Doubt in Topic based quant quizzes I Probability 2

Hi! Doubt in this question-

2 doubts here:

  1. How did Greg know immediately that taking 5 is gonna be the way here? I had to try all options before hitting 5
  2. Isn’t the probability that Greg calculated in the answer the probability of “receiving credit”? When I calculated, I solved it like “1- this favourable case”- which would give us probability of NOT receiving credit- as that is what is asked in the question.

Can you please guide me on this?
Thanks!

Do you agree that the corresponding probability is simply \frac{1}{{\operatorname{C}} (10,t)}? If so, it seems pretty clear that {\operatorname{C}} (10,t) is maximized when t = 5.

I understand that 1/ C(10,t) is probability of student getting a RIGHT answer, i.e. getting a credit. But the questions says, value of t when student is “least” likely to get credit. Therefore, I was taking these cases to be like 1- 1/C(10, t).

Can you please explain why this is wrong?

\frac{1}{\operatorname{C} (10,t)} is actually the probability that the student gets the right answer. For example, if t = 2, then \frac{1}{\operatorname{C} (10,2)} = \frac{2 \cdot 1}{10 \cdot 9} is the probability of picking the right answer. This means selecting both of the correct options.

the “least likely” bit is just concerned with the value of t such that \frac{1}{\operatorname{C} (10,t)} is minimized.

Subtracting our corresponding probability from 1 gives the “complementary probability”. In this case, this is the probability that the student does NOT pick the correct set of t options.

Since the question is about picking the correct set of t options, this isn’t directly relevant unless we wanted to go on a detour.

The way I thought about this question is that once you must select over half of the total options to get the question right, you can instead guess which options are the wrong ones. For example when you need to correctly guess 7 out of the 10 total options, you can instead view it as guessing which 3 out of the 10 are incorrect, which is easier than guessing 5 out of 10. Therefore when you have a problem like this, the probability of correctly guessing all the right options is the lowest when you must guess a value near half of the total number of options and is highest when the number of options you must guess is near 0 or the total number of options.

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Aah okay, understood now. Thanks for this detailed explanation. Was getting tricked by the wording of the question- clear now, thanks!

Did not quite get your thought process, but what Cylverixxx explained is clear. Thanks for contributing though!