• Welcome! The new FreeMathHelp.com forum is live. We've moved from VB4 to Xenforo 2.1 as our underlying software. Hopefully you find the upgrade to be a positive change. Please feel free to reach out as issues arise -- things will be a little different, and minor issues will no doubt crop up.

Binomial Distribution?

PedroFr

New member
Joined
Nov 1, 2015
Messages
3
So I'm having trouble solving this exercise: The question is the following, John probability of getting to work late each day is 0.25 . And I'm asked to say how many days there has to be for the probability of John reaching work late once or more is equal to 0.95. I tried doing Binomial Distribution, by doing P(X>=1)=1-[P(X=0)+P(X=1)] and then doing the following equation : 1-[P(X=0)+P(X=1)] =0.95 and then using the Binomial Distribution formula to find the number of days need for the probability to be equal to 0.95. But with no luck, and here I am stuck in this exercise.
 

ksdhart

Full Member
Joined
Aug 1, 2014
Messages
384
Okay, so you're trying to find the number of days (n) when John's probability of being late on at least 1 of those n days 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. Let n be the total number of days. And let x be 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.
 

PedroFr

New member
Joined
Nov 1, 2015
Messages
3
Okay, so you're trying to find the number of days (n) when John's probability of being late on at least 1 of those n days 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. Let n be the total number of days. And let x be 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.
Ah, I see where my error was (I think :p) , 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?
 

Ishuda

Elite Member
Joined
Jul 30, 2014
Messages
3,345
Ah, I see where my error was (I think :p) , 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?
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?
 

PedroFr

New member
Joined
Nov 1, 2015
Messages
3
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?
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.
Thank you all for your answers, I got alot more enlighted now :)
 

ksdhart

Full Member
Joined
Aug 1, 2014
Messages
384
I'd just like to add a bit of clarification regarding the binomial distribution formula. My apologies if you already know this. The binomial distribution formula works because it's essentially asking "What are the probabilities of [the event] not happening?" And since this problem asks for the probability of John being late at least once, the only way he can not be late even once is to always be on time (in other words x=0). So we find that probability and subtract it from 1.
 
Top