Question

A lottery payoff. A $1 bet in a state lottery's Pick 3 game pays $500 if the three-digit number you choose exactly matches the winning number, which is drawn at random. Here is the distribution of the payoff X: Payoff $0; probability: 0.999. Payoff $500; Probability: 0.001. Each day's drawing is independent of other drawings. (a) Hoe buys a Pick 3 ticket twice a week. The number of times he wins follows a B(104, 0.001) distribution. Using the Poisson approximation to the binomial, what is the probability that he wins at least once? (b) the exact binomial probability is 0.9888. How accurate is the Poisson approximation here? (c) if Joe pays $5 a ticket, he needs to win at least twice a year to come out ahead. using Poisson approximation, what is the probability that Joe comes out ahead?

Answer 1

Part (a)

B(n, p) is approximated by Poisson (λ), where λ = np.

So, in the given case, λ = 104 x 0.001 = 0.104.

Probability that he wins at least once = 1 – Probability he does not win at all

= 1 – P(Y = 0), where Y ~ Poisson (0.104)

= 1 – 0.901225[using Excel Function POISSON(x,Mean, Cumulative)]

= 0.098775 ANSWER

Part (b)

Using Excel Function: BINOMDIST(Number_s:Trials:Probability_s:Cumulative),

Probability that he wins at least once = 1 – Probability he does not win at all

= 1 – 0.901225

= 0.098775, which is identical to the above Poisson approximation. ANSWER

Part (c)

Probability that Joe comes out ahead = Probability that he wins at least twice

= 1 – P(Y = 0, 1), where Y ~ Poisson (0.104)

= 1 – (0.901225 + 0.093727)[using Excel Function POISSON(x,Mean, Cumulative)]

= 0.005047 ANSWER

DONE