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Question: A box \({{B}_{1}}\) contains 1 white ball, 3 red balls and 2 black balls. Another box \({{B}_{2}}\) ...

A box B1{{B}_{1}} contains 1 white ball, 3 red balls and 2 black balls. Another box B2{{B}_{2}} contains 2 white balls, 3 red balls and 4 black balls. A third box B3{{B}_{3}} contains 3 white balls, 4 red balls and 5 black balls.
If 1 ball is drawn from each of the boxes B1,B2 and B3{{B}_{1}},{{B}_{2}}\ and\ {{B}_{3}}, the probability that all 3 drawn balls are of the same colour is,
A. 82648\dfrac{82}{648}
B. 90648\dfrac{90}{648}
C. 558648\dfrac{558}{648}
D. 566648\dfrac{566}{648}

Explanation

Solution

Hint: We will be using the concept of permutation and combination to find all the possible combinations, then we will use the concept of probability to find the final answer.

Complete step-by-step answer:
Now, we have been given that a box B1{{B}_{1}} contains 1 white ball, 3 red balls and 2 black balls.
A box B2{{B}_{2}} contains 2 white balls, 3 red balls and 4 black balls.
A box B3{{B}_{3}} contains 3 white balls, 4 red balls and 5 black balls.
Now, we have to find the probability that all 3 drawn balls are of the same colour.
Now, there are three possible cases that all the drawn balls are white or all the drawn balls are red or all the drawn balls are black.
So, we have to find the three probabilities and sum them up to find the total probability.
Now, the probability that all the balls are white can be found by using compound probability. Since the event of choosing a ball from each box is an independent event. Therefore, we multiply all the independent probability.
P(all white balls)=16×29×312P\left( all\ white\ balls \right)=\dfrac{1}{6}\times \dfrac{2}{9}\times \dfrac{3}{12}
Similarly, we have,
P(all red balls)=36×39×412 P(all black balls)=26×49×512 \begin{aligned} & P\left( all\ red\ balls \right)=\dfrac{3}{6}\times \dfrac{3}{9}\times \dfrac{4}{12} \\\ & P\left( all\ black\ balls \right)=\dfrac{2}{6}\times \dfrac{4}{9}\times \dfrac{5}{12} \\\ \end{aligned}
So, the probability that all 3 balls are of same colour is,
=6648+36648+40648 =82648 \begin{aligned} & =\dfrac{6}{648}+\dfrac{36}{648}+\dfrac{40}{648} \\\ & =\dfrac{82}{648} \\\ \end{aligned}
Hence, the correct option is (A).

Note: To solve these types of questions it is important to note that we have multiplied the independent probability of having a ball of identical colour in a box with other boxes to find the probability of having all the three balls of same colour.