite" probability: that 3 or fewer marbles are red, so we should subtract the 11/(3^5) from 1. 3^5 = 243, and 1 - 11/243 = 232/243.

[@jeffthrone]

Damn, these questions are hard.

[@aytanhamidli9659]

I couldn't understand number of 6. Because d and e are these correct?

[@jeffreyvollmer8806]

Hi Aytan! Which problem(s) was/were confusing for you? Problem 6's answer is D. Problem 7's answer is E.

[@aytanhamidli9659]

@@jeffreyvollmer8806 problem 6's I also see answer e. I didn't get it exactly.

[@jeffreyvollmer8806]

@@aytanhamidli9659 In Problem 6, answer E is the chance the batter hits exactly one homerun in the four plate appearances. But the problem asks for the probability that the batter hits at least one homerun. So we have to allow for the possibilities that the batter hits 2, 3, and 4 homers in the four plate appearances. The way to see how answer E represents the chance of hitting exactly one homerun is to ask how many ways can this outcome happen, and what are the probabilities for each of those ways. The batter could hit a homerun on the first attempt, and then fail on the subsequent three attempts. The chance of this sequence occurring would be (1/10) x (9/10) x (9/10) x (9/10), or 9^3 / 10^4. But equally likely is that the batter hits the homerun on the second attempt and fails on the other first, third, and fourth attempts. The chance of this sequence occurring would be (9/10) x (1/10) x (9/10) x (9/10), or, again, 9^3 / 10^4. The other ways to hit exactly one homerun happen when only the third plate appearance results in a homerun, or only the final plate appearance results in a homerun. Both of those probabilities are each 9^3 / 10^4. So there are four ways to hit exactly one homerun, and each way have a probability of 9^3 / 10^4. So the total probability of hitting exactly one homerun in the four plate appearances is (4 x 9^3) / 10^4.

[@aytanhamidli9659]

@@jeffreyvollmer8806 thanks!

[@zoezhuyan8527]

probability is so tough

[@pinkandgreen7681]

i agree, i feel like it's such an infinite thing that I will never know it well to master it.

[@truthbespokentwicetruthbes9918]

Why is he running with answers and not clearly mentioning the answer after so much effort in solution, these are the kind of tutors i dont like who keep answers in their head imagining their student also knows what he knows after solving same question 10 times.

[@jeffreyvollmer8806]

Thanks for the feedback. Honestly, it's never my intent to "hide" the answers from the group. These recordings are done live and are unedited, and thus there are inevitably some moments that I would redo if I were given the chance! I agree that I could have done a better job of specifically calling out the correct answer in Problem 7. I don't think I explicitly say that the answer is E, but 1:16:29 - 1:20:11 does explain why the answer should be 232/243. Is there another problem that you felt had a less-than clear answer? I'll try to clear it up for you here!