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### Choosing Chocolate Bars

```
Date: 12/04/2001 at 10:49:13
From: Tracie Berlin
Subject: Probability

If I have 3 Mr. Goodbars, 13 Hershey milk chocolate bars, 7 Special
Dark bars, and 3 Krackel bars, what is the probability that I will get
at least 1 Krackel if I pull two out of the bag at the same time
(together, not looking at one first and then looking at the other)?

I know the probability of pulling a Krackel out if I pick one piece
of candy is 3/26, but I don't know if I can even figure out how you'd
do it if you had two.

You know the chances are greater, but I'm not sure if you can just
take 3/26 + 3/26, because there would only be 25 after the first has
been taken out. You have no way of knowing what that first bar is
because you are looking at them both at the same time.

I'm stumped. Help!
```

```
Date: 12/04/2001 at 12:21:12
From: Doctor Greenie
Subject: Re: Probability

Hello, Tracie -

This is a nice problem to show two alternative ways to attack the same
problem. These two alternative approaches are based on the following
two different ways of rephrasing "... at least one Krackel..."

(1) "at least one Krackel" = "either one Krackel or two Krackels"

(2) "at least one Krackel" = "not zero Krackels"

With the first interpretation, the "... either ... or ..." means you
find the answer by adding the probabilities of "one Krackel" and "two
Krackels." With the second interpretation, you find the probability of
"zero Krackels", and the answer to the problem is "everything else" -
i.e., 1 minus the probability of zero Krackels.

Let's solve the problem both ways. You are right about not being able
to use 3/26 twice, because once one bar is selected there are no
longer 26 bars to choose from for the second bar. (Note that even
though you say you choose two at the same time, mathematically we
still consider the two choices as occurring one after the other....)
So, for example, if we consider the case of getting two Krackels, the
probability of getting a Krackel with the first choice is 3/26, and
the probability of getting another Krackel with the second choice is
2/25.

First interpretation.

In this interpretation, "at least one Krackel" is thought of as
"either one or two Krackels"; thinking of the two selections as
occurring one after the other, this is then equivalent to "either
Krackel and then another Krackel, or Krackel and then something else,
or something else and then Krackel."

3    2
P(Krackel, Krackel) = -- * --
26   25

3   23
P(Krackel, other)   = -- * --
26   25

23    3
P(other, Krackel)   = -- * --
26   25

So we then have

3    2    3   23   23    3
P(at least one Krackel) = -- * -- + -- * -- + -- * --
26   25   26   25   26   25

6+69+69   144
= ------- = ---
650     650

Second interpretation.

In this interpretation, "at least one Krackel" is thought of as "not
zero Krackels".

23   22   506
P(other, other) = -- * -- = ---
26   25   650

So we then have

P(at least one Krackel) = 1 - P(other, other)

506   144
= 1 - --- = ---
650   650

I hope this helps. Write back if you have further questions about any
of this.

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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