b)

Let \( S\) be the sum assured, from a). Firstly, define \( L_0\) for each value of \( K_{[41]}\):

$$L_0 = Sv^{K_{[41]} +1 } - 350 \ddot{a}_\overline{{K_{[41]} +1}|} \text{if} K_{[41]} \leqslant 2$$ \( = -350 \ddot{a}_\overline{{3}|} \text{if} K_{[41]} \geqslant 3\).

Since we are assuming the equivalence principle holds,\( \mathbb{E}(L_0) = 0 \). So \( Var(L_0) = \mathbb{E}(L_0^2) \).

Now find \( \mathbb{E}(L_0^2) \) like you would for any ordinary discrete random variable:

$$Var(L_0) = \mathbb{E}(L_0^2) = \sum_{k=0}^{\infty} l_0^2 \Pr(K_{[41]} = k).$$ Note that this will be a sum of four terms, one for each of \(K_{[41]} = 0,1,2 \) and \(K_{[41]}\geqslant 3 \) respectively. \( l_0 \) are the values which \( L_0\) takes on.

c)

We require \( \Pr(L_0) > 0\) i.e of the loss being positive, and as can be seen from the definition of \( L_0\) above, this occurs when \(K_{[41]} = 0,1,2 \). So the required probability is equivalent to the probability of the life dying in the first three years. If the life survives to the end of the three years, we would have made a profit, not a loss.

So the required probability is \( 1 - {}_3 p_{[41]} = 0.0052\).

Hope this helps :)