A Surprisingly Simple Trick to Solve the Toughest GRE Probability Question 

By far the best underrated maths youtuber

You are super intelligent, you have that vibe

Just started reading your most recent paper, generalized busebann functions, and wow there's some super interesting stuff there, but wow 51 pages it was a very intense read. I'm personally studying weiner prosses at the moment, and how it specifically relates to financial maths, so I loved how much was relatable to that. Again I love your videos and I really hope you keep it up, there's definitely a market for educational videos on yt and I know you can tap into it and make maths more appealing to a wider range of people!

A similar question was discussed in my maths class and it's pretty basic (I am in 12 grade, for reference)

is it following the hypergeometric law ?

Great thumbnail! I'm a math major in American speak but I can confidently say someone without a mathematical background would understand this. Just out of curiosity is there a link to any of your papers i could have a look at? Cheers and keep it up!

I got the answer wrong at first bc I accidently solved for the wrong thing. My first solution calculated the probablity of selecting at least one working phone. But here is how I fixed it. My solution is much more complicated than yours though 😅: - 12 choose 2 is 66. The probability of choosing the two broken phones is 1/66. - Then, the probability of choosing exactly one broken phone is 20/66, as there are ten working phones, and two broken phones, i.e., the probability of choosing broken phone #1 and a working phone is 10/66, and the probability of choosing broken phone #2 and one working phone is 10/66. 10/66 + 10/66 = 20/66. - Thus, the probability of choosing at least one broken phone is: 10/66 + 10/66 + 1/66 = 21/66 - Finally, the probability of choosing two working phones is: 1 - (21/66) = 45/66 = .6818181818... = (10 choose 2) / (12 choose 2)

Do you believe that 10/12 times 9/11 is not the expected way to get to the answer?