If E_1 and E_2 are two events of a random exp | vedclass

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

Question

(C) Given $P(E_1) = \frac{1}{8}$,$P(E_1 \mid E_2) = \frac{1}{3}$,and $P(E_2 \mid E_1) = \frac{1}{4}$.Using the definition of conditional probability,$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $P(E_1) = \frac{1}{8}$,$P(E_1 \mid E_2) = \frac{1}{3}$,and $P(E_2 \mid E_1) = \frac{1}{4}$.Using the definition of conditional probability,$P(E_1 \cap E_2) = P(E_2 \mid E_1) 	imes P(E_1) = \frac{1}{4} 	imes \frac{1}{8} = \frac{1}{32}$.Then,$P(E_2) = \frac{P(E_1 \cap E_2)}{P(E_1 \mid E_2)} = \frac{1/32}{1/3} = \frac{3}{32}$. (Matches $IV$)Now,$P(E_1 \cup E_2) = P(E_1)   P(E_2) - P(E_1 \cap E_2) = \frac{1}{8}   \frac{3}{32} - \frac{1}{32} = \frac{4 3-1}{32} = \frac{6}{32} = \frac{3}{16}$. (Matches $III$)We know $P(\bar{E}_2) = 1 - P(E_2) = 1 - \frac{3}{32} = \frac{29}{32}$.Also,$P(E_1 \cap \bar{E}_2) = P(E_1) - P(E_1 \cap E_2) = \frac{1}{8} - \frac{1}{32} = \frac{3}{32}$.Thus,$P(E_1 \mid \bar{E}_2) = \frac{P(E_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{3/32}{29/32} = \frac{3}{29}$. (Matches $I$)Finally,$P(\bar{E}_1 \mid \bar{E}_2) = 1 - P(E_1 \mid \bar{E}_2) = 1 - \frac{3}{29} = \frac{26}{29}$. (Matches $II$)The correct matching is $A-III, B-IV, C-I, D-II$.

List-$I$	List-$II$
$A. P(E_1 \cup E_2)$	$I. \frac{3}{29}$
$B. P(E_2)$	$II. \frac{26}{29}$
$C. P(E_1 \mid \bar{E}_2)$	$III. \frac{3}{16}$
$D. P(\bar{E}_1 \mid \bar{E}_2)$	$IV. \frac{3}{32}$

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.1$	$0.32$	$0.21$	$0.15$	$0.05$	$0.17$

Similar Questions

In a class consisting of $40$ boys and $30$ girls,$30 \%$ of the boys and $40 \%$ of the girls are good at Mathematics. If a student selected at random from that class is found to be a girl,then the probability that she is not good at Mathematics is

For a biased die,the probabilities for different faces to turn up are given below:

$Face$ $1$ $2$ $3$ $4$ $5$ $6$

$Probability$ $0.1$ $0.32$ $0.21$ $0.15$ $0.05$ $0.17$

The die is tossed and you are told that either face $1$ or $2$ has turned up. Then the probability that it is face $1$ is:

If a die is rolled twice and the sum of the numbers appearing on them is observed to be $6$, then the probability that the number $1$ appears at least once on them is

One ticket is selected at random from $50$ tickets numbered $00, 01, 02, \ldots, 49$. The probability that the sum of the digits on the selected ticket is $8$,given that the product of these digits is zero,equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module