Download Pascal`s triangle

Bowdoin Math 1300, Spring 2016 – Handout #4
Repeated and independent trials, the binomial coefficients, and Pascal’s triangle
February 29, 2016
If we have n repeated and independent trials, with P (success) = p and P (failure) = q = 1 − p. For example,
we may be tossing a biased coin with P (heads) = p and P (tails) = q = 1 − p. In this case,
n k n−k
P (k heads out of n tosses) =
p q
,
k = 0, 1, 2, 3, . . . , n.
(1)
k
The tree diagram for the first two tosses looks as part (a) of Figure 1 (here, Hi is the event “heads at ith
toss, while Ti is the event “tails at ith toss”). Assuming that heads is a success and tails is a failure, if we
want to compute the probability of having exactly one success and one failure (out of two trials) where the
order does not matter we may merge the (two) branches that correspond to this event, and see that we that
P (one success out of two trials) = 21 pq = 2pq, as illustrated in part (b) of the same figure.
(a)
(b)
Figure 1: Merging the branches that correspond to the outcomes with the same number successes (out of n).
This “merging” of the outcomes with the same number of successes out of n trials (with n fixed) can be performed
at all the levels of the tree, and the picture we get is the one illustrated in Figure 2. Take, for example, the circle
that is highlighted by a star (∗): it corresponds to the event E = “1 success (heads) out of three trials (tosses)”.
The event is comprised of the union of three disjoint events: H1 ∩ T2 ∩ T3 , or T1 ∩ H2 ∩ T3 , or T1 ∩ T2 ∩ H3 .
Each has the same probability q 2 p, and it’s three of them, whence P (E) = 3q 2 p.
Figure 2: Merging the branches that correspond to the outcomes with the same number successes (out of n).
1
Also note that the integer number in each circle in Figure 2 (e.g. the number “3” in the case denoted by the
star) is the sum of the numbers in the “parent” circles (in this case, 1 + 2 = 3): that is because the number
of ways to get to a circle of interest (from the top of the tree) with a sequence of heads and tails is the sum of
(i) the number of ways to get to one of the two parents, plus (ii) the number of ways to get to one of the second
parent. This justifies why the integer numbers in the circles are the same numbers that appear in Pascal’s
triangle (see Figure 3) where each of the numbers is obtained by summing the two “parent” numbers above it.
Figure 3: Pascal’s triangle: each number s computed as the sum of the two parents.
Finally, since we know that the probability of having k successes (heads) out of n trials (tosses) is given by the
formula (1), we conclude that the numbers in Pascal’s triangle are actually equal to the binomial coefficients,
as illustrated in Figure 4.
Figure 4: The binomial coefficients may be computed using Pascal’s triangle!
Note the symmetry
of the
binomial coefficients. For example, 43 = 41 , and
n
the case that nk = n−k
, for any integers n and k such that 0 ≤ k ≤ n.
2
5
2
=
5
3
. In general, it is always