Question

Suppose we have four boxes A, B, C and D containing coloured marbles as given below

$\begin{array}{cccc}& & Marblecolour& \\ Box& Red& White& Black\\ A& 1& 6& 3\\ B& 6& 2& 2\\ C& 8& 1& 1\\ D& 0& 6& 4\end{array}$

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red what is the probability that it was drawn from box A? box B? , box C?

Answer 1

Let $E_1$ : the event that box A is selected , $E_2$ the event that box B is selected,

$E_3$ : the event that box C is selected and $E_4$ : the event that box D is selceted,

$\therefore E_1,E_2,E_3,E_4$ are mutually exclusive and exhaustive . Moreover,

$P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=P\left(E_4\right)=\frac{1}{4}.$

Let E: ball drawn is red,

$\therefore P\left(\frac{E}{E_1}\right)=\frac{1}{1+6+3}=\frac{1}{10}P\left(\frac{E}{E_2}\right)=\frac{6}{6+2+2}=\frac{6}{10}P\left(\frac{E}{E_3}\right)=\frac{8}{8+1+1}=\frac{8}{10}andP\left(\frac{E}{E_4}\right)=\frac{0}{0+6+4}=0$

By using Baye's theorem

P(ball is drawn from box A)=$P\left(\frac{E}{E_1}\right)$

$=\frac{P\left(\frac{E}{E_1}\right)P\left(E_1\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)+P\left(\frac{E}{E_3}\right)P\left(E_3\right)+P\left(\frac{E}{E_4}\right)P\left(E_4\right)}=\frac{\frac{1}{10}\times \frac{1}{4}}{\frac{1}{10}\times \frac{1}{4}+\frac{6}{10}\times \frac{1}{4}+\frac{8}{10}\times \frac{1}{4}+0\times \frac{1}{4}}=\frac{\frac{1}{40}}{\frac{1}{4}(\frac{1}{10}+\frac{6}{10}+\frac{8}{10}+0)}=\frac{\frac{1}{40}}{\frac{1}{4}\times \frac{15}{10}}=\frac{\frac{1}{40}}{\frac{15}{40}}=\frac{1}{15}$

P(ball is drawn from box B)
$P\left(\frac{E}{E_2}\right)=\frac{P\left(\frac{E}{E_2}\right)P\left(E_2\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)+P\left(\frac{E}{E_3}\right)P\left(E_3\right)+P\left(\frac{E}{E_4}\right)P\left(E_4\right)}=\frac{\frac{6}{10}\times \frac{1}{4}}{\frac{1}{10}\times \frac{1}{4}+\frac{6}{10}\times \frac{1}{4}+\frac{8}{10}\times \frac{1}{4}+0\times \frac{1}{4}}=\frac{\frac{6}{40}}{\frac{1}{4}(\frac{1}{10}+\frac{6}{10}+\frac{8}{10}+0)}=\frac{\frac{6}{40}}{\frac{1}{4}\times \frac{15}{10}}=\frac{\frac{6}{40}}{\frac{15}{40}}=\frac{6}{40}\times \frac{40}{15}=\frac{2}{5}$

P(ball is drawn from box C)

$P\left(\frac{E}{E_2}\right)=\frac{P\left(\frac{E}{E_3}\right)P\left(E_3\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)+P\left(\frac{E}{E_3}\right)P\left(E_3\right)+P\left(\frac{E}{E_4}\right)P\left(E_4\right)}=\frac{\frac{8}{10}\times \frac{1}{4}}{\frac{1}{10}\times \frac{1}{4}+\frac{6}{10}\times \frac{1}{4}+\frac{8}{10}\times \frac{1}{4}+0\times \frac{1}{4}}=\frac{\frac{8}{40}}{\frac{1}{4}(\frac{1}{10}+\frac{6}{10}+\frac{8}{10}+0)}=\frac{\frac{8}{40}}{\frac{1}{4}\times \frac{15}{10}}=\frac{\frac{8}{40}}{\frac{15}{40}}=\frac{8}{40}\times \frac{40}{15}=\frac{8}{15}$