The content of the 3 boxes are as follows. If one box is chosen at random and three balls are drawn from it and they are all of different colours, find the probability that they come from box 2 . Box 1 contains 1 black, 2 white, 3 red balls, and Box 2 contains 1 black, I white, 2 red balls, and…

Question

The content of the 3 boxes are as follows. If one box is chosen at random and three balls are drawn from it and they are all of different colours, find the probability that they come from box 2 . Box 1 contains 1 black, 2 white, 3 red balls, and Box 2 contains 1 black, I white, 2 red balls, and Box 3 has 5 black, 4 white, 1 red balls.

Aicharya · Accepted Answer

This is a problem of Baye's theorem,

Let \(A=\) All 3 Ball's are of different colours. \(E_1=\) Box 1 is Choosen \(E_2=\) Box 2 is Choosen \(E_3=\) Box 3 is Choosen \(\frac{A}{E_i}=3\) different colour balls comes from a particular box. Now, \(P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}\) \(\begin{aligned} & P\left(\frac{A}{E_1}\right)=\frac{{ }^1 C_1 \times{ }^2 C_1 \times{ }^3 C_1}{{ }^6 C_3}=\frac{3}{10} \\ & P\left(\frac{A}{E_2}\right)=\frac{{ }^1 C_1 \times{ }^1 C_1 \times{ }^2 C_1}{{ }^4 C_3}=\frac{1}{2} \\ & P\left(\frac{A}{E_3}\right)=\frac{{ }^5 C_1 \times{ }^4 C_1 \times{ }^1 C_1}{{ }^{10} C_3}=\frac{1}{6} \end{aligned}\) Now by Baye's theorem, \(\begin{aligned} P\left(\frac{E_2}{A}\right) & =\frac{P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)}{P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)+P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_3\right) \cdot P\left(\frac{A}{E_3}\right)} \\ & =\frac{\frac{1}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{1}{2}+\frac{1}{3} \times \frac{3}{10}+\frac{1}{3} \times \frac{1}{6}}=\frac{15}{29} \end{aligned}\)

The content of the 3 boxes are as follows. If one box is chosen at random and three balls are drawn from it…

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