Q:

Probability of getting no worms in your box

I am trying to work out the below fabricated scenario:

Boxes of apples are checked for worms one at a time, and the probability that a given box of apples has worms is $p = 0.1$, independently of the others. If $6$ boxes of apples are tested, what is the probability that there are no boxes of apples infested with worms?

Now I know that we have two outcomes for each event, $\Omega =$ {$S,F$}.
So with six tests we have $2^6$ sets of outcomes. Now I want to find the number of sets of outcomes that have atleast one fail, which is $2^6-1$ since there is only one set with {$S,S,S,S,S,S$} 
Probability of getting no worm boxes is, $1 - pr(\text{getting worms}) \longrightarrow$ $ 1 - \frac{63}{64}$ if they have an even chance of being success or fail(not sure if this is correct), so to factor in the $.1$ of being infested I am unsure what to do. I thought originally you would times the $63/64$ by .1, but that seems to increase the probability of getting all 6 non-defective over even one trial.
Any tips would be appreciated! 

A:

The chance of getting no worms in one box is $0.9$. The chance of getting no worms in two boxes is $0.9 \times 0.9$, since 90% of the time you will be lucky on the first box and 90% of that time you will be lucky on the second box. You can probably figure out where this is going.

