Question

Ana dashes playing cube. If falls number lower than 4, she chooses random point (x,y) from square apexes in (0,0), (0,1), (1,0), (1,1). In contrary, she chooses random point (x,y) from the unit circle.
a) Which is probability that x+y>1 (x and y are coordinates of chosen point )?
b) If x+y>1, which is probability that the point is chosen from circle?

Answer 1

a point (x,y) such that
\[x+y>1\] is the region shaded above. i graphed
\[y=1-x\] on unit square. the probability of picking (x,y) in that region is
\[\frac{1}{2}\] because it represents half the area

Answer 2

wait

Answer 3

stop writing. I'll tell u what I don't understand, u don't have to write it all

Answer 4

ok i wait

Answer 5

well my answer was wrong anyway, so i should wait!

Answer 6

this is how I was doing this
1. I find probabilty that number is lower than 4 and probability that point is chosen above y=1-x
2. I find probabiliti that number 4, 5 or 6 and probability that point is chosen above y=1-x and under y=sqrt(1-x^2)

but since that part of circle is in square too, I don't know do I have to subtract something

Answer 7

ok i messed up so lets go slow
1: probability that number is lower than 4 and the point is chosen above y = 1 - x is 1/4
is that part ok?

Answer 8

yes

Answer 9

2; probability that number chosen is 4, 5, or 6 is 1/2, so now we only need to find the probability that a point chosen at random from the unit circle is above the line y = 1-x

Answer 10

why under y=1-x?

Answer 11

oh, sorry :D

Answer 12

this amounts to computing the area in the unit circle above the line y = 1 - x and then dividing by the area of the unit circle. is that ok?

Answer 13

it is

Answer 14

so area of the unit circle is
\[\pi\] and area of the shaded region we can find by taking
\[\frac{\pi}{4}\] the area of the unit circle in quadrant I and subtracting
\[\frac{1}{2}\] the area of the 1 by 1 triangle we don't want

Answer 15

so i get for the area of the shaded region
\[\frac{\pi}{4}-\frac{1}{2}\] is that part ok?

Answer 16

hahha, great :D I was doing it with integral :D

Answer 17

oh well that is fancy and will probably work, but i am rather simple minded

Answer 18

of course we are not done. we have to divide
\[\frac{\frac{\pi}{4}-\frac{1}{2}}{\pi}\]

Answer 19

sure

Answer 20

then multiply the result by
\[\frac{1}{2}\] and finally add 1/4 to the whole thing to answer the question. i sense that this is ok now right? will do the arithmetic and see what i get

Answer 21

which result do I have to divide by 2? the result for circle? and why to divide?

Answer 22

oh i didn't mean to confuse. i meant for "final answer"
we are just computing the probability you end up above y = 1-x if pick inside the unit circle. then we have to multiply by 1/2, the probability you got 4, 5,or 6
and then add 1/4 from part 1

Answer 23

i just meant to get the actual answer to question one, those are the steps

Answer 24

aha, I forgot it :) but don't I have to subtract something since the part in circle is in square too?

Answer 25

no i don't think so because the event are disjoint

Answer 26

either you get 1,2,3 OR you get 4,5,6

Answer 27

then either you are above or below . so we can just add the probabilities, unless i am missing something.
put
\[A=\{1,2,3\}\] then
\[A^c=\{4,5,6\}\]
\[B=\{(x,y)|y>1-x\}\]
\[B^c=\{(x,y)|y<1-x\}\] and we want
\[P(B)=P(B\cap A)+P(B\cap A^c)=P(A)P(B|A)+P(A^c)P(B|A^c)\]

Answer 28

the point being that A and A^c are disjoint, i.e. there is no overlap,so you can just add up the probabilities

Answer 29

it seems it's good :) thank u very much :) I'm very happy now because did it the same way :)

Answer 30

can u see one problem more?

Answer 31

great. let me know what answer you get
sure post it or ask here. probably better to just post new in case i cannot do it!

Answer 32

i posted :)