We calculate this as 25% chance of flipping two heads on flips one and two. Same thing for
flips three and four.
Let’s now call the first event
(F1&2) and the second one
build the table
. Notice that all the % contributions have to add up to 1.
Table of contributions to total probability:
law of total probability
Total probability for two heads then two heads is (.50 x .25) + (.50 x .25) =
Q3: What is the probability that you will get two heads on flips three and four,
two heads on flips one and two?
Probability of two heads on
(F3&4) conditional on two heads for p(F1&2).
Using Bayes Theorem, we determine that it is not very useful in this form. Here, we are
looking for the probability of the second item,
(F3&4), given the first,
vertical bar means “given.”
Which still is not very helpful. We did not calculate it this way. We did not calculate
(F1&2|F3&4). If we had, we would know the answer is just 1 minus the result. But we
calculate each part’s relation to the whole.
use the form of Bayes’ theorem that
uses the law of total probability as the denominator
to the total probability. From the table we made, we
already know all of those pieces on the right-hand side:
that you will flip two heads in a row after
previously flipping two heads in a row is 50%. Bear in mind that the
of flipping two