The figure below shows how Pascal's Triangle is related to the expansion of a binomial.

• If the rows of the triangle are numbered 0, 1, 2, ..., then row n will contain the coefficients obtained from the expansion of (p + q) ^ n.

We see that when we flip 3 coins, 3 of the 8 possible outcomes give us 2 heads, so that the probability of 2 heads is 3/8.

We could reason out this result without having to list all the possibilities:

• A given outcome consists of 3 'heads' or 'tails', each occurring with probability 1/2.
• It follows that the probability of a given outcome is (1/2)^3 = 1/8.

• There are C(3, 2) = 3 outcomes with 2 'heads'.

• The probability of obtaining 2 Heads on 3 flips is therefore C(3,2) = 3 times the probability 1/8 of any given outcome.

`06

Now if we roll a single die, the probability of any given outcome is 1/6.

We wish to determine the probability of obtaining two 1's when rolling 3 dice.

• The probability of obtaining 1 on each of the first two rolls and something else on the third is the product of three factors, 1/6 for each of the first two rolls representing the probability of obtaining 1 on each, and 5/6 for the third roll, where we obtain something other than 1.
• We can also obtain two 1's by obtaining the 1's on the first and third rolls and something else on the second; we see that we obtain the probability of this outcome by multiplying the same factors as before but in a different order.
• We would also obtain the same result for the probability of obtaining 1's on the second and third rolls and something different on the first.

We see that the probability of a given occurrence having exactly two 1's is 1/6 * 1/6 * 5/6 = 5 / 216.

• Since there are C(3,2) = 3 ways of obtaining exactly two 1's, we see that the probability we seek is 3 * 1/6 * 1/6 * 5/6 = 3 * (5 / 216).
• We can write this result as C(3,2) * (1/6)^2 * (5/6).

If we regard 1 as a 'success', then the probability of 2 successes on 3 trials is C(3, 2) * p^2 * q, where p is the probability of success on a given trial (e.g., 1/6 in the previous example) and q is the probability of failure (1 - 1/6 = 5/6, or q = 1 - p).

• The probability of 5 successes on 12 rolls of the die is the product of the C(5,12) ways in which the 5 successes could occur, and the probability (1/6)^5 * (5/6)^7 of any one of these occurrences.

We generalize this expression to C(n, r) * p^r * q^(n-r), which is the probability of r successes and n-r failures on n events.

To estimate the area of a picture of a mastodon on a 4 * 8 sheet of plywood, we could move a long distance from the sheet and begin firing an inaccurate shotgun in the general direction of the picture.

• The pellets would hit pretty much random over the entire sheet of plywood.
• If we count the number of pellets which have hit the plywood and the number which have hit within the picture of the mastodon, we will be able to obtain a reasonably good estimate of the ratio of the picture's area to that of the plywood.
• If we multiply this ratio by the area of the plywood we should get the area of the picture.

We ask a related question: How many of 500 points distributed randomly over a square should be expected to lie within an inscribed circle.

• The circle has some radius r, and the side of the square therefore has length equal to 2 r, the diameter of the circle.
• The ratio of circle area to square area is therefore `pi r^2 / (2r)^2 = `pi r^2 / (4 r^2) = `pi / 4.
• We therefore expect that `pi/4 * 500 of the randomly positioned points will lie within the circle.