An urn contains 5 red and 2 green balls. A ba | vedclass

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

Question

(B) Let $E_1$ be the event that the first ball drawn is red. Then a green ball is added to the urn. The probability $P(E_1) = \frac{5}{7}$. After draw

Vedclass · Accepted Answer

$\frac{32}{49}$

Vedclass · Answer

(B) Let $E_1$ be the event that the first ball drawn is red. Then a green ball is added to the urn. The probability $P(E_1) = \frac{5}{7}$. After drawing one red ball,the urn contains $4$ red and $3$ green balls. Adding a green ball results in $4$ red and $4$ green balls. The probability of drawing a red ball in the second draw given $E_1$ is $P(E|E_1) = \frac{4}{8} = \frac{1}{2}$.Let $E_2$ be the event that the first ball drawn is green. Then a red ball is added to the urn. The probability $P(E_2) = \frac{2}{7}$. After drawing one green ball,the urn contains $5$ red and $1$ green ball. Adding a red ball results in $6$ red and $1$ green ball. The probability of drawing a red ball in the second draw given $E_2$ is $P(E|E_2) = \frac{6}{7}$.Using the law of total probability,the probability that the second ball is red is:$P(E) = P(E_1) 	imes P(E|E_1) + P(E_2) 	imes P(E|E_2)$$P(E) = \left(\frac{5}{7} 	imes \frac{4}{8}ight) + \left(\frac{2}{7} 	imes \frac{6}{7}ight)$$P(E) = \frac{20}{56} + \frac{12}{49} = \frac{5}{14} + \frac{12}{49} = \frac{35 + 24}{98} = \frac{59}{98}$.Wait,re-evaluating the urn contents: If $E_1$ occurs,we have $4$ red and $2$ green left,then add $1$ green,total $4$ red and $3$ green. Total balls $= 7$. Probability $P(E|E_1) = \frac{4}{7}$.If $E_2$ occurs,we have $5$ red and $1$ green left,then add $1$ red,total $6$ red and $1$ green. Total balls $= 7$. Probability $P(E|E_2) = \frac{6}{7}$.$P(E) = \frac{5}{7} 	imes \frac{4}{7} + \frac{2}{7} 	imes \frac{6}{7} = \frac{20}{49} + \frac{12}{49} = \frac{32}{49}$.

Similar Questions

Let $A$ and $B$ be two events such that $P(A) = \frac{3}{8}$,$P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$. Then $P(A'|B) - P(A|B) =$ . . . . . .

If $A$ and $B$ are two independent events such that $P(A^{\prime}) = 0.75$,$P(A \cup B) = 0.65$ and $P(B) = p$,then the value of $p$ is

Two dice are thrown and the sum of the numbers appearing on the dice is observed to be a multiple of $4$. If $p$ is the conditional probability that number $4$ has appeared at least once,then $3p + 2 =$

Two cards are drawn randomly from a pack of $52$ playing cards one after the other with replacement. If $A$ is the event of drawing a face card in the first draw and $B$ is the event of drawing a club card in the second draw,then $P(\overline{B}|A) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module