Independent event
Two events are said to be independent of each other if the probability that one event occurs on any given trial of an experiment is not influenced by the occurrence of the other event under any condition.Alternatively, two events are considered disjoint if they have no outcomes in common and they can never happen together.
Therefore, two events A and B are independent if knowing that one occurs does not change the probability that the other occurs. If A and B are independent,
P(A ∩ B) = P(A) P(B)
P (A│B) = P (A); P (B) > 0
P (B│A) = P (B); P(A) > 0
Likewise, events E1, E2……En are independent iff for any subset of these events: Ei1, Ei2…….Eik are:
P(Ei1 ∩ Ei2∩ …∩Eik = P(Ei1)P(Ei2)…P(Eik)
Let event X, Y and Z be mutually independent, then
P(X∩Y) = P(X)P(Y)
P(Y∩Z) = P(Y)P(Z)
P(X∩Z) = P(X)P(Z)
P(X∩Y∩Z) = P(X)P(Y)P(Z)
Example: Prove that if X and Z are independent events, then so are the events X and Z′.
We know that P(X∩Z) = P(X)P(Z)
Also, P(Z’) = 1 – P(Z)
Therefore,
For X and Z to be independent, P(X∩Z’) = P(X)P(Z’)
We can also write: P(X∩Z’) = P(X) – P(X∩Z)
P(X∩Z’) = P(X) – P(X∩Z)
P(X∩Z’) = P(X) – P(X)P(Z)
P(X∩Z’) = P(X)(1- P(Z))
Hence, P(X∩Z’) = P(X)P(Z’)
In a similar manner, it can be shown that if the events X and Z are independent, then
(a) X′ and Z are independent,
(b) X′ and Z′ are independent
Law of total probability
Let A and B be any event in a sample space S, then
P(A) = P(A∩B) +P(A∩B’) = P(A│B)P(B) + P(A│B’)P(A’)
For multiple events, the marginal probability for event A=
Where:
Example: A person has undertaken a construction job. The probability for a strike to occur is 0.50 and 0.70 that the construction job will be completed on time if there is no strike. The probability is 0.30 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job completes on time.
Solution: Let E be the event that the construction job will be completed on time, and F be the event that there will be a strike. We have to find P(E).
P(F) = 0.50, P(no strike) = P(F′) = 1 − P(F) = 1 − 0.50 = 0.50
P(E|F) = 0.30, P(E|F′) = 0.70
Since events F and F′ form a partition of the sample space S, therefore, by theorem on total probability, we have
P(E) = P(F) P(E|F) + P(F′) P(E|F′)
= 0.50 × 0.30 + 0.50 × 0.70
= 0.15 + 0.35 = 0.50
The probability that the construction job completes in time is 0.50
