Hey everyone, I think there’s an error in the solution for Question 10 of the Quant Section in Mini Quiz #15.
It seems the solution didn’t account for the scenario where all three numbers are 1 (111), which affects the final probability. I calculated the answer as 2/9 (option B) by subtracting 9 unwanted cases from 27 (8 from combinations of 0’s and 2’s plus 1 for 111). However, in the video, only 8 was subtracted, leading to option D as the answer.
You are correct. I’d also like to add that combinatorics does not seem to be the best approach to this question. Here’s my solution: we’re looking for the probability that exactly 1 out of the 2 remaining draws is a one. By independence, this is the product of the probabilities that 1 of the draws is a one (probability 1/3) and the other draw is not a one (probability 1 - 1/3 = 2/3). Answer: 1/3 * 2/3 = 2/9.
“if one of the numbers is a 1” could’ve been better phrased as “if at least one of the numbers is a 1”
Anyway, the answer of \frac{6}{19} given in the video is correct.
As for the mathematical explanation, you’d seek out conditional probability:
P(\text{exactly two 1's} |\text{at least one copy of 1}) = \frac{P(\text{exactly two 1's})}{P(\text{at least one copy of 1})} = \frac{\frac{6}{27}}{\frac{19}{27}} = \frac{6}{19}
Guiiii please don’t spread misinformation; also your explanation reads like something an AI would say. This is the second time you’re saying the given answer in the video is wrong, and yours is right with no proper justification.
@ganesh Could you please look into this? I’m not sure if u want a vast chunk of people being misinformed because someone decided to post an AI-generated solution.
