A and B are two independent events. P(A)=frac | vedclass

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

Question

(D) Given that $A$ and $B$ are independent events, $P(A)=\frac{2}{5}$ and $P(B)=\frac{1}{3}$.Thus, $P(\overline{A}) = 1 - \frac{2}{5} = \frac{3}{5}$ a

Vedclass · Accepted Answer

$A-II, B-IV, C-III, D-I$

Vedclass · Answer

(D) Given that $A$ and $B$ are independent events, $P(A)=\frac{2}{5}$ and $P(B)=\frac{1}{3}$.Thus, $P(\overline{A}) = 1 - \frac{2}{5} = \frac{3}{5}$ and $P(\overline{B}) = 1 - \frac{1}{3} = \frac{2}{3}$.Since they are independent, $P(A \cap B) = P(A) 	imes P(B) = \frac{2}{5} 	imes \frac{1}{3} = \frac{2}{15}$.$(A) P(\overline{A} \cup B) = P(\overline{A}) + P(B) - P(\overline{A} \cap B)$.Since $A$ and $B$ are independent, $\overline{A}$ and $B$ are also independent.$P(\overline{A} \cup B) = \frac{3}{5} + \frac{1}{3} - (\frac{3}{5} 	imes \frac{1}{3}) = \frac{9+5-3}{15} = \frac{11}{15}$. This matches $(II)$.$(B) P(\frac{A}{\overline{B}}) = P(A) = \frac{2}{5}$ (since $A$ and $\overline{B}$ are independent).Wait, checking the provided options and the image, let's re-evaluate.$P(\frac{A}{\overline{B}}) = \frac{P(A \cap \overline{B})}{P(\overline{B})} = \frac{P(A)P(\overline{B})}{P(\overline{B})} = P(A) = \frac{2}{5}$.Looking at the options, there seems to be a mismatch in the provided list. Let's calculate $(C) P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{2}{5} + \frac{1}{3} - \frac{2}{15} = \frac{6+5-2}{15} = \frac{9}{15} = \frac{3}{5}$. This matches $(III)$.Based on standard matching, $(A)-(II)$ and $(C)-(III)$ are correct. Option $(D)$ matches this pattern.

List-$I$	List-$II$
$(A) P(\overline{A} \cup B)$	$(I) \frac{2}{3}$
$(B) P(\frac{A}{\overline{B}})$	$(II) \frac{11}{15}$
$(C) P(A \cup B)$	$(III) \frac{3}{5}$

$A$ and $B$ are two independent events. $P(A)=\frac{2}{5}, P(B)=\frac{1}{3}$. Match the following List-$I$ with List-$II$.
List-$I$ List-$II$
$(A) P(\overline{A} \cup B)$ $(I) \frac{2}{3}$
$(B) P(\frac{A}{\overline{B}})$ $(II) \frac{11}{15}$
$(C) P(A \cup B)$ $(III) \frac{3}{5}$

Similar Questions

Two players $A$ and $B$ are alternately throwing a coin and a die together. $A$ player who first throws a head and a $6$ wins the game. If $A$ starts the game, then the probability that $B$ wins the game is:

$A$ fair die is rolled indefinitely. Player $A$ wins if two consecutive rolls show $3$ or $5$, and player $B$ wins if two consecutive rolls show $1$ or $2$ or $4$ or $6$. The probability that player $A$ wins is:

In a throw of a dice,the probability of getting a $1$ in an even number of throws is:

There are four machines and it is known that exactly two of them are faulty. They are tested one by one, in a random order until both the faulty machines are identified. The probability that only two tests are needed is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ and $B$ are two independent events. $P(A)=\frac{2}{5}, P(B)=\frac{1}{3}$. Match the following List-$I$ with List-$II$.List-$I$List-$II$$(A) P(\overline{A} \cup B)$$(I) \frac{2}{3}$$(B) P(\frac{A}{\overline{B}})$$(II) \frac{11}{15}$$(C) P(A \cup B)$$(III) \frac{3}{5}$