Let us now find the probability of event `"‘A or B’"`, i.e., `P (A ∪ B)`
Let `A = {HHT, HTH, THH}` and `B = {HTH, THH, HHH} ` be two events associated with ‘tossing of a coin thrice’
Clearly `A ∪ B = {HHT, HTH, THH, HHH}`
Now `P (A ∪ B) = P(HHT) + P(HTH) + P(THH) + P(HHH)`
If all the outcomes are equally likely, then
`P (A ∪ B) = 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2`
Also` P(A) = P(HHT) + P(HTH) + P(THH) = 3/8`
and `P(B) = P(HTH) + P(THH) + P(HHH) = 3/8`
Therefore` P(A) + P(B) = 3/8 + 3/8 = 6/8`
It is clear that` P(A∪ B) ≠ P(A) + P(B)`
The points `HTH` and `THH ` are common to both `A` and `B .`
In the computation of `P(A) + P(B)` the probabilities of points `HTH` and `THH`, i.e., the elements of `A ∩B` are included twice.
Thus to get the probability `P(A ∪ B) `we have to subtract the probabilities of the sample points in `A ∩ B` from `P(A) + P(B)`
i.e. `P(A∪B) = P(A) + P(B) − ΣP(ω_i ),∀ ω_i ∈ A∩ B`
`= P(A) + P(B) − P(A ∩B)`
Thus we observe that, `P(A∪ B) = P(A) + P(B) − P(A ∩B)`
In general, if A and B are any two events associated with a random experiment, then by the definition of probability of an event, we have
` P(A ∪ B ) = Σ p(ω_i ),∀ ω_i ∈ A ∪ B`.
Since `A∪B = (A–B) ∪ (A ∩B)∪(B–A)` ,
we have
`P(A ∪ B) = [ Σ P(ω_i ) ∀ ω_i ∈ (A–B)] + [ Σ P(ω_i ) ∀ ω_i ∈ A ∩ B ] + [ ΣP ( ω_i ) ∀ ω_i ∈ B – A]`
(because `A–B, A ∩ B` and `B – A `are mutually exclusive) ... (1)
Also `P(A) + P(B) = [Σ p(ω_i ) ∀ω_i ∈A]+[ Σ p(ω_i ) ∀ω_i ∈B]`
`= [ Σ P (ω_i )∀ω_i ∈(A–B)∪(A∩B)] + [ Σ P (ω_i )∀ω_ i ∈(B – A)∪(A∩B)]`
`= [ Σ P (ω_i )∀ω_i ∈(A – B ) ] + [ Σ P ( ω_i ) ∀ ω_i ∈(A∩B)] + [ Σ P (ω_i ) ∀ω_i ∈ (B–A)] +
[ Σ P(ω_i )∀ω_ i ∈(A∩ B)]`
`= P(A∪B) + [ Σ P (ω_i )∀ ω_i ∈A∩B]` [using (1)]
`= P(A∪ B)+P(A ∩B) `.
Hence `P(A∪B)=P (A)+P(B) – P(A∩B)` .
Alternatively, it can also be proved as follows:
`A ∪ B = A ∪ (B – A)`, where A and `B – A` are mutually exclusive,
and `B = (A ∩ B) ∪ (B – A)`, where `A ∩ B` and B – A are mutually exclsuive.
Using Axiom (iii) of probability, we get
`P (A ∪B) = P (A) + P (B – A)` ... (2)
and `P(B) = P ( A ∩ B) + P (B – A)` ... (3)
Subtracting (3) from (2) gives
`P (A ∪ B) – P(B) = P(A) – P (A ∩ B)`
or `P(A ∪ B) = P(A) + P (B) – P (A ∩ B)`
The above result can further be verified by observing the Venn Diagram (Fig 16.1)
`color(red)(=> P(A ∪B) = P(A) + P(B)) `,
`color(blue)("If A and B are disjoint sets")`, i.e., they are mutually exclusive events, then `color(red)(A ∩ B = φ)`
Therefore `color(blue)(P(A∩B) = P (φ) = 0)`
Thus, for mutually exclusive events `A` and `B`, we have
`P(A ∪B) = P(A) + P(B) `,
which is Axiom (iii) of probability.
