A bag contains 4 red and 3 black balls. A sec | vedclass

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(D) red ball can be drawn in two mutually exclusive ways.$(i)$ Selecting bag $I$ and then drawing a red ball from it.(ii) Selecting bag $II$ and then

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$\frac{17}{35}$

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(D) red ball can be drawn in two mutually exclusive ways.$(i)$ Selecting bag $I$ and then drawing a red ball from it.(ii) Selecting bag $II$ and then drawing a red ball from it.Let $E_1$,$E_2$ and $A$ denote the events defined as follows:$E_1 = 	ext{Selecting bag } I$$E_2 = 	ext{Selecting bag } II$Since one of the two bags is selected randomly,we have:$P(E_1) = \frac{1}{2}$ and $P(E_2) = \frac{1}{2}$Now,$P(A|E_1) = 	ext{Probability of drawing a red ball when the first bag is selected} = \frac{4}{7}$$P(A|E_2) = 	ext{Probability of drawing a red ball when the second bag is selected} = \frac{2}{5}$Using the law of total probability:$P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)$$P(A) = \frac{1}{2} 	imes \frac{4}{7} + \frac{1}{2} 	imes \frac{2}{5}$$P(A) = \frac{2}{7} + \frac{1}{5} = \frac{10 + 7}{35} = \frac{17}{35}$

$A$ bag contains $4$ red and $3$ black balls. $A$ second bag contains $2$ red and $3$ black balls. One bag is selected at random. If from the selected bag,one ball is drawn at random,then the probability that the ball drawn is red,is

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