Math question?

[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?

[Nifster]

No I did not.

“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”). “

nice try though

[Jewbacca]

Yes, actually you did. Read his follow up questions.

[Benito Cereno]

Big Dick from Boston

[Tanniker Smith]

If you think that's bad, I knew there'd be a typo. In the first case, I calculated the odds of not getting a single five -- which was the opposite of what was wanted.

The answer has to be subtracted from 100% to get the probability that one or more will show up.

[SAJ]

Probably better to ask Tom Lehrer about THAT one, m’FRiend!

Sorry, haven’t a link, but the song begins: “You can’t take seven from three, so you look at the four in the tens’ place...”

[jjotto]

https://www.youtube.com/watch?v=8wHDn8LDks8

[SAJ]

Tanniker --
For your future convenience, the probability of an event occurring (or not occurring) is always a real number between 0 and 1. Therefore, in any discussion of probabilities, any time I see figures stated as percentages, any time I see a '%' sign, I've simply nothing else to do except...sigh.

In fact, as MNdude ultimately explained his problem, he wanted the calculation of HOW MANY of 1000 'callees' he would ultimately reach, given a 0.16 (or so...he was vague) probability of reaching someone on a given call, and given that he would call them no more than six times total, and given (reasonably) that he would not call them again, having reached them once.

Well, of course, this is a different problem from the "fives" problem being thrown around here (and, in an AWFUL lot of cases, thrown around hilariously badly), but the solution to his actual problem is simple, if one is familiar with the actual discipline of probability.

Just FYI, I learned probability from Don Feder and Pete Weiner at Yale in 1970-1972, and have employed the science in my business enterprises ever since. I believe that I am quite competent offering a solution to simple problems such as these...and am entirely willing to back that view with some reasonable wager with ANYONE here. Proceeds to Free Republic, of course.

It really does become tiresome when FReepers either don't understand the question being asked or go off on some tangent or other when they post a 'solution'. Hence my 'sigh...' in the previous post.

You will have seen, I'm sure, the assorted calculations of 6^6 on this thread. That's fine, and they're accurate, but utterly irrelevant, and 6^6 plays no part whatever in ANY of the possible iterations of MNdude's original question or his later revision of what he 'really wanted to know'.

Best to you, and FReegards!

[SAJ]

Thanks for posting that, m’FRiend! Anyone smart enough to get the jokes within Lehrer’s little send-up, as most FReepers doubtless are, must know straightaway what a disaster that Kommie Kore is.