# Probability That None of Independent Events Occurs

• Sep 1st 2009, 07:06 PM
azdang
Probability That None of Independent Events Occurs
The problem says:

Let $\displaystyle A_1, A_2, A_3,...,A_n$ be independent events. Show that the probability that none of the $\displaystyle A_i's$ occur is less than or equal to exp(-$\displaystyle \Sigma_{i=1}^{n}P(A_i))$.

I'm confused at where the inequality and the exponent are coming from.

Here is what i've done.

I figured that the P(none happening) = 1 - P(one happening)

P(one happening) = $\displaystyle \Sigma_{k=1}^{n}P(A_k)$.

Then, P(none) = 1 - $\displaystyle \Sigma_{k=1}^{n}P(A_k)$.

I think this is right, but where to go from here, I'm not sure.

Thank you!
• Sep 1st 2009, 10:46 PM
matheagle
First of all
P(none happening) = 1 - P(at least one happening)

Then P(at least one happening)

$\displaystyle =P(\cup A_i)\le\sum_{i=1}P(A_i)$

BUT by independence this is an equality (which is why it works).

What you seem to need is that $\displaystyle 1-x\le e^{-x}$

Use calculus to prove that.

Let $\displaystyle f(x)=e^{-x}+x-1$

then obtain the first two derivatives and show that the min is 0 at 0, so $\displaystyle f(x)\ge 0$.
• Sep 2nd 2009, 05:13 AM
azdang
So, the only thing I had wrong was that it should be P(at least one happening) rather than just P(one happening), but that should still be equal to http://www.mathhelpforum.com/math-he...7be7147f-1.gif, correct?

I will try to work out the other stuff today. Thank you!
• Sep 2nd 2009, 09:06 AM
pedrosorio
That is correct.

Now all you need to show is that

$\displaystyle 1- \sum_{k=1}^{n}P(A_k) \leq e^{-\sum_{k=1}^{n}P(A_k)}$

which is equivalent to proving, as matheagle explained:

$\displaystyle 1- x \leq e^{-x}$

EDIT: Thanks matheagle, I'm always messing up =X
• Sep 2nd 2009, 09:56 AM
matheagle
you're missing the minus in the exponent
but that's true too

$\displaystyle 1- x \leq e^x$

we want $\displaystyle 1- x \leq e^{-x}$

Quote:

Originally Posted by pedrosorio
That is correct.

Now all you need to show is that

$\displaystyle 1- \sum_{k=1}^{n}P(A_k) \leq e^{\sum_{k=1}^{n}P(A_k)}$

which is equivalent to proving, as matheagle explained:

$\displaystyle 1- x \leq e^x$

• Sep 2nd 2009, 10:36 AM
azdang
Thank you both very much! I got it now :)