The kicker: How do you explain the difference between 4c and 5c? Can you relate this to a
larger context about conditional probability and making decisions?
Problem Hint:
Compute the new probability for T1
Derived the total probability using the new value of T1
Use Bayes theorem with the updated values to compute new conditional probability of
passing T3 given you have passed T1 and T2
Consider conditional probability and how T1, T2 and T3 are considered a systems
a)
1
−
0.25
¿
¿
1
−
0.25
¿
¿
P
(
k successes
∈
ntrials
)=
(
2
1
)
0.25
1
¿
= 0.375 = 37.5% =
38%
b)
Total probability = (0.38 x 0.33) + (0.22 x 0.33) + (0.38 x 0.33) = 0.1254 + 0.0726 + 0.1254 =
0.3234 = 32.34% =
32%
c)
Probability of passing T1 and T2 = (0.38 x .50) + (0.22 x 0.50) = 0.30 = 30%
p
(
T
3

T
1
∧
T
2
)
=
p
(
T
1
∧
T
2
)
∙ p
(
T
3
)
p
(
T otal
)
=
0.30
∙
0.38
0.32
= 0.356 = 35.6% =
37%
d)
One would expect to see an increase in the conditional probability of passing test number three
given that you passed the first two tests, if the probability of passing either of those tests
independently increases. In 5c, the probability of passing test number one increased, thus
resulting in the increased conditional probability.
Conditional probability can be used for decision making in processes that involve multiple steps.
In multiple system deployments, for example, organizations can use conditional probability to
determine how successful each deployment will be given the success of the previous
deployments. If deployment A is successful, how likely is it that deployment B will be successful
as well?