12 comes from here, "A person looks at a clock 12 times every day."
The time is included in this:
I am sure you are using other resouces online, but if you are pursuing greater education or solutions to problems - I would investigate in purchasing an Introduction to Probability textbook!
Stop cowardice, ban guns!
Absolutely
Stop cowardice, ban guns!
bruin (11-05-2014)
So now the problem is like:
You have a deck of 60 distinct cards, and you draw cards from it without replacement. What is the probability of drawing a particular card?
I use here to make sure that you will understand the "structure" of the problem does not change as . Of course the whole deck will be exhausted when and thus you cannot choose more than that.
The final issue here is that you have multiple decks, and drawing them independently. In such case you will still need to define which event you are concerning. If you want to have the particular card to be drawn in all decks, then you will be multiplying the probabilities as you did. But if you want the card to be drawn in at least 1 deck, then it will be different. The way to calculate that is similar to the formula I posted.
In a deck of 60 cards, if you mark one of the cards with a star, and n = 1, the probability of drawing the card with the star on it is 1/60. If n = 12, the probability is 1/60 + 1/60 + 1/60... 12 times, and .016 ^12 = .192 ONLY IF you replace the card. If you don't replace the card then with each event the probability of drawing the star card increases slightly:
1. p = .016
2. There's 59 cards in the deck. p = 1/59 or .0169
3. There 's 58 cards in the deck. p = 1/58 or .0172
4. There's 57 cards in the deck. p = 1/57 or .0175
5. There's 56 cards in the deck. p = 1/56 or .0178
6. There's 55 cards in the deck. p = 1/55 or .0181
7. There's 54 cards in the deck. p = 1/54 or .0185
8. There's 53 cards in the deck. p = 1/53 or .0188
9 There's 52 cards in the deck. p = 1/52 or .0192
10 There's 51 cards in the deck. p = 1/51 or .0196
11. There's 50 cards in the deck. p = 1/50 or .02
12. There's 49 cards in the deck. p = 1/49 or .0204
So add these probabilities and you get .204, and the probability is higher at 20.4% because NOT replacing the cards increased the chances of drawing the card with the star. However, the case in the question at hand is about time. With consecutive time, what happens is the equivalent of endlessly replacing a minute with another minute. You don't take the minute out of the hour or the day (as if a card) and hold it then go back to the clock and look (as if drawing from a deck of cards). The analogy of time with a deck of cards would mean a new card must appear automatically with every minute. That's the equivalent of still having 60 cards in the deck.
So if we're talking about random times (as long as it's not looking at the clock twice in the same minute), it looks to me like each of the 12 chances are still mutually exclusive events, and you would get .192 like the first example above. This is what I did at post #6.
The final issue at post#8 is a probability based on consecutive events. It's given that each 1440 minutes (a day) has only 12 chances in it to see 13 minutes past the hour, which can occur 24 different ways (am and pm), and each day has the same condition applied to it. Thus, each day the probability is the same. If you want to use the card deck analogy again you're just repeating a draw of 12 cards from a new deck, and 12 cards is all you get in a day. In a deck of 60 cards you're successfully drawing the card with the star at least once after you draw 12 cards WITH replacement (like time contains another minute the deck contains another card). But we're talking about a consecutive event occurring (you draw a star in all 7 decks of cards, not at least 1). So you can't just add the probabilities of each day, you multiply. And it makes sense because seeing 13 minutes past the hour every day for 7 consecutive days even if you look at the clock 12 times a day would have a very low probability of occurring.
Am I not indicating that I understand the "structure" of the problem?
Last edited by Petros; 11-03-2014 at 08:43 PM.
I'm having a problem with these two propositions regarding time:
Proposition 1
Think of an hour of time as a continuous cycle that moves in such a way that 60 minutes is never depleted to 59. This would be the equivalent of a deck of 60 cards from which drawing a card still leaves you with 60 cards given you haven't made another draw within the same minute. It's like drawing a card WITH automatic replacement.
Every 60 minutes 13 minutes past the hour occurs one time, from no matter where you start. To say every time you look at the clock within 60 minutes the P = .016 would mean each time the conditions are exactly the same, and as long as you do not look twice in the same MINUTE, this is true.
The other way, to look at time as 60 minutes in which you "draw" the minute as if taken away, which would be analogous to drawing a card from a deck and NOT replacing it, is not consistent with the nature of time as a continuous cycle.
So in trying to understand the P of seeing 13 minutes after the hour given you have 12 chances per day to see it, we're essentially drawing a card from a deck of 1440 cards WITH replacement.
Proposition 2
A finite amount of time is like a deck of cards, and time can "run out" just like cards in a deck. If you have 60 minutes and you have one chance to look at a clock, it's the same thing as having a deck of 60 cards and drawing one card WITHOUT replacement. After you look at the clock during one minute you can't give this minute back because it's one less minute remaining in the 60 minutes.
