Probability of Repeated Events

[pdunn]

Hello my question is not academic based just personal.

I wanted to know if an event as a probability of P of occurring per day and it occurs everyday for a week. Then, for the week the probability is P?

i.e.

The probability of picking the correct number in a dice roll is 1/6. Then at the end the wee, the probability of having picked the correct number is still 1/6.

I have been out of college for years and out of probability even longer. So when answering with mathematics, please explain. I would like a written and a mathematic answer.

Thank you very much,

 

[HallsofIvy]

"I wanted to know if an event as a probability of P of occurring per day and it occurs everyday for a week. Then, for the week the probability is P?"

I'm not sure I understand the question. Whatever the probability that the event will occur each day, if it did occur every day for a week, the the probability that it did occur each day is 1!

If the probability that the even will occur each day is P, then the probability it will occur every day for a week (7 days) is P⁷.

However, and I think this is what you are really asking: If the event has occurred every day for a week then the probability that it will occur tomorrow is still P. That's what saying "the probability the event will occur on each day is P" means!

 

[pdunn]

Thank you for your reply, but I was not clear on the situation.

Event: Dice roll
Probability of picking the correct number (P): 1/6
Goal: Pick the correct number once (i.e. for just one day)

The event will happen for the each day of the week, each day the probability of me picking the correct number is 1/6. At the end of the week, what is the probability that I had achieved my goal.

I believe the answer would be 1/6 and the rolls are independent of each other. Since I only want to win one, it doesn't matter how many times the event happens. However, I don't have the mathematical logic behind this.

 

[HallsofIvy]

Thank you for your reply, but I was not clear on the situation.

Event: Dice roll
Probability of picking the correct number (P): 1/6
Goal: Pick the correct number once (i.e. for just one day)

The event will happen for the each day of the week, each day the probability of me picking the correct number is 1/6. At the end of the week, what is the probability that I had achieved my goal.

I believe the answer would be 1/6 and the rolls are independent of each other. Since I only want to win one, it doesn't matter how many times the event happens. However, I don't have the mathematical logic behind this.

You write down a number ("pick a number") then roll a die. If the die comes up the same as the number you picked, you have succeeded. However, here it seems that you are saying ("Since I only want to win one, it doesn't matter how many times the event happens.") that you will have achieved your goal if you pick correctly at least once. That is different from "occurs everyday for a week" as you first said. I take it now that you mean rolling the die occurs every day for a week and you want the probability that you will get the right number at least once.

It is simpler to look at the reverse: On each day the probability you do not get the number you picked is [itex]1- \frac{1}{6}= \frac{5}{6}[/itex]. The only way you could not succeed in getting your number at least once is to fail to get it every day- the probability of that is [itex](\frac{5}{6})^7[/itex].

The probability that you do get your number at least once is, then,
[itex]1-(\frac{5}{6})^7[/itex] which is about 0.7209.

 

[EnumaElish]

I think OP meant something along the lines of "if the probability of rain on Sat. is 50% and on Sun. it's 50% then the prob. of weekend rain is 50% (not 100%, or 25%)."

Prob(Sat & Sun) = 0.25
Prob(Sat or Sun) = Prob(Sat & not Sun) + Prob(Sun & not Sat) + Prob(Sat & Sun) = 3 (0.5)² = 0.75
Prob(either Sat or Sun, not both) = 2 (0.5)² = 0.5

So, the weekend rain prob. depends on the definition of "weekend rain." If it means "and/or" then rain prob = 0.75. If it means "strict or (mutually exclusive)" then rain prob = 0.5.

Somewhat counterintuitively, in the "and/or" case, the prob. of a rainy weekend approaches 1 as the weekend is divided into finer subintervals and the prob of rain during each subinterval remains 0.5. Increasing the number of trials increases the prob. that a certain outcome will be observed at least once. In the limit case, when the number of trials is infinite, "weekend rain" is a certainty (although in this case each time it rains, it'll stop after an infinitesimal duration of time, so that should not be a problem for the weekend barbecue, tennis, or pool party.)

I haven't really thought whether the same calculation applies if the weather announcer was "betting on" a certain OUTCOME ("rain" or "no rain," and nothing in between) for each day, rather than announcing probabilities.

 

[pdunn]

Thank you all for your help and patience.
I do believe EnumaElish is correct.

It wouldn't matter if it is 'or' or 'exclusive or' (+). I believe I solved the (+) case.

