If A and B are two independent events such th | vedclass

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

Question

(D) Given $A$ and $B$ are independent events,$P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{5}$.$1$. For option $A$: Since $A$ and $B$ are independent,$P(A

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

(D) Given $A$ and $B$ are independent events,$P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{5}$.$1$. For option $A$: Since $A$ and $B$ are independent,$P(A/B) = P(A) = \frac{1}{2}$. Thus,option $A$ is correct.$2$. For option $B$: $P(A/(A \cup B)) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A) + P(B) - P(A \cap B)}$.Since $P(A \cap B) = P(A)P(B) = \frac{1}{2} 	imes \frac{1}{5} = \frac{1}{10}$,$P(A \cup B) = \frac{1}{2} + \frac{1}{5} - \frac{1}{10} = \frac{5+2-1}{10} = \frac{6}{10} = \frac{3}{5}$.So,$P(A/(A \cup B)) = \frac{1/2}{3/5} = \frac{1}{2} 	imes \frac{5}{3} = \frac{5}{6}$. Thus,option $B$ is correct.$3$. For option $C$: $P((A \cap B)/(A' \cup B')) = \frac{P((A \cap B) \cap (A' \cup B'))}{P(A' \cup B')}$.By De Morgan's Law,$A' \cup B' = (A \cap B)'$.So,$(A \cap B) \cap (A \cap B)' = \emptyset$.Therefore,the numerator is $P(\emptyset) = 0$. Thus,option $C$ is correct.Since all options are correct,the answer is $D$.

If $A$ and $B$ are two independent events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{5}$,then which of the following is true?

Similar Questions

Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are the same. If the probability of a random toss resulting in a head is $\frac{1}{3}$,then the probability that the experiment stops with heads is:

If three numbers are randomly selected from the set $\{1, 2, 3, \ldots, 50\}$, then the probability that they are in arithmetic progression is

Two squares are chosen one by one on a chessboard. The probability that they have a side in common is

The probability that a randomly chosen $2 \times 2$ matrix with all the entries from the set of first $10$ primes is singular,is equal to

There are $3$ houses available in a residential area. $3$ people apply for the houses. If each person applies for a house without consulting others,what is the probability that all $3$ apply for the same house (in $/9$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module