<span>The mathematical theory of probability assumes that we have a well defined repeatable (in principle) experiment, which has as its outcome a set of well defined, mutually exclusive, events.</span>
If we assume that each individual coin is equally likely to come up heads or tails, then each of the above 16 outcomes to 4 flips is equally likely. Each occurs a fraction one out of 16 times, or each has a probability of 1/16.
Alternatively, we could argue that the 1st coin has probability 1/2 to come up heads or tails, the 2nd coin has probability 1/2 to come up heads or tails, and so on for the 3rd and 4th coins, so that the probability for any one particular sequence of heads and tails is just (1/2)x(1/2)x(1/2)x(1/2)=(1/16).
Now lets ask: what is the probability that in 4 flips, one gets N heads, where N=0, 1, 2, 3, or 4. We can get this just by counting the number of outcomes above which have the desired number of heads, and dividing by the total number of possible outcomes, 16.
<span>N # outcomes with N heads probability to get N heads</span>
0 1 1/16 = 0.0625
1 4 4/16 = 1/4 = 0.25
2 6 6/16 = 3/8 = 0.375
3 4 4/16 = 1/4 = 0.25
4 1 1/16 = 0.0625
We can plot these results on a graph as shown below.
The dashed line is shown just as a guide to the eye. Notice that the curve has a "bell" shape. The most likely outcome is for N=2 heads, where the curve reaches its maximum value. This is just what you would expect: if each coin is equally likely to land heads as tails, in four flips, half should come up heads, that is N = 4x(1/2) = 2 is the most likely outcome. Note however that an occurrence of N = 1 or N = 3 is not so unlikely - they occur 1/4 or 25% of the time. To have an occurrence of only N = 0, or N = 4 (no heads, or all heads) is much less likely - they occur only 1/16 or 6.25% of the time.
The above procedure is in principle the way to solve all problems in probability. Define the experiment, enumerate all possible mutually exclusive outcomes (which are usually assumed to be each equally likely), and then count the number of these outcomes which have the particular property being tested for (here for example, the number of heads). Dividing this number by the total number of possible outcomes then gives the probability of the system to have that particular property.
Often, however, the number of possible outcomes may be so large that an explicit enumeration would become very tedious. In such cases, one can resort to more subtle thinking to arrive at the desired probabilities. For example, we can deduce the probabilities to get N heads in 4 flips as follows:
N=0: There is only one possible outcome that gives 0 heads, namely when each flip results in a tail. The probability is therefore 1/16.
N=4: There is only one possible outcome that gives 4 heads, namely when each flip results in a head. The probability is therefore 1/16.
N=1: There are 4 possible outcomes which will have only one coin heads. It may be that the 1st coin is heads, and all others are tails; or it may be that the 2nd coin is heads, and all others are tails; or it may be that the 3rd (or the 4th) coin is heads, and all others are tails. Since there are 4 possible outcomes with one head only, the probability is 4/16 = 1/4.
N=3: To get 3 heads, means that one gets only one tail. This tail can be either the 1st coin, the 2nd coin, the 3rd, or the 4th coin. Thus there are only 4 outcomes which have three heads. The probability is 4/16 = 1/4.
N=2: To enumerate directly all the possible outcomes which have exactly 2 heads only, is a bit trickier than the other cases. We will come to it shortly. But we can get the desired probability for N=2 the following way: We have already enumerated all possible outcomes with either N = 0, 1, 3, or 4 heads. These account for 1 + 4 + 4 + 1 = 10 possible outcomes. The only outcomes not include in these 10 are those with exactly N=2 heads. Since there are 16 possible outcomes, and 10 do not have N=2 heads, there must therefore be exactly 16 - 10 = 6 outcomes which do have exactly N=2 heads. The probability for N=2 is therefore 6/16 = 3/8.
2) Consider the experiment of rolling 3 dice, each of which has 6 sides.
What is the probability that no two dice land with the same number side up, i.e. each of the three dice rolls a different number?
Since each die has 6 possible outcomes, the number of possible outcomes for the roll of three dice is 6x6x6 = 216. We could enumerate all these 216 possibilities, and then count the number of outcomes in which each die has a different number. This is clearly too tedious! Instead we reason as follows:
is the intersection of two circles, both of diameter 10. Each petal in turn is twice the area of a circular segment bounded by a chord of length 
, which implies the segment is subtended by an angle of 
. This means the area of the segment is
, and the area of 
.
is simply the area of the square minus the area of 
.
(provided these regions are indeed disjoint; it's hard to tell from the picture)
