# Thread: Probability That None of Independent Events Occurs

1. ## Probability That None of Independent Events Occurs

The problem says:

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

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

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

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

Thank you!

2. First of all
P(none happening) = 1 - P(at least one happening)

Then P(at least one happening)

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

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

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

Use calculus to prove that.

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

then obtain the first two derivatives and show that the min is 0 at 0, so $f(x)\ge 0$.

3. 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 , correct?

I will try to work out the other stuff today. Thank you!

4. That is correct.

Now all you need to show is that

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

which is equivalent to proving, as matheagle explained:

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

EDIT: Thanks matheagle, I'm always messing up =X

5. you're missing the minus in the exponent
but that's true too

$1- x \leq e^x$

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

Originally Posted by pedrosorio
That is correct.

Now all you need to show is that

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

which is equivalent to proving, as matheagle explained:

$1- x \leq e^x$

6. Thank you both very much! I got it now

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