Finding probability of at least one event with symmetric conditions

[erisedk]

Homework Statement

For the three events A, B and C, P (exactly one of the events A or B occurs) = P (exactly one of the events B or C occurs) = P (exactly one of the events C or A occurs) = p ,
and P (all the three events occur simultaneously) = p² ,
where 0 < p < ½ .
Then the probability of at least one of the three events A, B and C occurring is:

Ans: $\dfrac{3p+2p^2}{2}$

Homework Equations

The Attempt at a Solution

http://s3.amazonaws.com/minglebox-photo/core-0000-c88370190d4b414d010d4b415d220010.data-0000-fdbffe7622c53ecd0122c5c50d0b0334.gif
NOTE: The regions shown do not overlap with each other, i.e. P(A) ≠ region 1, instead P(A) = region ( 1 + 2 + 4 + 5).
By symmetry, I assume region 1 = 3 = 7 = p/2
Region 5 = p²
Again by symmetry, region 2 = 4 = 6 = x.
I need to find x because the probability that I got to calculate is the sum of all the regions, ie.
$\dfrac{3p}{2} + 3x + p^2$
I don't know how. I initially though that the sum of all probabilities might be 1, but then it doesn't say that this is the case. For all we know, there might be 10 more events. So, I dismissed that option. I don't know how to proceed from here.

 

[PeroK]

Why not just write down the equations you know in terms of the areas 1-7? You don't have to assume symmetry.

 

[erisedk]

How does that help?
Here:
1 + 3 = 3 + 7 = 7 + 1 = p
5 = p²
I still don't know 2, 4, 6
Also, from the answer, I just noticed that 2, 4 and 6 must be 0. Which is very surprising.

 

[PeroK]

How does that help?
Here:
1 + 3 = 3 + 7 = 7 + 1 = p
5 = p²
I still don't know 2, 4, 6
Also, from the answer, I just noticed that 2, 4 and 6 must be 0. Which is very surprising.

Those equations are not right. Look more carefully at your diagram. What areas represent "exactly one of A or B"?

 

[erisedk]

Got it. 1 + 4 + 3 + 6 represents exactly one of A and B. Thank you so much :)

 

[TheMathNoob]

Homework Statement

For the three events A, B and C, P (exactly one of the events A or B occurs) = P (exactly one of the events B or C occurs) = P (exactly one of the events C or A occurs) = p ,
and P (all the three events occur simultaneously) = p² ,
where 0 < p < ½ .
Then the probability of at least one of the three events A, B and C occurring is:

Ans: $\dfrac{3p+2p^2}{2}$

Homework Equations

The Attempt at a Solution

http://s3.amazonaws.com/minglebox-photo/core-0000-c88370190d4b414d010d4b415d220010.data-0000-fdbffe7622c53ecd0122c5c50d0b0334.gif
NOTE: The regions shown do not overlap with each other, i.e. P(A) ≠ region 1, instead P(A) = region ( 1 + 2 + 4 + 5).
By symmetry, I assume region 1 = 3 = 7 = p/2
Region 5 = p²
Again by symmetry, region 2 = 4 = 6 = x.
I need to find x because the probability that I got to calculate is the sum of all the regions, ie.
$\dfrac{3p}{2} + 3x + p^2$
I don't know how. I initially though that the sum of all probabilities might be 1, but then it doesn't say that this is the case. For all we know, there might be 10 more events. So, I dismissed that option. I don't know how to proceed from here.

how did you come up with that Venn's diagram? is it given or you came up with it?. If it's given, then just write down the equation of probability of the union of events.

 

[erisedk]

Oh no, it wasn't given. I got the answer now.

 

[TheMathNoob]

Got it. 1 + 4 + 3 + 6 represents exactly one of A and B. Thank you so much :)

My intuition tells me that you have to use negation to find AnB which corresponds to the areas 2 and 5. I don't know if that's correct

 

[TheMathNoob]

m

 

[TheMathNoob]

Oh no, it wasn't given. I got the answer now.

Ummm I would be glad to know the way you came up with the answer because I don't know XD.

 

[erisedk]

Oh sure!
1 + 4 + 3 + 6 = p
2 + 3 + 4 + 7 = p
1 + 2 + 7 + 6 = p
Adding them,
2 ( 1 + 2 + 3 + 4 + 6 + 7 ) = 3p
(1+2+3+4+6+7) = 3p/2
5 = p²
Hence, 1 + 2 + 3 + 4 + 5 + 6 + 7 = 3p/2 + p²