A acevipa Feb 2008 297 7 Jun 1, 2010 #1 Use the additive law of probability to establish, using mathematical induction, Boole's Law: \(\displaystyle P(A_{1}\cup A_{2}\cup...\cup A_{n})\leq P(A_{1}) + P(A_{2})+...+P(A_{n})\)

Use the additive law of probability to establish, using mathematical induction, Boole's Law: \(\displaystyle P(A_{1}\cup A_{2}\cup...\cup A_{n})\leq P(A_{1}) + P(A_{2})+...+P(A_{n})\)

Anonymous1 Nov 2009 517 130 Big Red, NY Jun 1, 2010 #2 acevipa said: Use the additive law of probability to establish, using mathematical induction, Boole's Law: \(\displaystyle P(A_{1}\cup A_{2}\cup...\cup A_{n})\leq P(A_{1}) + P(A_{2})+...+P(A_{n})\) Click to expand... Common proofs like this one can be found online. Here is one: http://www.andrew.cmu.edu/course/21-228/lec7.pdf There are many others that may better suit your needs, though.

acevipa said: Use the additive law of probability to establish, using mathematical induction, Boole's Law: \(\displaystyle P(A_{1}\cup A_{2}\cup...\cup A_{n})\leq P(A_{1}) + P(A_{2})+...+P(A_{n})\) Click to expand... Common proofs like this one can be found online. Here is one: http://www.andrew.cmu.edu/course/21-228/lec7.pdf There are many others that may better suit your needs, though.

Grandad MHF Hall of Honor Dec 2008 2,570 1,416 South Coast of England Jun 2, 2010 #3 Hello acevipa acevipa said: Use the additive law of probability to establish, using mathematical induction, Boole's Law: \(\displaystyle P(A_{1}\cup A_{2}\cup...\cup A_{n})\leq P(A_{1}) + P(A_{2})+...+P(A_{n})\) Click to expand... Since you are asked to use induction, the proof will be something like this: Suppose that the proposition is true for \(\displaystyle n = k\). So\(\displaystyle P(A_{1}\cup A_{2}\cup...\cup A_{k})\leq P(A_{1}) + P(A_{2})+...+P(A_{k})\) Then\(\displaystyle P\Big((A_{1}\cup A_{2}\cup...\cup A_{k})\cup A_{k+1}\Big)\) \(\displaystyle =P(A_{1}\cup A_{2}\cup...\cup A_{k})+P(A_{k+1})-P\Big((A_{1}\cup A_{2}\cup...\cup A_{k})\cap A_{k+1}\Big)\), using the addition law of probability\(\displaystyle \leq P(A_{1}\cup A_{2}\cup...\cup A_{k})+P(A_{k+1})\), since \(\displaystyle P\Big((A_{1}\cup A_{2}\cup...\cup A_{k})\cap A_{k+1}\Big) \geq 0\) \(\displaystyle \Rightarrow P(A_{1}\cup A_{2}\cup...\cup A_{k}\cup A_{k+1})\) \(\displaystyle \leq P(A_{1}) + P(A_{2})+...+P(A_{k})+P(A_{k+1})\), using the Induction Hypothesis When \(\displaystyle n = 1\), the hypothesis is clearly true. So it is true for all \(\displaystyle n \ge 1\). Grandad Last edited: Jun 2, 2010

Hello acevipa acevipa said: Use the additive law of probability to establish, using mathematical induction, Boole's Law: \(\displaystyle P(A_{1}\cup A_{2}\cup...\cup A_{n})\leq P(A_{1}) + P(A_{2})+...+P(A_{n})\) Click to expand... Since you are asked to use induction, the proof will be something like this: Suppose that the proposition is true for \(\displaystyle n = k\). So\(\displaystyle P(A_{1}\cup A_{2}\cup...\cup A_{k})\leq P(A_{1}) + P(A_{2})+...+P(A_{k})\) Then\(\displaystyle P\Big((A_{1}\cup A_{2}\cup...\cup A_{k})\cup A_{k+1}\Big)\) \(\displaystyle =P(A_{1}\cup A_{2}\cup...\cup A_{k})+P(A_{k+1})-P\Big((A_{1}\cup A_{2}\cup...\cup A_{k})\cap A_{k+1}\Big)\), using the addition law of probability\(\displaystyle \leq P(A_{1}\cup A_{2}\cup...\cup A_{k})+P(A_{k+1})\), since \(\displaystyle P\Big((A_{1}\cup A_{2}\cup...\cup A_{k})\cap A_{k+1}\Big) \geq 0\) \(\displaystyle \Rightarrow P(A_{1}\cup A_{2}\cup...\cup A_{k}\cup A_{k+1})\) \(\displaystyle \leq P(A_{1}) + P(A_{2})+...+P(A_{k})+P(A_{k+1})\), using the Induction Hypothesis When \(\displaystyle n = 1\), the hypothesis is clearly true. So it is true for all \(\displaystyle n \ge 1\). Grandad