V. Theorem of Total Probability
The theorem of total probability, also known as Theorem of Elimination, is employed to calculate the probability of an event whose occurrence is dependent on the occurrence (or non-occurrence) of some intermediate events in the experimental process. If an event E is associated with other intermediate, mutually exclusive events H1, H2, …., Hn, then the total probability of its occurrence will be,
P(E) = P(H1 ⋂ E) + P(H2 ⋂ E) + …. + P(Hn ⋂ E)
P(E) = P(H1) x P(E/H1) + P(H2) x P(E/H2) + ……… + P(Hn) x P(E/Hn)
Example:
Suppose that the accounting manager of a company wants to introduce a new policy of assessment of employees of the company for their promotion and the policy is subject to clearance from the general manager (GM). At present, the post of GM is vacant and is likely to be filled soon by appointing one of the three deputy GMs. The chances of policy in question being implemented are dependent on who is appointed the GM. Now, the chances of managers A, B, and C being appointed GM are 0.4, 0.5, and 0.1 respectively. While the likelihood of policy implementation is 0.2, 0.6, and 0.4, respectively, with the three. What is the probability that the policy will eventually be implemented?
Solution:
Let’s say H1, H2, and H3 represent the respective events that A, B, and C are promoted, and E be the event that policy is implemented. Now,
P(H1) = 0.4
P(H2) = 0.5
P(H3) = 0.1
P(E/H1) = 0.2
P(E/H2) = 0.6
P(E/H3) = 0.4
Now, P(E) = P(H1⋂E) + P(H2⋂E) + P(H3⋂E)
P(E) = P(H1).P(E/H1) + P(H2).P(E/H2) + P(H3).P(E/H3)
P(E) = (0.4)(0.2) + (0.5)(0.6) + (0.1)(0.4)
P(E) = 0.08 + 0.3 + 0.04
P(E) = 0.42
Therefore, the overall chances that the policy will be introduced are 42%.