## 3 important Theorem of Probability

## Conditional Probability:

**Theorem:Prove that if E and F are independent events, then so are the events E and F**.

**Proof**: Since E and F are independent

So, P(E ∩ F) = P(E) . P(F) ....**eq-(1)**

From the venn diagram

E ∩ F and E ∩ F′ are **mutually exclusive events**.

E =(E ∩ F) ∪ (E ∩ F′).

Therefore P(E) = P(E ∩ F) + P(E ∩ F′)

P(E ∩ F′) = P(E) − P(E ∩ F)

= P(E) − P(E) . P(F) **(by eq-(1))**

= P(E) (1−P(F))

= P(E). P(F′)

**Hence, E and F′ are independent**

In a similar manner, it can be shown that if the events E and F are

independent, then

(a) E′ and F are independent,

(b) E′ and F′ are independent

**Theorem:Prove that if E and F are independent events, then so are the events E’ and F**.

**Proof**: Since E and F are independent

So, P(E ∩ F) = P(E) . P(F) ....**eq-(1)**

From the venn diagram

E ∩ F and E' ∩ F are **mutually exclusive events**.

F =(E ∩ F) ∪ (E' ∩ F).

Therefore, P(F) = P(E ∩ F) + P(E' ∩ F)

P(E' ∩ F) = P(F) − P(E ∩ F)

= P(F) − P(E) . P(F) **(by eq-(1))**

= P(F) (1−P(E))

P(E' ∩ F) = P(F). P(E′)

P(E' ∩ F) = P(E'). P(F)

**Hence, E' and F are independent**

**Theorem:Prove that if E and F are independent events, then so are the events E’ and F’**.

**Proof: **We know that $E$ and $F$ are independent, so $P(E∩F)=P(E)⋅P(F)$ (eq-(1)).

Now, let's consider the events $E_{′}$ and $F_{′}$. Since $E_{′}$ is the complement of $E$ and $F_{′}$ is the complement of $F$, we have:

$P(E_{′}∩F_{′})=1−P(E∪F)$

$=1−[P(E)+P(F)−P(E∩F)]$

$P(E) –P(F) +P(E∩F)$

$P(E) –P(F) +P(E)P(F)$

$P(E') –P(F)( 1−P(E))$

$=P(E') –P(F)P(E')$

$=P(E') (1–P(F))$

**$(_{′}∩_{′})(E')P(F')proved$**

**Theorem: A and B are two independent events, then the probability of occurrence of at least one of A and B is given by 1– P(A′) P(B′)**

**Solution** We have

P(at least one of A and B) = P(A ∪ B)

= P(A) + P(B) − P(A ∩ B)

= P(A) + P(B) − P(A) P(B)

= P(A) + P(B) [1−P(A)]

= P(A) + P(B). P(A′)

= 1− P(A′) + P(B) P(A′)

= 1− P(A′) [1− P(B)]

** P(at least one of A and B) = 1− P(A′) P (B′)**

**Important Question:**The probability of simultaneous occurrence of at least one of two events A and B is p. If the probability that exactly one of A, B occurs is q, then **prove that P (A′) + P (B′) = 2 – 2p + q**.

**Solution** Since P (exactly one of A, B occurs) = q (given)

P (A∪B) – P ( A∩B) = q

⇒ p – P (A∩B) = q

⇒ P (A∩B) = p – q

⇒ 1 – P (A′∪ B′) = p – q

⇒ P (A′∪ B′) = 1 – p + q

⇒ P (A′) + P (B′) – P (A′∩ B′) = 1 – p + q

⇒ P (A′) + P (B′) = (1 – p + q) + P (A′ ∩ B′)

= (1 – p + q) + (1 – P (A ∪ B))

= (1 – p + q) + (1 – p)

= 2 – 2p + q. **Proved**

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12th class : Probability formula notes