Four linked problems drawn from this topic’s subtopics. 12 parts · untimed · answers and worked solutions at the end.
Example 1
The table gives the loss L in dollars on one policy. The insurer pays 50% of the loss above a $2 deductible: payout Y=0.5×max(L−2,0). There are no other costs.
Given that a loss occurs, find P(L=90∣L>0). Give your answer to 4 decimal places.
Loss distribution
Loss L($)
Probability
0
0.5
30
0.4
90
0.1
Show worked answer
Answer: 0.2000
Condition on the two positive-loss outcomes: P(L>0)=0.5.
P(L=90∣L>0)=0.50.1=0.2.
Example 2
An unknown number n of parcels is dispatched independently, each with the same unknown probability p of arriving late. The number X arriving late has E(X)=3.6 and Var(X)=2.52. A reserve team can handle at most m late parcels.
A service succeeds with probability p. Its payoff Y is $5 on success and −$3 on failure. The mean payoff is $1. A proposed improvement raises the success probability by 0.05 and costs $0.4 on every attempt.
a Find p. Give your answer to 2 decimal places.
b Find the variance of the payoff Y, in dollars squared. Give your answer to 4 decimal places.
Should the improvement be adopted if only expected net payoff matters?
Single-attempt payoffs
Outcome
Payoff Y($)
Success
5
Failure
−3
Show worked answer
Answer: Either: the expected gains are equal
Increasing p by 0.05 increases expected gross payoff by (5+3)×0.05=0.4 dollars.
After the cost, the change is 0 dollars.
The cost is paid on every attempt, including failures.
Discrete Random Variables Review Practice | Year 12 Methods | Ascent Arcade