
A bag contains 3 white balls 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
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Hint: Here, we are asked to find the probability of drawing white ball. There are two mutually exclusive ways of drawing a white ball. First, selecting the first bag and then drawing the white ball from it and second, selecting the second bag and drawing white ball from it. We will consider both the ways and then find the required probability.
Complete step-by-step answer:
Let $ {B_1} $ be the event of selecting the first bag, $ {B_2} $ be the event of selecting the second bag and $ W $ be the event of drawing white ball.
There are two bags and therefore, probability of selecting any one bag is given by
$ P\left( {{B_1}} \right) = P\left( {{B_2}} \right) = \dfrac{1}{2} $
The first bag contains 3 white balls and 2 black balls, thus, total 5 balls. Hence the probability of drawing the white ball from the first bag is:
$ P\left( {W/{B_1}} \right) = \dfrac{3}{5} $
Similarly, the second bag contains 2 white and 4 black balls, thus, total 6 balls. Hence the probability of drawing the white ball from the second bag is:
$ P\left( {W/{B_2}} \right) = \dfrac{2}{6} = \dfrac{1}{3} $
Now, to find the required probability $ P\left( W \right) $ , we will use the law of total probability by which, we can say that
$
P\left( W \right) = P\left( {{B_1}} \right)P\left( {W/{B_1}} \right) + P\left( {{B_2}} \right)P\left( {W/{B_2}} \right) \\
\Rightarrow P\left( W \right) = \left( {\dfrac{1}{2} \times \dfrac{3}{5}} \right) + \left( {\dfrac{1}{2} \times \dfrac{1}{3}} \right) \\
\Rightarrow P\left( W \right) = \dfrac{3}{{10}} + \dfrac{1}{6} = \dfrac{{9 + 5}}{{30}} = \dfrac{{14}}{{30}} = \dfrac{7}{{15}} \\
$
Thus, the probability that the ball drawn is white is $ \dfrac{7}{{15}} $ .
So, the correct answer is “ $ \dfrac{7}{{15}} $ ”.
Note: In this type of question, it is important to keep in mind that before drawing the ball, we need to select the bag first. Therefore, we have determined the probability of selecting a bag and then probability of drawing the white ball from each individual bag. At the end, we have used total probability thermo to find our final answer.
Complete step-by-step answer:
Let $ {B_1} $ be the event of selecting the first bag, $ {B_2} $ be the event of selecting the second bag and $ W $ be the event of drawing white ball.
There are two bags and therefore, probability of selecting any one bag is given by
$ P\left( {{B_1}} \right) = P\left( {{B_2}} \right) = \dfrac{1}{2} $
The first bag contains 3 white balls and 2 black balls, thus, total 5 balls. Hence the probability of drawing the white ball from the first bag is:
$ P\left( {W/{B_1}} \right) = \dfrac{3}{5} $
Similarly, the second bag contains 2 white and 4 black balls, thus, total 6 balls. Hence the probability of drawing the white ball from the second bag is:
$ P\left( {W/{B_2}} \right) = \dfrac{2}{6} = \dfrac{1}{3} $
Now, to find the required probability $ P\left( W \right) $ , we will use the law of total probability by which, we can say that
$
P\left( W \right) = P\left( {{B_1}} \right)P\left( {W/{B_1}} \right) + P\left( {{B_2}} \right)P\left( {W/{B_2}} \right) \\
\Rightarrow P\left( W \right) = \left( {\dfrac{1}{2} \times \dfrac{3}{5}} \right) + \left( {\dfrac{1}{2} \times \dfrac{1}{3}} \right) \\
\Rightarrow P\left( W \right) = \dfrac{3}{{10}} + \dfrac{1}{6} = \dfrac{{9 + 5}}{{30}} = \dfrac{{14}}{{30}} = \dfrac{7}{{15}} \\
$
Thus, the probability that the ball drawn is white is $ \dfrac{7}{{15}} $ .
So, the correct answer is “ $ \dfrac{7}{{15}} $ ”.
Note: In this type of question, it is important to keep in mind that before drawing the ball, we need to select the bag first. Therefore, we have determined the probability of selecting a bag and then probability of drawing the white ball from each individual bag. At the end, we have used total probability thermo to find our final answer.
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