So if you have 1440 minutes, and select 12 of these minutes throughout the day like drawing 12 playing cards, the time is running out with each minute you select, which is essentially like drawing from a deck of 1440 cards WITHOUT replacement.
Are both of these true when trying to determine probability regarding time?
I'm just seeing this now.
Is this a more accurate way to get an answer to what I asked in post #6:
What's the probability of seeing 13 minutes past the hour (:13) at least once per day if you look at the clock at 12 different times throughout the day?
P of seeing :13 at least once in a day with one look = 24/1440 = 1/60 or .016
P of NOT seeing :13 at least once a day with one look = 59/60 = .983
(or you can simply subtract 1 - .016 = .984)
With one chance: P of NOT seeing :13 = .983
With two chances: P of NOT seeing :13 = .983 x .983 = .966
With three chances: P of NOT seeing :13 = .983 x .983 x .983 = .9498
...and so on to the point of 12 chances:
P of NOT seeing :13 = .983^12 = .814
which means the P of seeing :13 at least once in 12 chances = 1 - .814 = .186
This is the probability for seeing 13 minutes past the hour (:13) at least once per day if you look at the clock at 12 different times throughout the day, which I need for the final question in post #8:
What's the probability of seeing 13 minutes past the hour (:13) at least once per day for 7 consecutive days given you look at the clock 12 times each day?
And I believe the answer is:
.186 ^7 = .000007 or .0007%
--that's a slightly different number than what I got in post #8 (which was .000008 or .0008% ).
Last edited by Petros; 11-05-2014 at 03:22 PM.
It seems that you have clear up all the stuff and it looks fine to me.
Yes, I thought I had pretty well grounded myself in basic prob theory but when BGM
posted the complementary formula for independent events 1 - (59/60) ^ 12
something went Boing! I found elementary examples such as:
"If you toss a fair six sided die 4 times what's the probability of seeing a 3?"
The answer is, of course, 1 - (5/6) ^ 4
Random sampling of a clock to see if a target time exists or not, struck me
as the same kind of thing as the tosses of a die.
Art
One more question on interpreting the last answer for the probability that you will see 13 minutes after the hour at least once per day for 7 consecutive days given you look at the clock 12 times each day (never twice in the same minute).
This number in the last question, .000007 is .0007% or .0007/100.
How do you interpret this number in terms of odds if it's a decimal? Isn't it valid to say the probability indicates the event will occur 7 times out of 10,000?
The period of time in the question spans 7 days. So, this would mean 7 x 10000 = 70000, and 70,000/ 360 = 194.4 years. Hence, assuming a lifetime isn't usually longer than 100 years, isn't it safe to conclude not more than 3 times in a lifetime is it likely that a person will see 13 minutes after the hour at least once a day for 7 consecutive days given you look at the clock 12 times each day?
Last edited by Petros; 11-07-2014 at 09:02 AM.
If you have a sequence of independent and identical Bernoulli trials, and each of them has a success probability of , then by Law of Large Number, when the number of trials is sufficiently large, the proportion of successes will be close to in probability.
So without checking your numerical numbers, I guess you want to state a similar fact like above.
However, in your situation it will be more complicated, because you are looking at consecutive events, i.e. "run" of a sequence. You cannot partition the whole observation period into different weeks and establish the claim, because they are not independent; and you can think of a simple situation where there are no consecutive 7-days observation in each of the 2 successive weeks separately, but you do have a consecutive 7-days (or more) in these 2 weeks when they combined. And also you need to consider whether a consecutive 8 days observation counted as 2 times or 1.
After clarifying all these you may try to answer what is the expected number of such runs in a sufficiently long period.
How about just this question:
Isn't it valid to say the probability (.0007%) indicates the event will occur 7 times out of 10,000?
What we have calculated / discussed before is in a very specific situation where we just look at particular 7 days and see whether the event happen. So you may state something like this:
If you inspect a random group of 7 days independently 1 million times (am I counting the decimals correctly?), the expected number of events happening is approximately 7.
If you observe a period of consecutive 1 million days, then the expected number of events will be different from above.
Since Petros seems to be seeking a "will occur" kind of answer, I thought of doing a cumulative discreet distribution
and locating the p = 0.005 and p = 0.995 points so a statement could be made with 99% confidence. Instead of
one million, I raised n up to ten million for a mean (n * p) of 70. Both Binomial and Poisson gave me the same
results of 49 and 92. So reasonably close to "will occur" might be along these lines:
"If you inspect a random group of 7 days independently ten million times, you will see the event happening
49 to 92 times at a 99% confidence level".
Art
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