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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability1 Mark
Question
Coloured balls are distributed in four boxes as shown in the following table: 

Box Colour
Black White Red Blue
I 3 4 5 6
II 2 2 2 2
III 1 2 3 1
IV 4 3 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, E1, E2, E3 and E4 be the events as defined below :
A : black ball is selected
E1 : box I is selected
E2 : box II is selected
E3 : box III is selected
E4 : box IV is selected
Since the boxes are chosen at random,
Therefore, P(E1) = P(E2) = P(E3) = P(E4)
Also P(A|E1) = $\frac{3}{18}$, P(A|E2) = $\frac {2}{8}$, P(A|E3) = $\frac{1}{7}$ and P(A|E4) = $\frac{4}{13}$ 
P(box III is selected, given that the drawn ball is black) = P(E3|A). By Bayes' theorem,
P(E3|A) = $\frac{\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\mathrm{A|E}_{3}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A|E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A|E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \mathrm{P}\left(\mathrm{A|E}_{3}\right)+\mathrm{P}\left(\mathrm{E}_{4}\right) \mathrm{P}\left(\mathrm{A|E}_{4}\right)}$ 
= $\frac{\frac{1}{4} \times \frac{1}{7}}{\frac{1}{4} \times \frac{3}{18}+\frac{1}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{1}{7}+\frac{1}{4} \times \frac{4}{13}}$ = 0.165

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