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February 17

probability and habit.

Lets say that someone wants to break a habit , what is the best way of modeling the probabilities?

If I take avoiding Facebook as an example where I have 1/3 chance of logging in in 1 day and 2/3 chance of not logging in:

Using a binomial distribution, if I avoid Facebook for 365 days then I have (2/3)^365 which is 5 x 10^-65 which is smaller than the Planck length.

I am guessing that the binomial is not working for this case but would the modelling of breaking a simple habit cause such small numbers?

Ap-uk (talk) 12:35, 17 February 2013 (UTC)ap[reply]

Your analysis would be right if the probability were to remain constant day after day for the whole year. But in reality it would not. If you've made it for say 200 days without logging in, it's probably not by chance but rather it's that you have adjusted your behavioral tendencies -- that is, you've adjusted your log-in probability down to near zero. Duoduoduo (talk) 14:53, 17 February 2013 (UTC)[reply]

Agreed. In math-speak, the probabilities from one day to the next are not independent, which is required for the formula you used. So, you need to use P = P_day1P_day2P_day3P_day4... instead. Of course, actually determining P_day1, P_day2, P_day3, P_day4 ... is not easy. StuRat (talk) 17:14, 17 February 2013 (UTC)[reply]

You'd really need to use conditional probabilities:

\Pr[{\hbox{365 days}}]=\Pr[{\hbox{day 1}}]\cdot \Pr[{\hbox{day 2}}\mid {\hbox{day 1}}]\cdot \Pr[{\hbox{day 3}}\mid {\hbox{day 1 and day 2}}]\cdots \Pr[{\hbox{day 365}}\mid {\hbox{day 1 and day 2 and }}\ldots {\hbox{ and day 364}}]

The probabilities near the end are probably really close to 1; given that you have successfully avoided Facebook for the first 364 days, the probability that you'll avoid it on the 365th day is probably almost 100%.

Also, comparing a probability to the Planck length is meaningless; a probability is not a length. —Bkell (talk) 13:15, 19 February 2013 (UTC)[reply]

Analytical geometry question

I'm stuck trying to solve the following problem. Any help would be appreciated!

Which lines crossing point (2, 6) together with the x-axis and the y-axis border a triangle with an area of 25? I'm trying to solve both rising and falling lines.

I know that the equation for the line is $y-6=m(x-2)$ where m is the slope of the line. And I also know that I need to find the points where y=0 or x=0. Then the area would be the product of those two points divided by two.

So when the line crosses the x-axis y=0 and $mx-2m+6=0$ . And when it crosses the y-axis x=0 and $y-6+2m=0$ . But I think I'm doing something wrong here, any ideas? 82.181.210.44 (talk) 15:32, 17 February 2013 (UTC)[reply]

No hang on, I got right two of the lines.

x={\frac {2m-6}{m}}

and

y=6-2m

. So

{\frac {2m-6}{m}}*(6-2m)=50

and we can see that m is either

-2

or

{\frac {-9}{2}}

. And the equations of the two lines are

y=-2x+10

and

y={\frac {-9x}{2}}+15

.

But I still can't figure out the two remaining lines.82.181.210.44 (talk) 17:05, 17 February 2013 (UTC)[reply]

If the triangle is in the second quadrant its area is negative, so the corresponding value in your analysis is -50. This will give a second quadratic equation in m, with two positive roots.←86.186.142.172 (talk) 17:22, 17 February 2013 (UTC)[reply]

You found two triangles in the first quadrant, but there is also a triangle in the 2nd quadrant and one in the fourth quadrant. As 86.186.142.172 says, both those areas can be thought of as negative, so plug that into your above equation to find the two lines. BTW, your initial statement of the problem was quite hard to follow. Try this: "I need to find 4 lines in the XY plane, each of which pass through (2,6), and, in conjunction with the X and Y-axes, form a triangle with an area of 25". The "4" and "each" are the critical bits that were missing from your Q. StuRat (talk) 17:51, 17 February 2013 (UTC)[reply]

Yes, thanks to both of you I finally figured it out. And sorry about the unclear question. The problem was in different language and I had to translate it myself and I'm not good at that. And sorry about the math tags that were missing at first. I was too busy working with this problem so I couldn't be bothered to learn how to use them. But anyways, I think the issue was that I was thinking that area can't be negative, and didn't realise that the roots would be positive. Thanks again! 82.181.210.44 (talk) 18:13, 17 February 2013 (UTC)[reply]

You're quite welcome, and I will mark this Q resolved. StuRat (talk) 18:19, 17 February 2013 (UTC)[reply]

If I were writing the question, I wouldn't specify how many solutions there are. I'd phrase it something like: Find the equations of all possible lines passing through the point (2, 6) such that the line and the co-ordinate axes enclose a triangle of area 25 square units. That way, students who only found two lines by assuming the solution had to be first quadrant would be penalised relative to those who recognised that solutions in quadrants two and four are possible. A good approach to solving the problem that does not require 'negative' area is as follows: having found the two intercepts, we know the area of the triangle is half of their product, but we don't know whether the intercepts are positive or negative, so the equation that actually requires solving is

{\frac {1}{2}}*|{\frac {2m-6}{m}}|*|6-2m|=25

(2m-6)^{2}=50|m|

(m-3)^{2}={\frac {25|m|}{2}}

and this equation will give all four possible values of m (-2, -9/2, 1/2, 18) and hence the equations for all four possible lines (y = 10 - 2x, 9x + 2y - 30 = 0, x - 2y + 10 = 0, and y = 18x - 30). EdChem (talk) 02:28, 22 February 2013 (UTC)[reply]

Resolved