Posted on 05/07/2015 7:27:31 AM PDT by MNDude
I'm trying to figure something out for work, but my math skills are not very good. How do I calculate percentages if an event reoccurs multiple times?
Suppose there is an event that has the chance of occurring on in 6 times (like rolling a dice and getting a "five").
What are the chances I will get a "five" if I roll the dice six times?
Wrong. You apparently misread the question.
And where I actually posted the correct formula (my last post on this thread).
I just sent you a PM. Call center calculations are NOT based on probability. They are best calculated using an erlang calculator that assumes talk time, work time, and contact time. You have to figure out those numbers, then figure out how you dial and the contact rate.
Using probability to staff call centers is a BIG mistake, as you will ALWAYS come up short, and under budget your expenses.
Read my PM and drop me a note with some specifics.
I used to run call centers that did both inbound and outbound contacts—about 13 million contacts a month. Projecting this stuff is an art—and you need to make sure you are going at it from the right direction.
Actually the odds of contact are reduced with every iteration of the list.
This is a replacement problem. If you are calling at the same time every day you cannot expect the customer to be home tomorrow night if they were not home tonight.
So, you need to calculate the entire process, rotating contact times, etc. Also you need to clear your list from the do not call lists in your state. The issues brought about from calling when you shouldn’t are not worth the effort for a temporary measure.
There is a 5/6 chance of failure on each roll. Multiply that number by itself roll each successive roll.
EXACTLY 1 five in six rolls equals the prob of 1 success times the prob of fail/fail/fail/fail/fail.
that's (1/6)(5/6)(5/6)(5/6)(5/6)(5/6) X 6
the final X 6 is because you don't care which of the six rolls is a success.
apologies in advance for typos. I'm on my iPad. I hate typing on it.
There are 46656 total outcomes, WHEN CONSIDERING ALL 6 SIDES of the die. There are only 64 total outcomes when considering only 5s vs. non-5s...which we are here. Of those 64, exactly 6 meet our specification, to wit, exactly one 5, as I illustrated in a previous post.
Don't trust me, but I **should** trust Blaise Pascal, were I you. There is an excellent discussion of (essentially) this very problem in letters between Pascal and the Chevalier du Mere in Peter Bernstein's dandy book, Against The Gods, which is a fine treatise on the history and development of probability theory and risk analysis.
Sigh....
It’s called a Bernoulli trial.
If you look on wikipedia,
the probability of success is 1/6 and the number of trials is 6.
perhaps you can understand other variations by looking at the article.
Basically if you want to know the probability of rolling a “5” at least once when rolling a 6 sided die 6 times it goes like this:
The percentage for rolling anything except a 5 is 5/6 or 83.33...% probability. Now the probability of rolling every die roll NOT a 5 is 5/6 times 5/6 (6 times) or (5/6)^6 which comes to 33.49% probability that you will not roll even 1 “5” when rolling 6 times.
So the inverse (meaning you DO roll at least 1 “5”) is basically 100% minus the probability that you will NOT roll any “5”’s or 66.51% probability.
Remember though that there are lies, damn lies and then statistics. What I wrote above is considered statistics. :-)
using Common Core...I’m guessing two weeks for the answer
Your chances of getting a specific number at least once in 6 rolls of a die is 1-(5/6)^6 = ~66.5%
As specifically noted in #24.
I read the OP’s question and not any of the responses. Sorry if that bothers you.
The 36 is in superscript (ie to the 36th power)
1-(5/6)36 = 99.86%
Write it like this, with the carat to signify superscript
(5/6)^36
...two weeks.
What is the common core method of figuring this out? :)
“I’m trying to figure something out for work, but my math skills are not very good. How do I calculate percentages if an event reoccurs multiple times?
Suppose there is an event that has the chance of occurring on in 6 times (like rolling a dice and getting a “five”).
What are the chances I will get a “five” if I roll the dice six times?”
If the rolls are independent and by “get a five” you mean you will get at least one five then the answer is:
1-(5/6)^6 which is the probability of the complement of getting no fives.
If the rolls are independent and by “get a five” you mean you will exactly one five then the answer is:
6*((1/6)*(5/6)^5)= (5/6)^5 which 6 times the probability of getting just one five on any specific roll.
16% chance independent of time of day? or is the 16% the average for the entire day?
Freeping in an iPhone. No clue where that symbol is.
Are you rolling one die six times or a pair of dice six times?
If its a pair of dice, there are 36 total possible combinations in the first roll, only four of which will add up to ‘five’.
Your odds of getting a ‘five’ in that first roll are 4/36 or 11.1 percent.
Your odds of getting a ‘five’ in each subsequent roll are the same.
Your odds of getting at least one ‘five’ after six rolls are 24/36 or 66.67 percent.
Depends on what factors control the situation you want to predict.
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