Let A, B and C be three events,which are pair | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We need to find $P[(\bar{A} \cap \bar{B})|C]$.By the definition of conditional probability,$P[(\bar{A} \cap \bar{B})|C] = \frac{P(\bar{A} \cap \ba

Vedclass · Accepted Answer

$P(\bar{A}) - P(B)$

Vedclass · Answer

(C) We need to find $P[(\bar{A} \cap \bar{B})|C]$.By the definition of conditional probability,$P[(\bar{A} \cap \bar{B})|C] = \frac{P(\bar{A} \cap \bar{B} \cap C)}{P(C)}$.Using De Morgan's Law,$\bar{A} \cap \bar{B} = \overline{A \cup B}$.Thus,$\bar{A} \cap \bar{B} \cap C = C \setminus ((A \cap C) \cup (B \cap C))$.Therefore,$P(\bar{A} \cap \bar{B} \cap C) = P(C) - P((A \cap C) \cup (B \cap C))$.Using the inclusion-exclusion principle,$P((A \cap C) \cup (B \cap C)) = P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)$.Since $A, B, C$ are pairwise independent,$P(A \cap C) = P(A)P(C)$ and $P(B \cap C) = P(B)P(C)$.Given $P(A \cap B \cap C) = 0$,we have $P((A \cap C) \cup (B \cap C)) = P(A)P(C) + P(B)P(C) - 0 = P(C)(P(A) + P(B))$.Substituting this back,$P(\bar{A} \cap \bar{B} \cap C) = P(C) - P(C)(P(A) + P(B)) = P(C)(1 - P(A) - P(B))$.Finally,$P[(\bar{A} \cap \bar{B})|C] = \frac{P(C)(1 - P(A) - P(B))}{P(C)} = 1 - P(A) - P(B)$.Since $1 - P(A) = P(\bar{A})$,the expression becomes $P(\bar{A}) - P(B)$.

Let $A, B$ and $C$ be three events,which are pairwise independent and $\bar{E}$ denotes the complement of an event $E$. If $P(A \cap B \cap C) = 0$ and $P(C) > 0$,then $P[(\bar{A} \cap \bar{B})|C]$ is equal to

Similar Questions

If $E_1$ and $E_2$ are two events of the sample space such that $P(E_1) = \frac{1}{4}$, $P(E_1 | E_2) = \frac{1}{2}$ and $P(E_2 | E_1) = \frac{1}{3}$, then $P(E_1 | \bar{E}_2) = $

In a city,$40\%$ of the people have brown hair,$25\%$ have brown eyes,and $15\%$ have both brown hair and brown eyes. If a person is selected at random from those having brown hair,what is the probability that they also have brown eyes?

If $P(A) = \frac{6}{11}$,$P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$,then $P(A \mid B) = $ . . . . . . .

If $E$ and $F$ are events such that $P(\overline{F}) = 0.7$ and $P(E \cap F) = 0.2$,then $P(E \mid F)$ is

$A$ parent has two children. If at least one of them is a boy,then the probability that the other is also a boy is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module