# A bag contains 3 white and 2 black balls and another bag contains 2 white 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.

Last updated date: 24th Mar 2023

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Answer

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Hint- Look for the probability of happening of each event separately and see which all are simultaneous to perform addition or multiplication.

Here, there are two events

Let \[{E_1}\] be the event of selecting the first bag and ${E_2}$ be the event of selecting the second bag.

$ \Rightarrow {E_1} = $ Choosing first bag

And ${E_2} = $ Choosing second bag

And we have to find the probability that the ball drawn is white.

Let $W$ be the event that the ball drawn is white.

$W = $ The ball drawn is white.

Now probability of choosing a bag at random$ = P\left( {{E_1}} \right) = P\left( {{E_2}} \right) = \dfrac{1}{2}$

As total number of outcomes=2 and Favourable outcomes=1

Now, Probability of white ball from the first bag $P\left( {\dfrac{A}{{{E_1}}}} \right)$ $ = \dfrac{3}{5}$

As there are 3 white balls and 2 blacks. Therefore, Number of favourable outcomes=3 and total outcomes=3+2=5.

And the Probability of white ball from second bag $P\left( {\dfrac{A}{{{E_2}}}} \right) = \dfrac{2}{6} = \dfrac{1}{3}$

As there are 2 white balls and 4 blacks. Therefore, Number of favourable outcomes=2 and total outcomes=2+4=6.

Now, The probability of one bag is chosen at random and the ball drawn is white$

P\left( A \right) = P\left( {{E_1}} \right) \times P\left( {\dfrac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right) \times P\left( {\dfrac{A}{{{E_2}}}} \right) \\

= \dfrac{1}{2} \times \dfrac{3}{5} + \dfrac{1}{2} \times \dfrac{1}{3} \\

= \dfrac{3}{{10}} + \dfrac{1}{6} \\

= \dfrac{{28}}{{60}} \\

= \dfrac{7}{{15}} \\

$

Thus, The probability that the ball drawn is white $ = \dfrac{7}{{15}}$.

Note- Whenever we face such types of questions the key concept is that we should write the events which are occurring separately and then we should find the probability of each event that are occurring like we did and then we get our desired answer.

Here, there are two events

Let \[{E_1}\] be the event of selecting the first bag and ${E_2}$ be the event of selecting the second bag.

$ \Rightarrow {E_1} = $ Choosing first bag

And ${E_2} = $ Choosing second bag

And we have to find the probability that the ball drawn is white.

Let $W$ be the event that the ball drawn is white.

$W = $ The ball drawn is white.

Now probability of choosing a bag at random$ = P\left( {{E_1}} \right) = P\left( {{E_2}} \right) = \dfrac{1}{2}$

As total number of outcomes=2 and Favourable outcomes=1

Now, Probability of white ball from the first bag $P\left( {\dfrac{A}{{{E_1}}}} \right)$ $ = \dfrac{3}{5}$

As there are 3 white balls and 2 blacks. Therefore, Number of favourable outcomes=3 and total outcomes=3+2=5.

And the Probability of white ball from second bag $P\left( {\dfrac{A}{{{E_2}}}} \right) = \dfrac{2}{6} = \dfrac{1}{3}$

As there are 2 white balls and 4 blacks. Therefore, Number of favourable outcomes=2 and total outcomes=2+4=6.

Now, The probability of one bag is chosen at random and the ball drawn is white$

P\left( A \right) = P\left( {{E_1}} \right) \times P\left( {\dfrac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right) \times P\left( {\dfrac{A}{{{E_2}}}} \right) \\

= \dfrac{1}{2} \times \dfrac{3}{5} + \dfrac{1}{2} \times \dfrac{1}{3} \\

= \dfrac{3}{{10}} + \dfrac{1}{6} \\

= \dfrac{{28}}{{60}} \\

= \dfrac{7}{{15}} \\

$

Thus, The probability that the ball drawn is white $ = \dfrac{7}{{15}}$.

Note- Whenever we face such types of questions the key concept is that we should write the events which are occurring separately and then we should find the probability of each event that are occurring like we did and then we get our desired answer.

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