this post was submitted on 21 Dec 2023

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What are the odds of getting a passing grade by sheer guessing here? (lemmy.world)

submitted 10 months ago* (last edited 10 months ago) by [email protected] to c/[email protected]

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Hi,

I just did a test which had two multiple choice questions. Each question was worth one point. Getting them both right would result in getting a 100% score. Suffice it to say, getting just one question right would give you 50% and with that a passing grade.

So you have two multiple choice questions. Both of which are unrelated to the other. Each question has four possible answers. When you finish the test. You get to have one more try. The questions and possible answers remain the same.

Let's say you use both tries and you remember your previous two respected answers. What would your odds be, if you were to brute force guess your way through this test, to get a passing grade or a 100%?

Edit: Both questions only have one correct answer.

IMPORTANT EDIT: YOU DO NOT KNOW WHICH ANSWER YOU HAD RIGHT OR WRONG THE SECOND TIME AROUND. You only know how many questions you got right. But you don't know which. Sorry for the confusion!

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[–] [email protected] 2 points 10 months ago* (last edited 10 months ago) (3 children)

~IIRC from my one stats class almost a decade ago, the math is pretty simple. If it’s truly a random guess then you have a 25% chance to get each question right, all you have to do is multiply them. (0.25)(0.25)=0.0625, you have a 6.25% of getting a 100. The other option to get a 50 is (0.25)(0.75)=0.1875, 18.75%. So there is a 75% chance of failing both.~

~If you get a second chance and can remember your two wrong answers, each problem now has only three options. To get each right, (0.33)(0.33)=0.1089, 10.89% chance. To get one right, (0.33)(0.67)=0.2211, 22.11% chance. 67% chance of failing both a second time.~

Something is amiss with my logic/math here but I’m too tired and am going to sleep now. With this logic, failing both the first time would be (0.75)(0.75)=0.5625 56.25%, the %s don’t add up to 100% so someone please correct me lol

Edit 2: thanks for the corrections everyone, I forgot that the order does matter in this equation

[–] [email protected] 3 points 10 months ago* (last edited 10 months ago)

There are a couple of errors, that I could spot, but I just woke up so my math might also be wrong hahaha.

(0.25)(0.25)=0.0625, you have a 6.25% of getting a 100.

Correct

The other option to get a 50 is (0.25)(0.75)=0.1875, 18.75%.

Almost, that's only true if the first one is the one you get right, but you can also get the second one right (3/4 * 1/4), meaning that it's double your answer, or 37.5% chance of getting at least one right.

So there is a 75% chance of failing both.~

56.25% as per above

~If you get a second chance and can remember your two wrong answers, each problem now has only three options. To get each right, (0.33)(0.33)=0.1089, 10.89% chance.

That assumes you got both wrong the first time around.

To get one right, (0.33)(0.67)=0.2211, 22.11% chance. 67% chance of failing both a second time.~

Same as above, also same as getting the second one right.

[–] [email protected] 2 points 10 months ago

You only accounted for the situations with one correct answer in the case where it is the first question.

[–] [email protected] 2 points 10 months ago* (last edited 10 months ago)

Your calculations to get 100% are right but you are off for the 50% and. You are only considering one specific outcome. But it doesn't matter if the first question is wrong or the second so the chance is 0.250.75+0.750.25 which is 37.5 or double your answer. We can double check it by looking at it from the other direction.

The chance of failure is 0.75*0.75= 56.25%.

So there is a 43.75% of passing the first go around. Split between a 6.25% to get 100 and 37.5% to get 50.

Same mistake for the second calculations. 44.22% is the chance to get 50%