Let A and E be any two events with positive p | vedclass

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

Question

(A) Let $A$ and $E$ be any two events with positive probabilities.Consider Statement $- 1$:$P(E/A) = \frac{P(E \cap A)}{P(A)}$. Since $P(A) \leq 1$,we

Vedclass · Accepted Answer

Both the statements are true

Vedclass · Answer

(A) Let $A$ and $E$ be any two events with positive probabilities.Consider Statement $- 1$:$P(E/A) = \frac{P(E \cap A)}{P(A)}$. Since $P(A) \leq 1$,we have $\frac{1}{P(A)} \geq 1$. Therefore,$P(E/A) = \frac{P(E \cap A)}{P(A)} \geq P(E \cap A)$.We know that $P(A/E)P(E) = P(A \cap E)$.Since $P(E \cap A) = P(A \cap E)$,it follows that $P(E/A) \geq P(A \cap E) = P(A/E)P(E)$.Thus,Statement $- 1$ is true.Consider Statement $- 2$:$P(A/E) = \frac{P(A \cap E)}{P(E)}$. Since $P(E) \leq 1$,we have $\frac{1}{P(E)} \geq 1$. Therefore,$P(A/E) = \frac{P(A \cap E)}{P(E)} \geq P(A \cap E)$.Thus,Statement $- 2$ is true.

Let $A$ and $E$ be any two events with positive probabilities:
Statement $- 1$: $P(E/A) \geq P(A/E)P(E)$
Statement $- 2$: $P(A/E) \geq P(A \cap E)$

Similar Questions

Two numbers are selected at random from the set $\{1, 2, 3, \ldots, 13\}$. If the sum of the selected numbers is even, what is the probability that both the numbers are odd?

Let $A$ and $B$ be two independent events such that $P(A) + P(B) = \frac{3}{4}$ and $P(\overline{A} | B) = \frac{2}{5}$. Then,$P(A \cap B)$ is -

$A$ die is thrown. If $E$ is the event 'the number appearing is a multiple of $3$' and $F$ is the event 'the number appearing is even',then find whether $E$ and $F$ are independent?

If two events $A$ and $B$ are such that $P(A^c) = 0.3$,$P(B) = 0.4$ and $P(A \cap B^c) = 0.5$,then $P(B | A \cup B^c)$ is equal to

$A$ family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $A$ and $E$ be any two events with positive probabilities:Statement $- 1$: $P(E/A) \geq P(A/E)P(E)$Statement $- 2$: $P(A/E) \geq P(A \cap E)$