#1 GMAT Combinatorics and Probability Tip

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[@saveliikholin3541]

for the last problem there is an even simpler explanation - the cases are symmetrical by “oddness” therefore ans is 1/2

[@TheTestedTutor]

I like the tip! I have responded with a video of my own, showing 2 common mistakes.

[@rahulsinha827]

Cleared some of my doubts on this topic really well. Thanks!

[@ronaksj]

Did we even need to solve the probability Question? Isn't it obvious fo the answer to be half?? There are equal chances of the sum being either odd or even.

[@ayushkumar-bg1xf]

Dude we Asian are different from westerner .we approach apti problems in different way..

[@charlesagha5261]

okay okay, after sitting still for a while, I finally got it, wheew! well not all of it. What I have found, at least for me, is that the permutation method is the most reliable way to solve a problem like this. I still don't get why prob(odd, odd, odd) is 1/2 ^3, but prob(even, even odd) is not the same straight calculation, however, when i do the permutation formula and list them in my head, i get the clearer picture that you can only have one "odd, odd, odd" mixture our of a possible 8 and only 3 even even odd mixtures out of a possible 8. This is as clear as night and day, but it just doesn't add up why the probability rules of independent events doesn't apply here as well. Prob(A and B and C) should be prob(A) X prob(B) X prob (C), which should be 1/2 ^3 in both cases, I just can't wrap my head around that still. what is the prob of picking even the first time? 1/2. Second time? 1/2! Odd the third time? Still 1/2! Multiply them all out and you get the same 1/8 for even even, odd. sigh, what am i missing

[@sakibchowdhury7957]

Instead of 'even, even, odd', it can be 'odd, even, even' and 'even, odd, even'. So, you have to count them too. So, the result is 3/8 instead of 1/8.

[@abdulkarim6664]

Are these GMAT tricks useful for someone studying for GRE?

[@cocossette1234]

Says in the beginning of the video that it's not about memorizing a formula. Then shortly after just uses the formula and doesn't explain anything.

[@ashishsinha9035]

Thank you!

[@neelgaganroy1741]

hi...as the balls are being replaced every time, the probability of either odd or even is 1/2..shouldn't this remain constant for any combination asked?? please comment on my logic..

[@theweareus]

I think so. The probability for odd-odd-odd is 1/2 * 1/2 * 1/2 = 1/8. For the 2 evens, 1 odd, there are 3 different ways we can get: 1. odd-even-even (1/8) 2. even-odd-even (1/8) 3. even-even-odd (1/8) it would be 1/8 + 1/8 + 1/8 = 3/8 so the total would be 1/8 + 3/8 = 4/8 = 1/2 cmiiw.

[@haydenbussard2783]

this is highly dependent on what the question is asking. For this question... yea theres also a 1/2 probability for an even summation. But consider there being 4 balls drawn, and each one replaced. Each one is going to have an equal probability of being even or odd, right? Now, there are 5 possible outcomes: 1) 4 odds... sum to even # 2) 1 even, 3 odds... sum to odd# 3) 2:2 split... sum to even # 4) 3 even, 1 odd... sum to odd # 5) 4 evens... sum to even # Now, you are guaranteed to have one of these possible outcomes. But in this scenario, you have a 2/5 probability to sum to an odd #, and 3/5 probability to sum to an even #.

[@NgocNguyen-lo9jq]

@@john72480 Hey, I'm still confused with the solution to this problem without calculation. So as I understood from your saying, as long as there is replacement for the balls, the probability to have an even/ odd sum (regardless of the number of balls being picked) is always 1/2 right?

[@NgocNguyen-lo9jq]

@@john72480 thanks a lot and have a good day

[@emmaramos8359]

Off chance that someone will answer this, but why did she do 4 choose 3 in question 1?

[@remydeandrade4525]

4 notebooks in sets of 3 - combination of 4,3 or 4 choose 3 hope it helps

[@pumarox95]

Umm.The 2nd question can be solved in 30 seconds without a formula. 😶 Pure logic: There's 4 scenarios.(all odd;all even ; 2 even, 1 odd ; 1 even , 2 odd) Out of these,2 yield odd.Thus 1/2.

[@mehdibennour8493]

Top 👍

[@mrnogot4251]

4:37, you forgot to mention that the intersection is empty.

[@nisaniquett]

Erika your new shorter haircut is awesome haahha. Great content too of course as always.

[@TyrantKaiser]

117907693425 i dont know how to do conbination with6

[@anuragkar7489]

how do we know that half of the one hundred numbers are odd?

[@maryymendes]

Because every other number is odd.

[@sayed1410]

Soory to say Your explanation does not elaborate enough ..

[@haydenbussard2783]

Hey! I see you commented 7 months ago... maybe the way I did it (2nd problem) will make more sense... and maybe not lol. So remember that know each number is being replaced once it is removed. Below are the 4 possible outcomes: 1) 3 odds = sum to odd # 2) 2 even, 1 odd = sum to odd # 3) 1 even, 2 odds = sum to even # 4) 3 evens = sum to even # Since there are only 4 possible outcomes, and each number is replaced after it's drawn... just evaluate your possible outcomes. 2 out of 4 will give you an odd number, so there is a 1/2 probability.

[@brendonpedro3448]

Thank you! I didn't understand why the order mattered.

[@sabinthapa7649]

U r soo beautiful...

[@tomleibowitz2515]

The way she does the second problem is absurd. The question is in effect saying, "what are the odds that when you add 3 integers, the sum will be odd?" The answer is obviously 50%. There's no math involved.

[@orangetrail]

My thoughts exactly, you have 4 posible outcomes when drawing 3 balls: 2 Odds 1 Even, 3 Odds, 3 Evens, 2 Evens 1 Odd 2 of them give you a Odd addition and 2 give you an Even addition so the probability to get an even addition is 2/4 or 1/2

[@p2dar2daag90]

This bitch is cute

[@babulalmarandi1243]

What is notepad? Must be common in US. One dislike