Bob is a recent law school graduate who intends to take the state bar exam. Acco
ID: 3368684 • Letter: B
Question
Bob is a recent law school graduate who intends to take the state bar exam. According to the National Conference on Bar Examiners, about 45% of all people who take the state bar exam pass. Let n = 1, 2, 3, ... represent the number of times a person takes the bar exam until the first pass.
(a) Write out a formula for the probability distribution of the random variable n. (Use p and n in your answer.)
P(n) =
(b) What is the probability that Bob first passes the bar exam on the second try (n = 2)? (Use 3 decimal places.)
(c) What is the probability that Bob needs three attempts to pass the bar exam? (Use 3 decimal places.)
(d) What is the probability that Bob needs more than three attempts to pass the bar exam? (Use 3 decimal places.)
(e) What is the expected number of attempts at the state bar exam Bob must make for his (first) pass? Hint: Use ? for the geometric distribution and round.
Question 2:
Approximately 11.4% of all (untreated) Jonathan apples had bitter pit in a study conducted by the botanists Ratkowsky and Martin. (Bitter pit is a disease of apples resulting in a soggy ore, which can be caused either by overwatering the apple tree or by a calcium deficiency in the soil.) Let n be a random variable that represents the first Jonathan apple chosen at random that has bitter pit.
(a) Write out a formula for the probability distribution of the random variable n. (Use p and n in your answer.)
P(n) =
(b) Find the probabilities that n = 3, n = 5, and n = 12. (Use 3 decimal places.)
(c) Find the probability that n ? 5. (Use 3 decimal places.)
(d) What is the expected number of apples that must be examined to find the first one with bitter pit? Hint: Use ? for the geometric distribution and roun
Explanation / Answer
a)P(n) = (0.55)n-1*0.45
b)
P(n=2)= (0.55)*0.45=0.248
c)
P(n=3)=(0.55)2*0.45=0.136
d)
P(>3)=(0.55)3 =0.166
e)
? =1/p=1/0.45 =2.222
2)
P(n) = (0.55)n-1*0.45
b)
P(3)=(1-0.114)2*0.114 =0.089
P(5)=(1-0.114)4*0.114 =0.070
P(12)=(1-0.114)11*0.114 =0.030
c)P(n>=5)= (1-0.114)4 =0.616
d)
expected number =1/p=1/0.114 =8.772