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 55% 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 μ for the geometric distribution and round.

Answer 1

Solution:

According to geometric distribution, probability that kth trial out of k independent trials is the first trial of success is given by,

P(X = k) = (1 - p)^{k - 1} p

Where, p is probability of success which remains constant in each trials.

a) Hence, the probability distribution of the random variable n is given by,

P(n) = (1 - p)^{n - 1} p ; n = 1, 2, 3, .................

b) Let us consider that "a person who take the state bar exam pass" as success.

Given that, about 55% of all people who take the state bar exam pass.

Hence, probability of success (p) = 0.55

Hence, the probability that Bob first passes the bar exam on the second try is,

On rounding to 3 decimal places we get, P(n = 2) = 0.247

The probability that Bob first passes the bar exam on the second try is 0.247.

c) Probability of success (p) = 0.55

Hence, the probability that Bob needs three attempts to pass the bar exam is given by,

On rounding to 3 decimal places we get, P(n = 3) = 0.111

The probability that Bob needs three attempts to pass the bar exam is 0.111.

d) Probability of success (p) = 0.55

Hence, the probability that Bob needs more than three attempts to pass the bar exam is given by,

On rounding to 3 decimal places we get, P(n > 3) = 0.091

The probability that Bob needs more than three attempts to pass the bar exam is 0.091.

e) The expected value of geometric mean having is given by,

μ = 1/p

Where, p is probability of success.

We have, probability of success (p) = 0.55

Hence, μ = 1/0.55 = 1.8182

On rounding to three decimal places we get, μ = 1.818

The expected number of attempts at the state bar exam Bob must make for his (first) pass is 1.818.

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