Addition Theorem of Probability
(i) If A and B be any two events in a sample space S, then the probability of occurrence of at least one of the event A and B is given by
P(AᴜB) = P(A) + P(B) – P(A∩B)
Note:
⇒If A and B are mutually exclusive events, then A∩B = Ф and hence P(A∩B) = 0.
P(AᴜB) = P(A) + P(B)
⇒Two events A and B are mutually exclusive
if P(AᴜB) = P(A) + P(B)
⇒ P(S) = P (A ᴜ A’) = P(A) + P(A’) = 1
(A ∩ A’ = Ф)
P(A’) = 1 – P(A)
(ii) If A, B and C are any three events in a sample space S, then
P(AᴜBᴜC) = P(A) + P(B) + P(C) – P(A∩B) – P(B∩C) – P(A∩C) + P(A∩B∩C)
Note:
⇒If A, B, C are mutually exclusive events then A∩B = Ф, B∩C=Ф, A∩C = Ф, A∩B∩C = Ф
P(AᴜBᴜC) = P(A) + P(B) + P(C)
⇒If A and B are any two events, then (A-B) ∩(A∩B) = Ф and A = (A – B) ∩ (A∩B) = Ф
P(A) = P (A – B) + P(A∩B) = P(A∩B’) + P(A∩B)
Similarly, P(B) – P(A∩B) = P (B – A) = P(B∩A’)
(iii) General form of addition theorem of probability
If A₁, A₂, …, An are n sample space, then
\(P\left( {{A}_{1}}\cup {{A}_{2}}\cup …\cup {{A}_{n}} \right)=\sum\limits_{i=1}^{n}{P\left( {{A}_{i}} \right)}-\sum\limits_{I<1}{P\left( {{A}_{i}}\cap {{A}_{j}} \right)}+\sum\limits_{i<j<k}{P\left( {{A}_{i}}\cap {{A}_{j}}\cap {{A}_{k}} \right)-}…+{{\left( -1 \right)}^{n-1}}P\left( {{A}_{1}}\cap {{A}_{2}}…\cap {{A}_{n}} \right)\).
Example: In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. Find the probability that the student opted for NCC or NSS.
Solution: Let A and B denoted the students in NCC and NSS, respectively
Here n (A) = 30
n (B) = 32
n (A∩B) = 24
24 students opted for both NCC and NSS i.e., that are common in both
P(A) = 30/ 60
P(b) = 32/ 60
And
P(A∩B) = 24/ 60
P (Students opted for NCC or NSS)
P (AᴜB) = P(A) + P(B) – P (A ∩ B) (∵ P(AᴜB) = P (A or B))
= 30/60 + 32/60 – 24/60
= (30 + 32 – 24)/ (60)
= (62 – 24)/ 60
= 38/ 60
= 19/ 30