The probability of [math]x[/math] losses occurring in year 1 is [math]0.5x[/math] for [math]x=0,1, 2,\ldots[/math]. The probability of [math]y[/math] losses in year 2 given [math]x[/math] losses in year 1 is given by the table:

Number of losses in year 1 (x)	Number of losses in year 2 (y) given x losses in year 1
	0	1	2	3	4+
0	0.60	0.25	0.05	0.05	0.05
1	0.45	0.30	0.10	0.10	0.05
2	0.25	0.30	0.20	0.20	0.05
3	0.15	0.20	0.20	0.30	0.15
4+	0.05	0.15	0.25	0.35	0.20

Calculate the probability of exactly 2 losses in 2 years.

• 0.025
• 0.031
• 0.075
• 0.100
• 0.131

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: E

[[math]] \begin{align*} P( x= 1, y= 1)= P( y= 1| x= 1) P( x= 1)= 0.3(0.5)^2= 0.0753 \\ P(x= 2, y= 0)= P(y= 0| x= 2) P(x= 2)= 0.25(0.5)^3=0.03125 \\ P(x=0,y=2) = P(y=2 | x = 0 ) P(x=0) = 0.05(0.5)^1 = 0.025 \end{align*} [[/math]]

The total is 0.13125.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.