Independent solution

How to solve this Sums of Independent Random Variables question

Setup

Setup

Let H and T be the ten-year counts for the two independent loss types.

HBinomial(10,0.2)H\sim\operatorname{Binomial}(10,0.2)
TBinomial(10,0.1)T\sim\operatorname{Binomial}(10,0.1)

Model

Model

The total is below two in exactly three disjoint count combinations.

Pr(H+T<2)=Pr(H=0,T=0)+Pr(H=1,T=0)+Pr(H=0,T=1)\Pr(H+T<2)=\Pr(H=0,T=0)+\Pr(H=1,T=0)+\Pr(H=0,T=1)

Compute

Compute

Use independence to multiply the component probabilities within each combination.

Pr(H=0,T=0)=0.8100.910=0.0374390624\Pr(H=0,T=0)=0.8^{10}0.9^{10}=0.0374390624
Pr(H=1,T=0)=10(0.2)0.890.910=0.0935976561\Pr(H=1,T=0)=10(0.2)0.8^9 0.9^{10}=0.0935976561
Pr(H=0,T=1)=0.81010(0.1)0.99=0.0415989583\Pr(H=0,T=1)=0.8^{10}10(0.1)0.9^9=0.0415989583
Pr(H+T<2)=0.1726356767\Pr(H+T<2)=0.1726356767

Answer

Answer

The probability rounds to 0.17.

0.17(A)\boxed{0.17\quad\text{(A)}}