You’re probably right, because I haven’t done a problem like this since forever, but help me figure out what I did wrong. I found a binomial distribution calculator (this is binomial distribution, right?), entered 30 trials, 21 “successes”, (counting a false answer as a success, and agreeing with you that 20 is break even so you need 21 to do worse than even) and .5 probability of success, and it said the cumulative probability was .9919… against, therefore <1%.
On this page, the cumulative refers to the probability of obtaining at most p successes. You want to run it with 30 and 9 which gives you the right answer, 2.14%
Or you could put in 30 and 20 which gives you the complement.
What is lower than 1% is the probability of getting 8 or less right answers.
You’re probably right, because I haven’t done a problem like this since forever, but help me figure out what I did wrong. I found a binomial distribution calculator (this is binomial distribution, right?), entered 30 trials, 21 “successes”, (counting a false answer as a success, and agreeing with you that 20 is break even so you need 21 to do worse than even) and .5 probability of success, and it said the cumulative probability was .9919… against, therefore <1%.
On this page, the cumulative refers to the probability of obtaining at most p successes. You want to run it with 30 and 9 which gives you the right answer, 2.14%
Or you could put in 30 and 20 which gives you the complement.
What is lower than 1% is the probability of getting 8 or less right answers.
Oh, I see how they did that. Thanks. Original post edited.