Bag I contains 3 red and 4 black balls and Ba | vedclass

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

Question

(A) Let $E_{1}$ be the event that a red ball is transferred from Bag $I$ to Bag $II$,and $E_{2}$ be the event that a black ball is transferred from Ba

Vedclass · Accepted Answer

$\frac{16}{31}$

Vedclass · Answer

(A) Let $E_{1}$ be the event that a red ball is transferred from Bag $I$ to Bag $II$,and $E_{2}$ be the event that a black ball is transferred from Bag $I$ to Bag $II$.The probabilities are $P(E_{1}) = \frac{3}{7}$ and $P(E_{2}) = \frac{4}{7}$.Let $A$ be the event that the ball drawn from Bag $II$ is red.If $E_{1}$ occurs,Bag $II$ now contains $5$ red and $5$ black balls. Thus,$P(A|E_{1}) = \frac{5}{10} = \frac{1}{2}$.If $E_{2}$ occurs,Bag $II$ now contains $4$ red and $6$ black balls. Thus,$P(A|E_{2}) = \frac{4}{10} = \frac{2}{5}$.We need to find $P(E_{2}|A)$. By Bayes' Theorem:$P(E_{2}|A) = \frac{P(E_{2})P(A|E_{2})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2})}$$P(E_{2}|A) = \frac{(\frac{4}{7}) 	imes (\frac{2}{5})}{(\frac{3}{7}) 	imes (\frac{1}{2}) + (\frac{4}{7}) 	imes (\frac{2}{5})}$$P(E_{2}|A) = \frac{\frac{8}{35}}{\frac{3}{14} + \frac{8}{35}} = \frac{\frac{8}{35}}{\frac{15+16}{70}} = \frac{8}{35} 	imes \frac{70}{31} = \frac{16}{31}$.

Bag $I$ contains $3$ red and $4$ black balls and Bag $II$ contains $4$ red and $5$ black balls. One ball is transferred from Bag $I$ to Bag $II$ and then a ball is drawn from Bag $II$. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Similar Questions

Let $H_1, H_2, \ldots, H_{n}$ be mutually exclusive and exhaustive events with $P(H_i) > 0, i = 1, 2, \ldots, n$. Let $E$ be any other event with $0 < P(E) < 1$.
$STATEMENT-1$: $P(H_i \mid E) > P(E \mid H_i) \cdot P(H_i)$ for $i = 1, 2, \ldots, n$.
$STATEMENT-2$: $\sum_{i=1}^{n} P(H_i) = 1$.

The probability that $A$ speaks the truth is $\frac{4}{5}$. $A$ coin is tossed. $A$ reports that a head appears. The probability that there was actually a head is

$A$ bag $P$ contains $3$ blue and $5$ red balls. Another bag $Q$ contains $4$ blue and $6$ red balls. $A$ ball is drawn at random from one of the bags and is found to be red. The probability that it is from bag $Q$ is

$A$ and $B$ are two events of a random experiment such that $P(B)=0.4$, $P(A \cap \bar{B})=0.5$, and $P(A \cup B) + P\left(\frac{B}{A \cup \bar{B}}\right) = 1.15$. Then $P(A) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module