Solved problems on probability and statistics

This document contains solutions to problems on probability and statistics, including binomial distribution, independence, mutually exclusive events, and more. It also mentions the document type as solved problems and does not specify any course code, course name, or college/university.

PROBLEM I
1) θ=0.2 and n=7
P ( X=3 ) =7C3
( θ ) 3 (1−θ)4
P ( X=3 ) Where X ~ Binomial (n=7, θ=0.2 ¿
Answer) a)
2) P ( A )=0.6∧P ( B )=0.1
P ( A ∩B )=0.08
P ( A ∪B ) =P ( A ) + P ( B ) −P ( A ∩ B )
P ( A ∪B ) =0.62
Since,
P ( A ∩B ) ≠ P ( A ) × P ( B )
And,
P ( A ∪ B ) ≠ P ( A )+ P ( B )
So it is neither independent nor mutually exclusive.
Answer) e)
3) P ( X ≥2 ) =1−P ( X=0 )−P ( X =1 )
P ( X ≥2 ) =1−0.1−0.3=0.6
Answer) d)
4) f X ( x )=
{ d
dx
x
4 0 ≤ x <1
d
dx
x4 + 4
20 1 ≤ x<2
d
dx 1 x ≥2
f X ( x ) =
{ 1
4 0 ≤ x <1
x3
5 1 ≤ x <2
0 x ≥2
Answer) b)

5) 6
20 × 5
19 × 4
18
(6
3 )
(20
3 )
Answer) d)
6) P ( Good Risk ) =0.4
P ( Bad Risk ) =0.6
P ( NoCitation∨Good Risk ) =1−0.1=0.9
P ( NoCitation∨Bad Risk )=1−0.5=0.5
P ( NoCitation ) =P ( No Citation∨Good Risk ) ×
P ( Good Risk ) +P ( No Citation∨Bad Risk ) × P ( Bad Risk ) = ( 0.4 ×0.9 ) + ( 0.6 × 0.5 ) =0.66
Answer) b)
7) f X ( x )= { x
2 0 ≤ x< 2
0 x ≥ 2
∫
0
x
x
2 dx=0.75
x2
4 =0.75
x= √4 ×0.75
Answer) c)
8) P ( X >Y )= 4
24 + 2
24 + 1
24 = 7
24
Answer) a)
PROBLEM II

1) P ( Goalkeeper∨Sam )=0.5
P ( Goalkeeper∨ Alex )=0.3
P ( NoGoalkeeper ∨Sam )=0.5
P ( NoGoalkeeper ∨Alex )=0.7
P ( S am ) =0. 6
P ( Alex ) =0. 4
P ( GoalKeeper ) =P ( Goalkeeper ∨Sam ) × P ( Sam ) + P (Goalkeeper ∨ Alex ) ×
P ( Alex ) = ( 0.5 ×0. 6 ) + ( 0.3 × 0. 4 ) =0.42
P ( NoGoalKeeper ) =P ( NoGoalkeeper ∨Sam ) × P ( Sam ) + P ( No Goalkeeper ∨Alex ) ×
P ( Alex ) = ( 0.5 ×0.6 )+ ( 0.7 ×0.4 )=0.58
P ( Alex∨NoGoalkeeper )= P ( No Goalkeeper∩ Alex )
P ( NoGoalKeeper )
P ( Alex∨NoGoalkeeper )= P ( No Goalkeeper∨ Alex ) × P ( Alex )
P ( NoGoalKeeper )
P ( Alex∨NoGoalkeeper )= 0.7 ×0.4
0.58 =0.28
0.58 = 14
29
2) f X Y (x , y )= { x + y
8 0≤ x <2 ,0 ≤ y <2
0 otherwise
f X ( x )=∫
0
2
x+ y
8 dy=¿∫
0
2
x
8 dy +∫
0
2
y
8 dy= x
4 + 1
4 ¿
f Y ( y)=∫
0
2
x + y
8 dx=¿∫
0
2
x
8 dx +∫
0
2
y
8 dx= y
4 + 1
4 ¿
Since,
f XY ( x , y ) ≠ f X ( x ) × f Y ( y )

We can say that X and Y are not independent.
3) a) If mutually exclusive then,
P ( A ∪ B ) =P ( A )+ P ( B )=0.5+0.3=0.8
b) If mutually independent then,
P ( A ∩B )=P ( A ) × P ( B ) =0.5 ×0.3=0.15
P ( A ∪ B ) =P ( A )+ P ( B )−P ( A ∩ B )=0.5+ 0.3−0.15=0.65
c) B⊂ A
Then we can find the value as,
P ( A ∪ B ) =P ( A ) =0.5
d) P ( A ∩B )=0.2
P ( A ∪ B ) =P ( A )+P ( B )−P ( A ∩ B )=0.5+0.3−0.2=0.6
4)
f X ( x ) =e− x
P ( Y < y )=P ( e5 x < y ) =P ¿
FY ( y ) = ∫
0
ln y
5
e−x dx=−e− x∨¿0
ln y
5 =1−e
−ln y
5 =1− y
−1
5 ¿
f Y ( y ) = d
dy FY ( y ) = d
dy ( 1− y
−1
5 ) =1
5 y
−6
5
Therefore the distribution of Y is
f Y ( y )= 1
5 y
−6
5