Would you please explain the math behind 'or'?

your example has

Prob(Sat or Sun) = Prob(Sat & not Sun) + Prob(Sun & not Sat) + Prob(Sat & Sun) = 3 (0.5)2 = 0.75


I believe the logic goes like this, but I could be wrong.

Since there is two trails, then there are 4 different outcomes, but the (not Sun ^ not Sat) case is false in the 'or' case so it is ignored. Thus leaving us with 3 viable situations. The squared is the number of times the event could occur. Therefore, for my daily dice roll,
Prob(M or T or W or Th or F or S or Su) = 127(1/6)^7 ??
I think I'm wrong, but like I said I've been out of college for years.

Again thanks for your help and patience.

p.s. I have heard of your name before in school, is it a book from the Middle East pre-Judaism?

 

[HallsofIvy]

You've got "or" and "exclusive or" reversed, I think. "Exclusive or" would be "It rains Saturday or Sunday but not both". That would be If the probability of rain each day is 0.5, then the probability of "rain Saturday but not Sunday" would be 0.5*0.5= 0.25. The probability of "rain Sunday but not Saturday" would also be 0.25.
The probability of rain "either Saturday or Sunday but not both" (exclusive or) would be
0.25+ 0.25= 0.50.

Simple "or" would be "it rains Saturday or Sunday or both days". The probability of rain both days is, similarly, 0.25 so the probability it "rains Saturday or Sunday or both days" would be 0.25+ 0.25+ 0.25= 0.75.

That worked out that simply because the probability of rain or not rain is 0.50.

More generally, it the probability of rain on any specific day is p, the probability it will not rain is 1- p.

The probability it will rain "Saturday but not Sunday" would be p(1-p). The probability it will rain "Sunday but not Sunday" would be (1-p)p. The probability that it will rain "Saturday or Sunday but not both" (exclusive or) would be p(1-p)+(1-p)p=
2p(1-p)= 2p- 2p².

The probability that it will rain "Saturday or Sunday or both" (inclusive or) would be
2p(1-p)+ p²= 2p- p².

 

[EnumaElish]

I do believe EnumaElish is correct.

And I believe HoI's posts are internally consistent.

your example has

Prob(Sat or Sun) = Prob(Sat & not Sun) + Prob(Sun & not Sat) + Prob(Sat & Sun) = 3 (0.5)2 = 0.75

I believe the logic goes like this, but I could be wrong.

Since there is two trails, then there are 4 different outcomes, but the (not Sun ^ not Sat) case is false in the 'or' case so it is ignored. Thus leaving us with 3 viable situations.

Correct, Prob(and/or)= 1 - Prob(neither Sat nor Sun) = 1 - (1-p)² = 1 - 0.25 = 0.75.

Therefore, for my daily dice roll,
Prob(M or T or W or Th or F or S or Su) = 127(1/6)^7 ??

Prob(one day only) =

Prob(M; not T nor W nor Th nor F nor S nor Su) = 1/6 (5/6)⁶

+

Prob(not M; T; not W nor Th nor F nor S nor Su) = 5/6 1/6 (5/6)⁵ = 1/6 (5/6)⁶

+

5 times 1/6 (5/6)⁶, for each of Wed. - Sun.

=

7(1/6)(5/6)⁶ = 7/6 (5/6)⁶ = 0.3907 approximately.

Die roll is distributed binomially with p = 1/6; hence I believe the formulaic solution is:
[tex]\left(\underset 1 {\overset n {\phantom x}} \right) p^1 (1-p)^{n-1}=n \times p(1-p)^{n-1}\text{ where } n = 7.[/tex]

p.s. I have heard of your name before in school, is it a book from the Middle East pre-Judaism?

As far as I know it is the earliest written creation myth in Mesopotamia.

 

[pdunn]

Thank you

Thank everyone for your help. My next question is more algebra than probability, but how do I solve for n.

Let's take the dice roll example.
I know p = (1/6), I want my total probability (P) to be 50%, so what does n equals?

np(1-p)^(n-1) = P

(n-1)*LN(np(1-p)) = LN(P),
(n-1) * (LN(n) + LN(p-p^2)) = LN(P),

How do I get n by itself?

Thank you

 

[EnumaElish]

[itex]\text {Mathematica}^\text{\copyright TM}[/itex] returns the attached output.