#### Answer

a) $0.0001$
b) $P(−1)=0.9999$, $P(4999)=0.0001$
c) $0.0365$
d) $0.0352$
e) $-0.5$ dollars

#### Work Step by Step

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:
$\frac{5000−10000}{10000}=−0.5$