## Elementary Statistics (12th Edition)

a) $0.0001$ b) $P(−1)=0.9999$, $P(4999)=0.0001$ c) $0.0365$ d) $0.0352$ e) $-0.5$ dollars
a) The robability is: $\frac{1}{10^4}=0.0001$ b) Using the value from above, we can see that the probability of losing a dollar is: $P(−1)=0.9999$. Hence the probability of winning $4999$ dollars is: $P(4,999)=1-0.9999=0.0001.$ c) We multiply the probability by $365$: $0.0001\cdot365=0.0365$. d) The probability of winning exactly once is equal to the probability of winning once minus the probability of winning a second time in the 365 days: $0.0365−0.03652=0.0352.$ e) Every $10000$ times, you win $5000$ dollars, but it costs $10000$ dollars to buy the tickets. Hence the expected value is: $\frac{5000−10000}{10000}=−0.5$