Page 26 - Statistics and Data Analysis in Geology

Elementary Statistics

second toss is not dependent in any way on the first. Likewise, the third toss is
independent of the two preceding tosses, and has an associated probability of 1 /2
for heads. So, we have "one-half of one-half of one-half" of a chance of getting all
three heads.
Suppose instead that we are interested in the probability of obtaining only one
head in three tosses. All possible outcomes, denoting heads as H and tails as T,
are:

HHH HTH TTT
HHT THH [THTI
[HTT] [TTH]

Bracketed combinations are those that satisfy our requirements that they con-
tain only one head. Because there are eight possible combinations, the probability
of getting only one head in three tosses is 3 /8.
What we have found is the number of possible combinations of three things
(either heads or tails), taken one item at a time. This can be generalized to the
number of possible combinations of n items taken Y at a time. Symbolically, this
is represented as (r) .
It can be demonstrated that the number of possible combinations of n items,
taken Y items at a time, is

The exclamation points stand for factorial and mean that the number preceding
the exclamation point is multiplied by the number less one, then by the number
less two, and so on:

n! = n (n- 1). (n- 2) ' (n- 3) - ... * (2.2)
*
The value of 3! is 3 .2 . 1 = 6. In our coin-flipping problem,

3! - 3-2.1 =-=3
6
-
(y) = 1!(3 - l)! 1 (2 * 1) 2
That is, there are three possible combinations that will contain one head. By
this equation, how many possible combinations are there that contain exactly two
heads?
3! - 3-2-1 6
--= 3
(z) = 2!(3 - 2)! 2 - l(1) 2
HHH [HTH] TTT
[HHT] [THH] THT
HTT TTH

These combinations are bracketed above in our collection of possible outcomes.
Next, how many possible combinations of three tosses contain exactly three
heads?
3! - 3.2.1
-
(i) = 3!(3 - 3)! 3 2 - l(1) =1

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