Math Is Fun Forum

Here's the answer to question (1):

We need to use the conditional probability formula:

(trivially provable  - http://www.stat.yale.edu/Courses/1997-9 … ndprob.htm)

Also obvious is that:

So lets go:

Let

P(S|G) = Probability that a person wears spectacles given that they are a girl
P(G|S) = Probability that a person  is a girl given that they wear spectacles
P(C) = Probability that a person picked at random is neither a girl nor wears spectacles

We are given:

It is plainly clear from a Venn diagram that:

Call this "Forumula A"

So lets keep our terminology simple and define:

D = P(G)
E = P(S)
F = P(GnS)

Note that it is F that we are asked to find.

Now here's where we use the conditional probability formulae:

So from Formula A we have:

Remember we know the numerical values of A and B, so let's put them in:

Solve this for F to get the desired:

(Further reading - http://en.wikipedia.org/wiki/Bayes%27_theorem)

Question (2)

Okay, so we want the probability of restoring bag A to its original contents after randomly taking two from bag A and placing them in B and then randomly taking two from B and placing them in A.

We could do this in only three ways:

Event A: Take two red from bag A and put them in bag B and then take two red from bag B and put the in bag A

OR

Event B: Take two green from bag A and put them in bag B and then take two green from bag B and put the in bag A

OR

Event C: Take one red and one green from bag A and put them in bag B and then take one red and one green from bag B and put them in bag A

So our answer is:

P(Restoring Bag A) = P(Event A) + P(Event B) + P(Event C)

Because we can restore bag A only by acheiving Event A OR Event B OR Event C.

Okay,

Event A:

Probability of taking a red from bag A = 3/5 and so remaining probability of taking a red = 2/4 and so probability of taking two reds is (3/5)*(2/4) = 3/10

Now bag B contains four reds and three greens. So probability of taking a red from bag B = 4/7 and so remaining probability of taking a red from bag B is 3/6 and so probability of taking two reds from B is 4/7 * 3/6 = 2/7

And so finally, probability of restoring bag A in this way is the probability of BOTH these things happening = 3/10 * 2/7 = 3/35

Event B:

This is calculated in exactly the same way but using green and gives 1/21

Event C:

Here, the probability of taking one red and one green from bag A is:

3/5 * 2/4 + 2/5 * 3/4  = 3/5 becasue we might take one red then one green OR one green then one red

These are put into B and in a similar way probability of taking one green and one red from B is now 3/7*4/6 + 4/7 * 3/6 = 4/7

Multiply these to give probability of restoring A in this way = 3/5 * 4/7 = 12/35

Finally add all the results to give:

3/35 + 1/21 + 12/35 = 10/21

let:

Using integration by parts we get:

Replacing

Finally on solving for

Here's a basic method:

Create a list of points initially empty.

For one rectangle check each side for intersection points with other rectangle's sides - put these points on the list.

For both rectangles check for corners which are inside the other rectangle - put these points on the list.

Triangluate the polygon formed by the points on the list. (One way - Sort the points in the list into cyclical order (clockwise or anti-clockwise) - then choise the first point and make triangle with 2nd and 3rd point, then with third and fourth point, and so on).

Sum the areas of these triangles to give area of overlap.

Mitch.

Two "lines" crossing. Each line has an assocaited thickness w - so each "line" can be thought of as two lines - parallel and a distance w apart.

So your two "lines" crossing is really four lines crossing (two pairs of parallel lines).

So in general the intersection will consist of four points defining a parallelogram.

You can work out the points of intersection of your two lines to get the coordinates of the corners of this parallelogram and then simply use:

http://jtaylor1142001.net/calcjat/Solut … logram.htm

to find the area.

We have:

and

Using the second equation gives:

Call the above equation (1)

Equation (1) can also be written:

Substituting these values for x and its square into the first of the original equations gives:

Each term has y as a factor, so y = 0 is a solution, putting this value for y into equation (1) gives x = 0, therefore (x = 0,y= 0) is one solution of the original equations. Now lets divde through by y giving:

divide through by 2 giving:

and so

giving

Now square both sides giving:

Expand to give:

which is to say:

This is quadratic in y cubed, let z = y^3 giving:

Using the quadratic formula this gives:

therefore

and so

and so, finally:

There are two other solutions here for y of:

and

But due to the fractional powers of the negative quantities, these are not real.

So in summary we have found solutions when:

Which gives our solution set as (Using equation (1) to find x for each y):

y = 2x-2 = 2(x-1) will pass through the point (3,4) and has a gradient of 2

Hi,

You could also make use of the more general product rule, for the case of three functions of x (as you have here) we have:

let y = uvw where each of u,v and w are functions of x then:

The extentsion to n functions of x should be quite obvious.

This is the way to go:

We simultainiously need:

From the first equation we have:

and so substituting this into the second we have:

So substituting this value for m into the first equation gives:

So finally, substituting these values for k and m into the desired formula we get:

Check by substituting in (2,10) and (5,60) and you will find they satisfy this equation.

We need to find:

And I have a series which gives me the value of:

Well, I notice that

which looks identical to my formula if only the 'x' in that formula were replaced by 0.05. And it's okay to use the formula because the modulus of 0.05 < 1.

Mitch.

#5:

#3: