Question 5.2: 5.2 Life Assurance Policy Surrender Study Global Insurance h......

(a) The random variable ‘number of policies surrendered’ is discrete, since there can be 0, 1, 2, 3, …, 9, 10 surrendered policies in the randomly selected sample of 10 policies. This random variable ‘fits’ the binomial probability distribution for the following reasons:

• The random variable is observed 10 times (10 policies were randomly sampled). Each policy is an observation of the random variable (i.e. policy surrender status). Hence n = 10.
• There are only two possible outcomes for each policy, namely:

– a policy is surrendered before maturity (the success outcome)

– a policy is not surrendered before maturity (the failure outcome).

• A probability can be assigned to each outcome for a policy, namely:

– p (= probability of a policy being surrendered) = 0.20

– (1 − p) (= probability of a policy not being surrendered) = 0.80

Note that the success outcome refers to surrendering a policy since the binomial question seeks probabilities for surrendered policies.

• The trials are independent. Each policy’s status (surrendered or not) is independent of every other policy’s status. Thus p = 0.20 is constant for each policy.

Since all the conditions for the binomial process have been satisfied, the binomial question can be addressed: ‘What is the probability that four of these 10 insurance policies will have been surrendered before maturity date?’

Find:       P(x = 4) when n = 10 and p = 0.2.

Then:      P(x = 4) = _{10}C_{4}(0.20)^4(1-0.20)^{10-4} = (210)(0.0016)(0.2621) = 0.088

Thus there is an 8.8% chance that four out of 10 randomly selected policies will have been surrendered before maturity.

(b) The binomial approach still applies. In terms of the binomial question, ‘no more than 3’ implies that either 0 or 1 or 2 or 3 of the sampled policies will be surrendered before maturity.

Thus find P(x ≤ 3). (This is a cumulative probability.)

Using the addition rule of probability for mutually exclusive events, the cumulative probability is:

P(x ≤ 3) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)

The three binomial probabilities must now be calculated separately and summed:

P(x = 0) = _{10}C_{0}(0.20)^0(1-0.20)^{10-0}=0.107

P(x = 1) = _{10}C_{1}(0.20)^1(1-0.20)^{10-1}=0.269

P(x = 2) = _{10}C_{2}(0.20)^2(1-0.20)^{10-2}=0.302

P(x = 3) = _{10}C_{3}(0.20)^3(1-0.20)^{10-3}=0.201

Then P(x ≤ 3) = 0.107 + 0.269 + 0.302 + 0.201 = 0.879

There is an 87.9% chance that no more than three out of the 10 policies randomly selected will be surrendered before their maturity date.

(c) This question translates into finding the cumulative probability of P(x ≥ 2)

i.e. P(x ≥ 2) = P(x = 2) + P(x = 3) + P(x = 4) + … + P(x = 10).

This requires that nine binomial calculations be performed. However, to avoid onerous calculations, the complementary law of probability can be used, as follows:

P(x ≥ 2) = 1 − P(x ≤ 1)
= 1 − [P(x = 0) + P(x = l)]
= 1 − [0.107 + 0.269] (from (b) above) = 1 – 0.376 = 0.624

Thus there is a 62.4% chance that at least 2 out of the 10 randomly selected policies will be surrendered before their maturity date.