E_1 and E_2 are two independent events of a r | vedclass

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

Question

(A) Given: $P(E_1) = \frac{1}{2}$ and $P(E_1 \cup E_2) = \frac{2}{3}$.Since $E_1$ and $E_2$ are independent events, $P(E_1 \cap E_2) = P(E_1) 	imes P

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given: $P(E_1) = \frac{1}{2}$ and $P(E_1 \cup E_2) = \frac{2}{3}$.Since $E_1$ and $E_2$ are independent events, $P(E_1 \cap E_2) = P(E_1) 	imes P(E_2)$.Let $P(E_2) = x$. Then $P(E_1 \cap E_2) = \frac{1}{2}x$.Using the formula $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$:$\frac{2}{3} = \frac{1}{2} + x - \frac{x}{2}$$\frac{2}{3} = \frac{1}{2} + \frac{x}{2}$$\frac{x}{2} = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6}$$x = \frac{1}{3}$. Thus, $P(E_2) = \frac{1}{3}$. $(A ightarrow III)$Now, $P(E_1 \cap E_2) = \frac{1}{2} 	imes \frac{1}{3} = \frac{1}{6}$.$P(E_1 | E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{1/6}{1/3} = \frac{1}{2}$. $(B ightarrow IV)$$P(\bar{E}_2 | E_1) = 1 - P(E_2 | E_1) = 1 - P(E_2) = 1 - \frac{1}{3} = \frac{2}{3}$. $(C ightarrow I)$$P(\bar{E}_1 \cup \bar{E}_2) = P(\overline{E_1 \cap E_2}) = 1 - P(E_1 \cap E_2) = 1 - \frac{1}{6} = \frac{5}{6}$. $(D ightarrow II)$Therefore, the correct match is $A-III, B-IV, C-I, D-II$.

List-$I$	List-$II$
$A. P(E_2) =$	$I. 2/3$
$B. P(E_1 \| E_2) =$	$II. 5/6$
$C. P(\bar{E}_2 \| E_1) =$	$III. 1/3$
$D. P(\bar{E}_1 \cup \bar{E}_2) =$	$IV. 1/2$

Similar Questions

If $l, m$ represent any two elements (identical or different) of the set $\{1, 2, 3, 4, 5, 6, 7\}$,then the probability that $lx^2 + mx + 1 > 0$ for all $x \in R$ is

If $A$ and $B$ are two events,then the probability of the event that at most one of $A$ and $B$ occurs is:

$A$ bag contains $2n$ coins,out of which $n-1$ are unfair with heads on both sides and the remaining are fair. One coin is picked from the bag at random and tossed. If the probability that a head appears in the toss is $\frac{41}{56}$,then the number of unfair coins in the bag is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module