# A cloak covered with patches

• Nov 16th 2012, 09:27 AM
wilhelm
A cloak covered with patches
Hi. Could you help me solve this problem. I've tried using inclusion–exclusion principle but it didn't work.

There is a cloak which area equals 1. It is completely covered with 5 patches, whose area is at least 0,5. Show that the area of the intersection of certain two patces is at least 0,2. ( Hint: It's not as easy as you think)
• Nov 16th 2012, 08:59 PM
chiro
Re: A cloak covered with patches
Hey wilhelm.

Can you relate the intersection of the two patches with Kolmgorovs axiom (i.e P(A OR B) = P(A) + P(B) - P(A and B)) where if A is patched part and B is the cloak the A is a subset of B which implies that P(A and B) = P(A)?
• Nov 18th 2012, 03:07 AM
wilhelm
Re: A cloak covered with patches
I've tried using the probability axioms, but it seems hopeless. There are five patches A, B, C, D, E. We have $\displaystyle 1=P(A _{1} \cup A _{2} \cup A_3 \cup A_4 \cup A_5) =$ $\displaystyle \sum_{I=1}^{5} P(A_i) - \sum_{1 \le i<j \le 5} P(A_{i} \cap A_j) + \sum_{1 \le i<j<k \le 5} P(A_{i} \cap A_j \cap A_k) -$$\displaystyle \sum_{1 \le i<j<k<l \le 5} P(A_{i} \cap A_j \cap A_k \cap A_l) +P(A_{1} \cap A_2 \cap A_3 \cap A_4 \cap A_5).$ Could you please give me a hint on how to estimate these values? I know the sum must equal one because the patches cover the whole cloak
• Nov 18th 2012, 05:19 PM
chiro
Re: A cloak covered with patches
One suggestion I have is if two things involve subset relations (i.e A is a subset of B) then P(A) <= P(B).

This gives you inequalities to use for the three, four, and five intersection terms amongst others and can put bounds on all subsets.

Recall that the intersection of two events in a subset of the union of those two events.