Could someone please answer this hypothetical question and provide the
formulas that you used? I never got the opportunity to study statisics
and these basic questions have bothered me for years.

A blindfolded batter is swings at a pitch. He has a 20% chance of
hitting that ball.

Questions:
If the pitcher throws 5 balls, what is the chance that the batter will
hit at least one of those balls? and what is the chance that he will
hit all 5?

Good question!

These "basic statistics" questions always seem simple 'til you start
thinking them through.  But here goes.

The second part of your question is actually the easier, so I'll
tackle that first.  For each pitch, the blindfolded batter has a
(rather uncanny) chance to hit the ball of 20%.  For two pitches, his
chances of hitting both are
.20 x .20 or .04, which is the same as saying a 4% chance.  For five
pitches, it's:

.20 x .20 x .20 x .20 x .20 = .00032 or .032%


The second (actually the first) part of your question requires a
reversal of this same method.  The odds of the batter missing any
given pitch is .80, so the odds that he (or she, to be fair) would
miss all five pitches is:

.80 x .80 x .80 x .80 x .80 = .32768 or 32.768%.  So the odds that at
least one pitch gets hit is 100% - 32.768% which gives us 67.232%. 
That is, the odds of hitting at least one pitch is a little better
than 67%, or put another way, about 2 out 3.

Thanks for getting me to dust off the statistical cobwebs.

Thanks for the quick and clearly stated answer. Much appreciated! :-)

Just a quick note -- this is more accurately a question of
probability, rather than statistics. Now on to the Monty Hall
question! ;-)

Oh, yes, the Monty Hall question.  It's **so** counterintuitive (not
the question, the answer).

run some simulations you will find it is easier to understand