Toss a fair coin three times, list whether a
head or tail occurs each time a coin is tossed and count the number of outcomes
of three tosses in which the order whether head or tail occurs first or last in
an outcome does not matter. Expand the formula to ‘n’ number of tosses.
Counting the total and favorable numbers of
outcomes constituting the specified number of objects selected for a binary
variable is important to calculate the probability of favorable events for the
Binomial Probability Distribution. Tree diagram is an important means to
visualize and count the number of outcomes. However, as the number of times the
objects are selected increases, the diagram becomes complicated and it will be
less possible to show all outcomes in the diagram. Thus, in that case one needs
to use the formula to count the number of outcomes. This article presents both
tree diagram and Binomial Distribution formula to exemplify the process of
counting.
Let X be a binary
variable that takes one of two values Head (H) or Tail (T) in any of three independent
tosses of a coin, also known as the sampling with replacement. There will be
altogether eight outcomes in three tosses, grouped into four combinations if
the order of the coin side whether H or T occurs does not matter. This is the
case of all objects sampled with replacement in which order does not matter, as
discussed in my previous statistical notes 10 and 15 (Diagram 1). Every toss is independent and in
every toss any of all possible values are likely to occur.
Diagram 1: Outcomes of All
Objects Selected With Replacement in which the Order of Objects does not Matter
Four combinations of three
objects include – one set of all three heads (HHH or H3), three sets
of two heads and one tail in any order (3HHT or 3H2T) three sets of
one head and two tails in any order (3HTT or 3HT3) or one set of
three tails (TTT or T3).
As the number of toss increases,
the diagram becomes complicated and it will be less possible to show all
outcomes in the diagram. Thus, in that case one needs to use the formula to
count the number of outcomes.
The total number of outcomes is,
in this case the sum of H3, 3H2T, 3HT3 and T3.
It is an expansion of H plus T cube, that is (H+T)3. Using the same
logic, the total number of outcomes in four tosses can be calculated using the formula
(H+T)4. If a coin is tossed ‘n’ number of times, the total number of
outcomes will be (H+T)n. The expanded form of this formula will look
like:
(H+T)n = C(n,0)Hn+C(n,1)Hn-1T+C(n,2)Hn-2T2+C(n,3)Hn-3T3+
C(n,x)Hn-xTx+……+ C(n,n)Tn
In the above formula, C(n,x)
is the combination of x tails in in ‘n’ tosses. It is termed as Binomial coefficient,
which is derived by number of tosses of a coin as exemplified in Diagram 2.
Diagram 2: Number of toss of a coin and Binomial Coefficient using Pascal’s Triangle
The total number of heads
or tails over the n independent tosses is a discrete random variable (X) that takes
the values from 0 to n. This random variable X is said to follow the binomial distribution.
The probability of ‘x’ tails (or ‘n-x’ heads in ‘n’ tosses, P (X=x), is given
by the expression C(n,x)Hn-xTx. This helps to calculate the
probability of all possible outcomes, for example, three heads in three tosses
of a coin.
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