Question

If from each of the three boxes containing \[3\] white and \[1\] black, \[2\] white and \[2\] black. \[1\] white and \[3\]black balls, one ball is drawn at random , then the probability that \[2\] white and \[1\] black ball will be drawn is
1. \[\dfrac{13}{32}\]
2. \[\dfrac{1}{4}\]
3. \[\dfrac{1}{32}\]
4. \[\dfrac{3}{16}\]

Answer 1

Hint: We will start the solution of this question by assuming that the events of drawing the balls from the box to be \[{{E}_{1}},{{E}_{2}},{{E}_{3}}\] respectively. Now we need to find the probability of each event and then with that we can use logics and make different cases to find the probability of \[2\] white and \[1\] black being drawn by you. We can solve this by the formula that
\[P(2\text{ }white\text{ }and\text{ }1\text{ }black)=P({{W}_{1}})P({{W}_{2}})P({{B}_{3}})+P({{W}_{1}})P({{B}_{2}})P({{W}_{3}})+P({{B}_{1}})P({{W}_{2}})P({{W}_{3}})\]

Complete step-by-step answer:
We start by assuming the event of drawing three balls to be \[{{E}_{1}},{{E}_{2}},{{E}_{3}}\] respectively. Now first we can see that there are different probabilities to get different balls in different cases
Let us start by the first case given which the probability is of getting white in all three boxes, starting with the first box.
Now probability of drawing a white ball we will get
\[P({{W}_{1}})=\dfrac{3}{4}\] which is the amount of white balls in the box to the amount of total balls
Similarly,
\[P({{W}_{2}})=\dfrac{2}{4}\] ; \[P({{W}_{3}})=\dfrac{1}{4}\]
Now the probability of picking the black ball in each of these three boxes will be.
\[P({{B}_{1}})=\dfrac{1}{4}\] ; \[P({{B}_{2}})=\dfrac{2}{4}\] ; \[P({{B}_{2}})=\dfrac{3}{4}\]
Now to draw 2 white balls and one white ball there will be three cases so we can multiply and add the probability to give us the answer to this question.
\[P(2\text{ }white\text{ }and\text{ }1\text{ }black)=P({{W}_{1}})P({{W}_{2}})P({{B}_{3}})+P({{W}_{1}})P({{B}_{2}})P({{W}_{3}})+P({{B}_{1}})P({{W}_{2}})P({{W}_{3}})\]
Now here substituting the values of probabilities
\[P(2\text{ }white\text{ }and\text{ }1\text{ }black)=\left( \dfrac{3}{4} \right)\left( \dfrac{2}{4} \right)\left( \dfrac{3}{4} \right)+\left( \dfrac{3}{4} \right)\left( \dfrac{2}{4} \right)\left( \dfrac{1}{4} \right)+\left( \dfrac{1}{4} \right)\left( \dfrac{2}{4} \right)\left( \dfrac{1}{4} \right)\]
Multiplying
\[P(2\text{ }white\text{ }and\text{ }1\text{ }black)=\dfrac{18}{64}+\dfrac{6}{64}+\dfrac{2}{64}\]
Simplifying
\[P(2\text{ }white\text{ }and\text{ }1\text{ }black)=\dfrac{13}{32}\]
Hence the probability that you will draw 2 white balls and 1 black ball is \[\dfrac{13}{32}\]
So, the correct answer is “Option 1”.

Note: We solve this question using the logic that we make different cases on how we can draw the given case. Like we know that to get 2 white balls and 1 black there can only be 3 cases where we draw black of any of the three boxes and then draw white in both others. As the requirements changes the cases will change too and so will the probability