The numbers racket is a well‑entrenched illegal gambling operation in most large cities. In one version, you choose one of the 1000 three‑digit numbers 000 to 999 and pay your local numbers runner a dollar to enter your bet. Each day, one three‑digit number is chosen at random and pays off $ 600 $600 . The mean payoff for the population of thousands of bets is μ = 60 μ=60 cents. Joe makes one bet every day for many years.

p(w) = p = 1/1000 = 0.001

q = 1-p = 0.999

reward for winning = 600$

payoff for attempt = 0.6$

a.

on an avg. jose loose 0.4$ per hand,

so equation will be to find out one win per x hands,

-0.4 x = -0.6x + 600

so x = 3000.

Hence 1 win out of 3000,

P(losing 0.4$) = \binom{3000}{1}(0.001)^1(0.999)^{2999}

P(losing 0.4$) = 0.1492

b.

Probability of winning 0.4$ per winning

equation,

0.4 x = -0.6x + 600

so ,

x = 600

so out of 600 game he have to win 1 game.

P(winning 0.4$) = \binom{600}{1}(0.001)^1(0.999)^{599}

P(winning 0.4$) = 0.3295

c.

Probability of winning 0.6$ per winning

equation,

0.6 x = -0.6x + 600

so ,

x = 500

so out of 500 game he have to win 1 game.

P(winning 0.6$) = \binom{500}{1}(0.001)^1(0.999)^{499}

P(winning 0.6$) = 0.3034

Attached is the same solutions incase the above isn't understandable.

yes                                            

explanation:

just cause                            

i am pretty sure it is d.