parts—18 pebbles split into two piles of 9, for exam-
ple. An odd pile of pebbles leaves an extra pebble when
one attempts to accomplish this feat: 23 pebbles splits
into two piles of 11, with one left over. This interpreta-
tion shows that the number zero is even: an empty pile
of pebbles can be split into two equal piles of nothing.
Combining together several piles of pebbles, each
of which can be evenly split, produces a large pile that
still can be so separated. This shows that the sum of
any collection of even numbers is even. Combining two
odd piles of pebbles produces a result that is even—the
two “errant” pebbles combine to produce a result that
can be evenly split. The sum of three odd numbers
leaves a single errant pebble, however, and so is odd. In
summary we have:
The sum of any number of even numbers
is even.
The sum of an even number of odd numbers
is even.
The sum of an odd number of odd numbers
is odd.
These simple ideas are quite powerful. For example, we
can quickly ascertain that:
The sum of the numbers 1 through 100 will
be even.
(This sum contains 50 even numbers and 50 odd
numbers.)
It is impossible to make change for a dollar
using a combination of 13 pennies, nickels and
quarters.
(Thirteen coins of odd denomination will sum to an odd
amount. It is impossible then to reach a sum of 100.)
If 15 arbitrary sheets are torn from a textbook,
then the sum of the missing page numbers is
necessarily odd.
(Each sheet contains an odd page number on one side
and an even page number on the other. The sum of 15
odd numbers and 15 even numbers is necessarily odd.)
Seventeen cups are placed upside-down on a
tabletop. Turning four cups over at a time, it is
impossible to reach a state in which every cup
is upright.
(To be left upright, each cup must be turned over an odd
number of times in the process of the game. Thus an odd
number of inversions must occur in all, being a sum of
17 odd numbers. But an odd total will never occur when
turning an even number of cups over at a time.)
See also
PARITY
.
event Any subset of the
SAMPLE SPACE
of all possible
outcomes of an experiment is called an event. For exam-
ple, the act of casting a die has sample space {1, 2, 3, 4,
5, 6}—all six possible scores—and the event “the score is
even” is the subset {2, 4, 6}. An event could be a single
outcome (“rolling a two,” for example, is the subset {2}),
the whole sample space (“rolling a number less then
10”), or the empty set (“rolling a multiple of seven”).
The probability of an event Eoccurring is the ratio
of the number of outcomes in that event to the total
number of outcomes possible. This ratio is denoted
P(E).For example, in casting a die, P(even) = = ,
P({5,6}) = = , and P(multiple of 7) = = 0.
S
ET THEORY
is useful for analyzing the chances of
combinations of events occurring. If Aand Brepresent
two events for an experiment, then:
The intersection A∩Brepresents the event
“both Aand Boccur.”
The union A∪Brepresents the event “either
Aor Boccurs.”
The complement A′represents the event “A
does not occur.”
To illustrate: if, in casting a die, Ais the event {2, 4, 6}
and Bthe event {5,6}, then A∩B= {6}, A∪B= {2, 4,
5, 6}, and A′= {1, 3, 5}.
The following probability laws hold for two events
Aand B:
i. P(A∩B) = P(A) + P(B) – P(A∪B).
ii. When Aand Bare disjoint events, that is have no
outcomes in common, then P(A∪B) = P(A) + P(B).
iii. P(A′) = 1 – P(A).
iv. When A and B are
INDEPENDENT EVENTS
, P(A∩B)
= P(A) ×P(B).
The study of
PROBABILITY
explains these rules.
See also
CONDITIONAL
;
MUTUALLY EXCLUSIVE
EVENTS
;
ODDS
.
0
–
6
1
–
3
2
–
6
1
–
2
3
–
6
178 event