Suppose we have four boxes A,B,C and D containing coloured marble… - Vidyadip

Question

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

Box	Marble colour
	Red	White	Black
A	1	6	3
B	6	2	2
C	8	1	1
D	0	6	4

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

Let R represents the drawing of red ball and the four boxes are represented by A, B, and D.
$\text{P}(\text{R}|\text{A})=\frac{1}{10}$
$\text{P}(\text{R}|\text{B})=\frac{6}{10}$
$\text{P}(\text{R}|\text{C})=\frac{8}{10}$
$\text{P}(\text{R}|\text{D})=\frac{0}{10}=0$
Since there are 4 bags.
Therefore, $\text{P}(\text{A})=\frac{1}{4},\ \text{P}(\text{B})=\frac{1}{4},\ \text{P}(\text{C})=\frac{1}{4},\ \text{P}(\text{D})=\frac{1}{4},$
$\text{P}(\text{A|R})=\frac{\text{P}(\text{A}).\text{P}(\text{A|R})}{\text{P}(\text{A})\text{P}(\text{R|A})+\text{P}(\text{B})\text{P}(\text{R|B})+\text{P}(\text{C})\text{P}(\text{R|C})+\text{P}(\text{D})\text{P}(\text{R|D})}$
$=\frac{\frac{1}{4}\times\frac{1}{10}}{\frac{1}{4}\times\frac{1}{10}+\frac{1}{4}\times\frac{6}{10}+\frac{1}{4}\times\frac{8}{10}+\frac{1}{4}\times0}$
$=\frac{\frac{1}{10}}{\frac{1}{10}+\frac{6}{10}+\frac{8}{10}}=\frac{1}{15}$
$\text{P}(\text{B|R})=\frac{\text{P}(\text{B}).\text{P}(\text{R|B})}{\text{P}(\text{A})\text{P}(\text{R|A})+\text{P}(\text{B})\text{P}(\text{R|B})+\text{P}(\text{C})\text{P}(\text{R|C})+\text{P}(\text{D})\text{P}(\text{R|D})}$
$=\frac{\frac{1}{4}\times\frac{6}{10}}{\frac{1}{4}\times\frac{1}{10}+\frac{1}{4}\times\frac{6}{10}+\frac{1}{4}\times\frac{8}{10}+\frac{1}{4}\times0}=\frac{6}{15}=\frac{2}{5}$
$\text{P}(\text{C|R})=\frac{\text{P}(\text{C}).\text{P}(\text{R|C})}{\text{P}(\text{A})\text{P}(\text{R|A})+\text{P}(\text{B})\text{P}(\text{R|B})+\text{P}(\text{C})\text{P}(\text{R|C})+\text{P}(\text{D})\text{P}(\text{R|D})}$

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