Let the events, B₁, B₂, B₃ and A be defined as
B₁ \(\rightarrow\) I box is selected
B₂ \(\rightarrow\) II box is selected
B₃ \(\rightarrow\) III box is selected
A \(\rightarrow\) Bulb drawn is defective
\(P({ B }_{ 1 })=P({ B }_{ 2 })=P({ B }_{ 3 })=\frac { 1 }{ 3 } \)
\(P({ A/B }_{ 1 })=\frac { 1 }{ 2 } ,P({ A/B }_{ 2 })=\frac { 1 }{ 8 } ,P({ A/B }_{ 3 })=\frac { 3 }{ 4 } \)
\(P({ B }_{ 1 }/A)=\frac { P({ B }_{ 1 }).P(A/{ B }_{ 1 }) }{ P({ B }_{ 1 }).P(A/{ B }_{ 1 })+P({ B }_{ 2 }).P(A/{ B }_{ 2 })+P({ B }_{ 3 }).P(A/{ B }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } \times \frac { 1 }{ 2 } }{ \frac { 1 }{ 3 } \times \frac { 1 }{ 2 } +\frac { 1 }{ 3 } \times \frac { 1 }{ 8 } +\frac { 1 }{ 3 } \times \frac { 3 }{ 4 } }\)
\( =\frac { \frac { 1 }{ 2 } }{ \frac { 1 }{ 2 } +\frac { 1 }{ 8 } +\frac { 3 }{ 4 } } = \frac{\frac{1}{2}}{\frac{4+1+6}{8}}\)
\(=\frac { 1 }{ 2 } \times \frac { 8 }{ 11 } =\frac { 4 }{ 11 } \)

Let E₁, E₂ and A be defined as
E₁ = First bag is drawn
E₂ = Second bag is drawn
A = Red ball is drawn
\(\therefore\) P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A/E₁) = \(\frac{3}{7}\), P(A/E₂) = \(\frac{5}{11}\)
By Baye's theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { 1/2\times 3/7 }{ 1/2\times 3/7+1/2\times 5/11 } \)
\(=\frac { 3/7 }{ 3/7+5/11 } =\frac { 3/7 }{ \frac { 33+35 }{ 77 } } =\frac { 3}{7 }\times \frac { 77 }{ 68 } =\frac { 33 }{ 68 } \)

Find out the coefficient of mean deviation about median

Intervals	f	c.f	Mid x	|D| = |x-45.14|	f|D|
0-10	8	8	5	40.14	321.12
10-20	12	20	15	30.14	361.68
20.30	16	36	25	20.14	322.24
30-40	20	56	35	10.14	202.8
40-50	37	93	45	0.14	5.18
50-60	25	118	55	9.86	246.50
60-70	19	137	65	19.86	377.34
70.80	13	150	75	29.86	388.18
	\( \sum f\) = 150				\( \sum { f|D| }\) = 2225.04

Mean deviation about median = \(\frac { \sum { f|D| } }{ \sum f } =\frac { 2225.04 }{ 150 } =14.83\)
Median = 45.14
MD = 14.83

x	f	c.f	|D| = |x-61|	f|D|
58	15	15	3	45
59	20	35	2	40
60	32	67	1	32
61	35	102	0	0
62	35	137	1	35
63	22	159	2	44
64	20	179	3	60
65	10	189	4	40
66	8	197	5	40
	N = 197			\(\Sigma f|D|= \) 336

Median = Size of\({ \left( \frac { N+1 }{ 2 } \right) }^{ th }\) item
= Size of\({ \left( \frac { 198 }{ 2 } \right) }^{ th }\) item
= Size of 99^th value = 61
Mean deviation about median = \(\frac { \sum { f|X-Median| } }{ \sum f } \)
= \(\frac { \sum { f|D| } }{ N } \)
= \(\frac { 336 }{ 197 } \) = 1.71

Age in Years (x)	f	c.f
20	13	13
30	61	74
40	47	121
50	15	136
60	10	146
70	18	164
80	36	200
	N = 200

\({ Q }_{ 1 }=size\ of{ \left( \frac { N+1 }{ 4 } \right) }^{ th }value\) 
= size of \({ \left( \frac { 201 }{ 4 } \right) }^{ th }value\)
= size of 50.25^th value = 30
Q₃ = Size of\({ \left( \frac { 3(N+1) }{ 4 } \right) }^{ th }\)value
= size of 3(50.25)^th value.
= size of 150.75^th value = 70
Q.D = \(\frac { 1 }{ 2 } ({ Q }_{ 3 }-{ Q }_{ 1 })= \frac{70-30}{2}=\frac{ 40}{2}=20\)
Co-efficient of Quartile deviation = \(\frac { { Q }_{ 3 }-{ Q }_{ 1 } }{ { Q }_{ 3 }+{ Q }_{ 1 } } =\frac { 70-30 }{ 70+30 } =\frac { 40 }{ 100 } =0.4\)