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A bag contains 3 white and 2 black balls and another bag contains 2 white and 4 black balls. One bag is chosen at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is white.

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Hint- Probability is the ratio of favourable number of outcomes and total number of outcomes.

First bag contains 3 white and 2 black balls.
Second bag contains 2 white and 4 black balls.
Consider the following events which is given
${A_1} = $Selecting first bag
${A_2} = $Selecting second bag
$x = $Ball drawn is white
Then, Probability of selecting the bag\[{\text{ = }}\dfrac{{{\text{Favorable bags}}}}{{{\text{total bags}}}}\]
\[ \Rightarrow P\left( {{A_1}} \right) = P\left( {{A_2}} \right) = \dfrac{1}{2}\]
$ \Rightarrow $Probability of getting white ball from first bag \[{\text{P}}\left( {\dfrac{x}{{{A_1}}}} \right){\text{ = }}\dfrac{{{\text{Favorable balls}}}}{{{\text{total balls}}}} = \dfrac{3}{5}\]
$ \Rightarrow $ Probability of getting white ball from second bag \[{\text{P}}\left( {\dfrac{x}{{{A_2}}}} \right){\text{ = }}\dfrac{{{\text{Favorable balls}}}}{{{\text{total balls}}}} = \dfrac{2}{6} = \dfrac{1}{3}\]
$ \Rightarrow $Probability that the ball drawn is white is \[{\text{P}}\left( x \right){\text{ = }}P\left( {{A_1}} \right){\text{ P}}\left( {\dfrac{x}{{{A_1}}}} \right) + P\left( {{A_2}} \right){\text{P}}\left( {\dfrac{x}{{{A_2}}}} \right)...............\left( 1 \right)\]
\[
  {\text{P}}\left( x \right) = \dfrac{1}{2} \times \dfrac{3}{5} + \dfrac{1}{2} \times \dfrac{1}{3} \\
  {\text{P}}\left( x \right) = \dfrac{3}{{10}} + \dfrac{1}{6} = \dfrac{7}{{15}} \\
\]
So, this is the required probability.

Note-In such types of questions first calculate the probability of selecting a bag and probability of getting white ball drawn from each bag using the formula which is stated above, then apply the formula which is written in equation (1) to get the required probability.
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