Question

A box $B_{1}$ contains $1$ white balls, $3$ red balls and $2$ black balls. Another box $B_{2}$ contains $2$ white balls, $3$ red balls and $4$ black balls. A third box $B_{3}$ contains $3$ white balls, $4$ red balls and $5$ black balls.If $1$ ball is drawn from each of the boxes $B_{1}$, $B_{2}$ and $B_{3}$ ,the probability that all $3$ drawn balls are of the same colour, is
1. $\dfrac{82}{648}$
2. $\dfrac{90}{648}$
3. $\dfrac{558}{648}$
4. $\dfrac{566}{648}$

Answer 1

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 we have to use logics and make different cases to find the probability that all three balls drawn are of all the same colour. It can be any colour white, red or black but it must be of the same colour. This sum can be solved using the formula made by logic that
$P(SameColour)=P(AllWhite)+P(AllRed)+P(AllBlack)$

Complete step by step answer:
To start the solution we can assume that the event of drawing three balls to be ${{E}_{1}},{{E}_{2}},{{E}_{3}}$ respectively. Now we can see that there are different probabilities to draw a certain ball from a certain box.
We can 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 in the first box the probability be
$P({{W}_{1}})=\dfrac{1}{6}$ which is the amount of white balls in the box to the amount of total balls in the first box
Similarly,
$P({{W}_{2}})=\dfrac{2}{9}$ ; $P({{W}_{3}})=\dfrac{3}{12}$
Now the probability of picking the black ball in each of these three boxes will be.
$P({{B}_{1}})=\dfrac{3}{6}$ ; $P({{B}_{2}})=\dfrac{3}{9}$ ; $P({{B}_{3}})=\dfrac{4}{12}$
Now the probability of picking the Red ball in each of these three boxes will be.
$P({{R}_{1}})=\dfrac{2}{6}$ , $P({{R}_{2}})=\dfrac{4}{9}$ , $P({{R}_{3}})=\dfrac{5}{12}$
Now to draw the same colour ball from all three boxes the probability will be
$P(SameColour)=P(AllWhite)+P(AllRed)+P(AllBlack)$
Now P(AllWhite) will be multiplying the probability of getting white ball from all three boxes similarly;
$P(SameColour)=\left( P({{W}_{1}})P({{W}_{2}})P({{W}_{3}}) \right)+(P({{R}_{1}})P({{R}_{2}})P({{R}_{3}}))+(P({{B}_{1}})P({{B}_{2}})P({{B}_{3}}))$
Now here substituting the values of probabilities
$P(SameColour)=\left( \dfrac{1}{6} \right)\left( \dfrac{2}{9} \right)\left( \dfrac{3}{12} \right)+\left( \dfrac{3}{3} \right)\left( \dfrac{3}{9} \right)\left( \dfrac{4}{12} \right)+\left( \dfrac{2}{6} \right)\left( \dfrac{4}{9} \right)\left( \dfrac{5}{12} \right)$
Multiplying
$P(SameColour)=\left( \dfrac{6}{648} \right)+\left( \dfrac{36}{648} \right)+\left( \dfrac{40}{648} \right)$
Simplifying
$P(SameColour)=\dfrac{82}{648}$
Hence the probability that you will draw balls of the same colour is $P(SameColour)=\dfrac{82}{648}$

So, the correct answer is “Option 1”.

Note:
We are solving this question by using the logic that we first start by making different cases on how we can draw the ball we need. We know that to draw the ball of the same color from all three boxes black there can only be 3 cases where we draw either black in all three of the boxes either red in all three of the boxes or white in all three of the boxes. As the requirements change the cases will change too and so will the probability.