Download 1) Once a Woman won $1 Million in scratch off game from a lottery

1) Once a Woman won $1 Million in scratch off game from a lottery. Some years later, she won $1
million in another scratch off game. In the first game, she beat odds of 1 in 5.2 million to win. In
the second, she beat odds of 1 in 805,600.
a) What is the probibility that an individual would win 1 million in both games if they bought one
scratch off ticket from each game?
b) What is the probability that an individual would win $1 million twice in the second scratch of
game?
Use Scientific Notation. Round to the nearest tenth as needed.
Answers:
a) (1/5200000)(1/805600) = 2.4 x 10-13
b) (1/805600)2 = 1.5 x 10-12
2) A bionomial probability experiment is conducted with the given parameters.
Compuete the probability of x successes in the n independent trials of the experiment.
n=9 p=.6 x greater/equal3 Round to four decimal places as needed.
Answer:
P(x≥3) = 0.9750
3) The number of hits to a website follows a poisson process. Hits occur at the rate of 3.3 per
minute between 7 PM and 11 PM. Given below are three scenarios for numbers of hits to the site.
Compute the probability of each scenario between 7:26 PM and 7:30 PM.
a) Exactly four
b)Fewer than four.
c) at least four.
Answers:
Between 7:26 and 7:30 the number of hits has a poisson distribution(X) with  = 4*3.3 = 13.2
a) P(x=4)= 0.0023
b) P(x<4)= P(x3)= 0.0009
c)P(x≥4)= 1-P(x3)= 0.0091
4) Suppose that a sample of peanut butter contains .3 insect fragments per gram. Compute the
probability that the number of insect fragments in a 4-gram sample of peanut butter is each of the
following values or range of values.
a)exactly four
b)fewer than four
c)at least four
d)at least one
Answers:
the number of insect fragments in a 4-gram sample of peanut butter has a poisson distribution(X)
with  = 4*0.3 = 1.2
a) P(x=4)= 0.0260
b) P(x<4)= P(x3)= 0.9662
c)P(x≥4)= 1-P(x3)= 0.0338
d) P(x≥1)= 1-P(x=0) =0.6988
5) Suppose a life insurance company sells a $270,000 one year term life insurance polict to a 24
year old female for $220. The probability that the female survives the year is .999593. Compute
and interpret the expected value of this policy to the insurance company. The expected value is $ ?
Answer:
The expected value is: 270,000(1-0.999593) = $109.89
On average the company will lose $109.89 per each 24 year old female so the expected earning
per person is: $220 - $109.89 = $110.11
6) The random variable X follows a poisson process with the given value of ^ and t Assuming ^=.14
and t =10 compute the following.
X has a poisson distribution with parameter  = 0.14(10) = 1.4
a) P(5) = P(X=5) = 0.0111
b) P(E<5) = P(X4)=0.9857
c)P(X  5)= 0.9968
d) P(3  X  5) = 0.1633
e) x = x = 1.4