- Thread starter PedroFr
- Start date

First, define our variables. Let

Ah, I see where my error was (I think ) , so I was doing the opposite of being late and being on time, so of course the outcome would be 0, since there's only 2 events (success or insuccess, being late or being on time) . So now I did the following: 1-P(X=0) = 0.95 (=) ... (=) and I got n = 10.4 , I don't know if I did a mistep or something, but my answer is far from the one from the teacher by 1 day ( My answer is 10, my teacher's is 11 days). Am I still missing something?at least1 of thosendays is 95%, correct? If that's the problem you're trying to solve, then your problem comes from using the binomial distribution formula. Think about this, and you'll see why your formula can't possibly be right (and hopefully what the correct formula is):

First, define our variables. Letnbe the total number of days. And letxbe the number of days John is late. So, when n=1, John can either be late (x=1), or on time (x=0). Using your formula, the probability that John is late at least 1 day is 1 - [P(X=0) + P(X=1)]. Plugging in the known values, that gives us: 1 - [0.75 + 0.25] = 0. So, that means the probability of John being late at least 1 day is 0%. Oops.

Well the 10.4 is correct if we could use fractional days but what is the probability for 10 days even? Slightly less than 0.95 isn't it?Ah, I see where my error was (I think ) , so I was doing the opposite of being late and being on time, so of course the outcome would be 0, since there's only 2 events (success or insuccess, being late or being on time) . So now I did the following: 1-P(X=0) = 0.95 (=) ... (=) and I got n = 10.4 , I don't know if I did a mistep or something, but my answer is far from the one from the teacher by 1 day ( My answer is 10, my teacher's is 11 days). Am I still missing something?

Yeah, I was just thinking in terms of 10.4 we wouldn't need 11 days to reach 0.95 of probability, but in the question it does say over 0.95 so I guess there's that.Well the 10.4 is correct if we could use fractional days but what is the probability for 10 days even? Slightly less than 0.95 isn't it?

Thank you all for your answers, I got alot more enlighted now