Let us now find the probability of event `"‘A or B’"`, i.e., `P (A ∪ B)`
Let `A = {HHT, HTH, THH}` and `B = {HTH, THH, HHH} ` be two events associated with ‘tossing of a coin thrice’
Clearly `A ∪ B = {HHT, HTH, THH, HHH}`
Now `P (A ∪ B) = P(HHT) + P(HTH) + P(THH) + P(HHH)`
If all the outcomes are equally likely, then
`P (A ∪ B) = 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2`
Also` P(A) = P(HHT) + P(HTH) + P(THH) = 3/8`
and `P(B) = P(HTH) + P(THH) + P(HHH) = 3/8`
Therefore` P(A) + P(B) = 3/8 + 3/8 = 6/8`
It is clear that` P(A∪ B) ≠ P(A) + P(B)`
The points `HTH` and `THH ` are common to both `A` and `B .`
In the computation of `P(A) + P(B)` the probabilities of points `HTH` and `THH`, i.e., the elements of `A ∩B` are included twice.
Thus to get the probability `P(A ∪ B) `we have to subtract the probabilities of the sample points in `A ∩ B` from `P(A) + P(B)`
i.e. `P(A∪B) = P(A) + P(B) − ΣP(ω_i ),∀ ω_i ∈ A∩ B`
`= P(A) + P(B) − P(A ∩B)`
Thus we observe that, `P(A∪ B) = P(A) + P(B) − P(A ∩B)`
In general, if A and B are any two events associated with a random experiment, then by the definition of probability of an event, we have
` P(A ∪ B ) = Σ p(ω_i ),∀ ω_i ∈ A ∪ B`.
Since `A∪B = (A–B) ∪ (A ∩B)∪(B–A)` ,
we have
`P(A ∪ B) = [ Σ P(ω_i ) ∀ ω_i ∈ (A–B)] + [ Σ P(ω_i ) ∀ ω_i ∈ A ∩ B ] + [ ΣP ( ω_i ) ∀ ω_i ∈ B – A]`
(because `A–B, A ∩ B` and `B – A `are mutually exclusive) ... (1)
Also `P(A) + P(B) = [Σ p(ω_i ) ∀ω_i ∈A]+[ Σ p(ω_i ) ∀ω_i ∈B]`
`= [ Σ P (ω_i )∀ω_i ∈(A–B)∪(A∩B)] + [ Σ P (ω_i )∀ω_ i ∈(B – A)∪(A∩B)]`
`= [ Σ P (ω_i )∀ω_i ∈(A – B ) ] + [ Σ P ( ω_i ) ∀ ω_i ∈(A∩B)] + [ Σ P (ω_i ) ∀ω_i ∈ (B–A)] +
[ Σ P(ω_i )∀ω_ i ∈(A∩ B)]`
`= P(A∪B) + [ Σ P (ω_i )∀ ω_i ∈A∩B]` [using (1)]
`= P(A∪ B)+P(A ∩B) `.
Hence `P(A∪B)=P (A)+P(B) – P(A∩B)` .
Alternatively, it can also be proved as follows:
`A ∪ B = A ∪ (B – A)`, where A and `B – A` are mutually exclusive,
and `B = (A ∩ B) ∪ (B – A)`, where `A ∩ B` and B – A are mutually exclsuive.
Using Axiom (iii) of probability, we get
`P (A ∪B) = P (A) + P (B – A)` ... (2)
and `P(B) = P ( A ∩ B) + P (B – A)` ... (3)
Subtracting (3) from (2) gives
`P (A ∪ B) – P(B) = P(A) – P (A ∩ B)`
or `P(A ∪ B) = P(A) + P (B) – P (A ∩ B)`
The above result can further be verified by observing the Venn Diagram (Fig 16.1)
`color(red)(=> P(A ∪B) = P(A) + P(B)) `,
`color(blue)("If A and B are disjoint sets")`, i.e., they are mutually exclusive events, then `color(red)(A ∩ B = φ)`
Therefore `color(blue)(P(A∩B) = P (φ) = 0)`
Thus, for mutually exclusive events `A` and `B`, we have
`P(A ∪B) = P(A) + P(B) `,
which is Axiom (iii) of probability.