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

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(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

$\frac{32}{49}$

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(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 $P(E|E_2) = \frac{6}{7}$.$P(E) = \frac{5}{7} 	imes \frac{4}{7} + \frac{2}{7} 	imes \frac{6}{7} = \frac{20}{49} + \frac{12}{49} = \frac{32}{49}$.

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