Sunday, April 24, 2011

Solution for probability Problem for area

Thanks everyone for trying the problem and posting in the comments, and congrats for those of you who got the right answer!


Problem:
Suppose a and b are single-digit positive integers chosen independently and at random. What is the probability that the point (a,b) lies above the parabola y=ax^2-bx.


Solution:
Because the point (a,b) needs to be above the parabola ax^2-bx, the value of y has to be greater than the function itself. Therefore we get y>ax^2-bx. Next, we're finding the probability that point (a,b) is above the parabola, so we just plug in a for x and b for y. Thus, we get b>a*a^2-b*a = a^3-ab. Now we can solve for b in terms of a.


b>a^3-ab 
b+ab>a^3
b(1+a)>a^3
b>(a^3)/(1+a) (We can divide by (1+a) because in the problem it states a is positive)


Now we have the inequality b>(a^3)/(1+a). From this you can see that (a^3)/(1+a) is an increasing function, since a^3 grows faster than 1+a. From this, you see that the largest integer you can plug in for a is 3 because plugging in a 4 for a requires b to be a double digit integer. We have restricted a to 1≤a≤3. Now, we plug in values of a and see what positive single digit integers of b can make the inequality true.


If we plug in 1 for a, we get b>1/4. All numbers 1-9 can be plugged in for b for the inequality to hold. Which gives 9 possible ordered pairs. 
If we plug in 2 for a, we get b>8/3. Numbers 3-9 can be plugged in for b which gives 7 ordered pairs.
If we plug in 3 for a, we get b>27/4. Number 7-9 can be plugged in for b which gives 3 ordered pairs.


In total, the number of possible ordered pairs is 9+7+3=19. This is out of the total number of 9*9=81 ordered pairs. Hence the probability is 19/81.


Here's the next problem:
Problem:
A square of side length 1 and a circle of radius (√3)/3 share the same center. What is the area inside the circle, but outside the square?


Good luck! and remember calculators are not allowed. 

12 comments:

  1. I think the first step would be to find the intersection points between the circle and the square, but i'm not sure where to go from there.

    ReplyDelete
  2. Ok, let's give this a shot. Man, I wish I could use my calculator.

    ReplyDelete
  3. i'll give it a try this time

    utubed.blogspot.com

    ReplyDelete
  4. Square is √(1/2) units away from center at all 4 points and (1/2) units on the center of sides. So the circle dips out of the square a little. We would need to find the points of intersection and and make triangles with the side length of the cords in the circle. That requires trig... Pretty sure calculators are allows. XD

    ReplyDelete
  5. I think I'm overanalysing this way too much. I'm on my second page already.

    ReplyDelete
  6. This is why I majored in polsci

    ReplyDelete
  7. i'll give a try in that one

    ReplyDelete
  8. Oh man I dOnt know if I want to do math in the summer!!!! I already got an 800 on the sat math hahah

    +follow

    ReplyDelete