Question

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.

Answer 1

Hint: Use the concept of theoretical probability which can be calculated by dividing the number of favourable outcomes by the total possible outcomes.

Complete step-by-step answer:
Let us consider the following that
${{\text{E}}_1} = $Probability of selecting the first bag. 
${{\text{E}}_2} = $ Probability of selecting a second bag.
${\text{A}} = $ Probability that the ball drawn is white.

Then calculating the probability, we get
\[
  {\text{P}}\left( {{{\text{E}}_1}} \right) = {\text{P}}\left( {{{\text{E}}_2}} \right) = \dfrac{1}{2}, \\
  {\text{Similarly}} \\
  {\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_1}}}} \right) = \dfrac{3}{5} \\
  {\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_2}}}} \right) = \dfrac{2}{6} \\
  {\text{Probibility that the ball drawn is white can be caluclated as }} \\
  {\text{P}}\left( {\text{A}} \right) = {\text{P}}\left( {{{\text{E}}_1}} \right){\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_1}}}} \right) + {\text{P}}\left( {{{\text{E}}_2}} \right){\text{P}}\left( {\dfrac{{\text{A}}}{{{{\text{E}}_2}}}} \right) \\
   = \dfrac{1}{2} \times \dfrac{3}{5} + \dfrac{1}{2} \times \dfrac{2}{6} \\
   = \dfrac{7}{{15}} \\
    \\
\]

Note: In such type of questions always keep in mind the formulae to calculate the probability of any event i.e.${\text{Probability = }}\dfrac{{{\text{Number of favorable outcomes }}}}{{{\text{Total number of possible outcomes }}}}$. Remember Probability of an event can neither be greater than $1$ nor it can be less than $0.$