The joint density equation is missing in the question. The equation is [tex]f(x,y)=\left\{\begin{matrix}2x^3, 0\leq x\leq 1,0\leq y\leq 2\\ 0, \text{ otherwise}\end{matrix}\right.[/tex]
Probability is 0.59
Step-by-step explanation:
We have
x -- denotes the annual claim health insurance
y -- denotes the annual claim life insurance
And joint density of x and y is given to us
[tex]f(x,y)=\left\{\begin{matrix}2x^3, 0\leq x\leq 1,0\leq y\leq 2\\ 0, \text{ otherwise}\end{matrix}\right.[/tex]
Event that the claim exceeds 0.5 but less than 2 can be written as 0.5≤ x+y ≤2
and claim of life insurance less than 1.5 can be written as 0 ≤ y ≤ 1.5
Required probability
= P( 0.5≤ x+y ≤2 ∩ 0 ≤ y ≤ 1.5 )
Lets plot this, to find the common region
For common region
when 0≤ x ≤ 0.5, then 0.5 ≤ y ≤ 1.5 ---- for region I
when 0.5≤ x ≤ 1, then 0 ≤ y ≤ 2-x ---- for region II
The required probability
= [tex]\int_{0}^{0.5} \int_{0.5-x}^{1.5}2x^3 dy dx+\int_{0.5}^{1} \int_{0}^{2-x}2x^3 dy dx[/tex]
= [tex]\int_{0}^{0.5} 2x^3[\int_{0.5-x}^{1.5}dy] dx+\int_{0.5}^{1} 2x^3[\int_{0}^{2-x}dy] dx[/tex]
= [tex]\int_{0}^{0.5} 2x^3(1.5-0.5+x) dx+\int_{0.5}^{1} 2x^3 (2-x) dx[/tex]
= [tex]\int_{0}^{0.5}2x^3 dx+\int_{0}^{0.5}2x^4 dx+\int_{0.5}^{1}4x^3 dx-\int_{0.5}^{1}2x^4[/tex]
= [tex][\frac{2x^4}{4}]_0^{0.5}+[\frac{2x^5}{5}]_0^{0.5}+[\frac{4x^4}{4}]_{0.5}^1-[\frac{2x^5}{5}]_{0.5}^1[/tex]
= [tex]$\frac{1}{32}+\frac{1}{80}+\frac{1}{16}-\frac{31}{20}$[/tex]
= 19/32
=0.59
