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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability5 Marks
Question
Coloured balls are distributed in four boxes as shown in the following table:
Box
Colour
Black
White
Red
Blue
I
II
III
IV
3
2
1
4
4
2
2
3
5
2
3
1
6
2
1
5
A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III.
✓

Answer

Let $A, E_1, E_2, E_3$ and $E_4$ denote the events that the ball is black, box I selected, box II selected, box III is selected and box IV is selected respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{1}{4}$
$\text{P}(\text{E}_2)=\frac{1}{4}$
$\text{P}(\text{E}_3)=\frac{1}{4}$
$\text{P}(\text{E}_3)=\frac{1}{4}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{3}{18}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{8}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{1}{7}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_4}\Big)=\frac{4}{13}$
Using Bayes' theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{4}\times\frac{1}{7}}{\frac{1}{4}\times\frac{3}{18}+\frac{1}{4}\times\frac{2}{8}+\frac{1}{4}\times\frac{1}{7}+\frac{1}{4}\times\frac{4}{13}}$
$=\frac{\frac{1}{7}}{\frac{1}{6}+\frac{1}{4}+\frac{1}{7}+\frac{1}{13}}=\frac{156}{947}$

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