Let X denote the net gain to the company from the sale of one such policy. Find the expected value to the company of a single policy if a person in this risk group has a 99.97% chance of surviving one year. The concept of expected value is also basic to the insurance industry, as the following simplified example illustrates.Ī life insurance company will sell a $200,000 one-year term life insurance policy to an individual in a particular risk group for a premium of $195. In particular, if someone were to buy tickets repeatedly, then although he would win now and then, on average he would lose 40 cents per ticket purchased. The negative value means that one loses money on the average. Using the formula in the definition of expected value, E ( X ) = 299 Let W denote the event that a ticket is selected to win one of the prizes. Applying the same “income minus outgo” principle to the second and third prize winners and to the 997 losing tickets yields the probability distribution: x 299 199 99 − 1 P ( x ) 0.001 0.001 0.001 0.997
There is one such ticket, so P(299) = 0.001. If a ticket is selected as the first prize winner, the net gain to the purchaser is the $300 prize less the $1 that was paid for the ticket, hence X = 300 − 1 = 299. Find the expected value of X, and interpret its meaning.Find the probability of winning any money in the purchase of one ticket.Let X denote the net gain from the purchase of one ticket. First prize is $300, second prize is $200, and third prize is $100. One thousand raffle tickets are sold for $1 each. We compute P ( X is even ) = P ( 2 ) + P ( 4 ) + P ( 6 ) + P ( 8 ) + P ( 10 ) + P ( 12 ) = 1 36 + 3 36 + 5 36 + 5 36 + 3 36 + 1 36 = 18 36 = 0.5Ī histogram that graphically illustrates the probability distribution is given in Figure 4.2 "Probability Distribution for Tossing Two Fair Dice".Ī service organization in a large town organizes a raffle each month. Thus P ( X ≥ 9 ) = P ( 9 ) + P ( 10 ) + P ( 11 ) + P ( 12 ) = 4 36 + 3 36 + 2 36 + 1 36 = 10 36 = 0.2 7 -īefore we immediately jump to the conclusion that the probability that X takes an even value must be 0.5, note that X takes six different even values but only five different odd values. The event X ≥ 9 is the union of the mutually exclusive events X = 9, X = 10, X = 11, and X = 12. This table is the probability distribution of X. X = 2 is the event, so P ( 3 ) = 2 ∕ 36. The possible values for X are the numbers 2 through 12. Find the probability that X takes an even value.Construct the probability distribution of X.Let X denote the sum of the number of dots on the top faces.