Multiple attempt percentages

Hi all,

I answered questions from set A and was correct 64% of the time. Every time I was incorrect I had to answer a different set of questions (set B) and of these I was correct 66% of the time. When I was incorrect with a question from set B I had to answer a question from set C of which I was correct 70% of the time.

I was wondering what the probability of getting all three questions wrong is?

How many questions would I get correct from set A,B and C before I got all three wrong (on average)?

How many times would I expect to get all three answers wrong if I answered 1000 questions from set A?

Thanks so much for your help!

Re: Multiple attempt percentages

So I've been going through all the treads trying to find something that may help me. Is it Binomial probability that I would need to use to find the answer?

Sorry for my ignorance... math was never my strong point!!

Re: Multiple attempt percentages

Quote:

Originally Posted by

**tjcoll** Hi all,

I answered questions from set A and was correct 64% of the time. Every time I was incorrect I had to answer a different set of questions (set B) and of these I was correct 66% of the time. When I was incorrect with a question from set B I had to answer a question from set C of which I was correct 70% of the time.

I was wondering what the probability of getting all three questions wrong is?

How many questions would I get correct from set A,B and C before I got all three wrong (on average)?

How many times would I expect to get all three answers wrong if I answered 1000 questions from set A?

Thanks so much for your help!

The probability of getting all three wrong is (1-0.64)(1-0.66)(1-0.70)

CB

Re: Multiple attempt percentages