# Solved: Bob is a recent law school graduate who intends to take the state bar exam. According to the National Conference on Bar Examiners, about

#### ByDr. Raju Chaudhari

Feb 24, 2021

Bob is a recent law school graduate who intends to take the state bar exam. According to the National Conference on Bar Examiners, about 57% of all people who take the state bar exam pass (Source: The Book of Odds by Shook and Shook, Signet). Let $n = 1, 2, 3,\cdots$ 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.
(b) What is the probability that Bob first passes the bar exam on the second try (n = 2)?
(c) What is the probability that Bob needs three attempts to pass the bar exam?
(d) What is the probability that Bob needs more than three attempts to pass the bar exam?
(e) What is the expected number of attempts at the state bar exam Bob must make for his (first) pass? Hint: Use m for the geometric distribution and round

#### Solution

Let $n$ denote the number of attempts to pass the state bar exam.

The probability that the people pass the state bar exam is $p=0.57$.

(a) The probability distribution of the random variable $N$ is

 \begin{aligned} P(N=n) =\left\{ \begin{array}{ll} (0.57)*(1-0.57)^{n-1}, & \hbox{n=1,2,\ldots} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{aligned}

(b) The probability that Bob first passes the bar exam on the second try (n = 2) is

 \begin{aligned} P(N=2) &= (0.57)*(1-0.57)^{2-1}\\ &= 0.2451 \end{aligned}

(c) The probability that Bob needs three attempts to pass the bar exam is

 \begin{aligned} P(N\leq 3) &= P(N=1) + P(N=2) +P(N=3)\\ &=0.57*(1-0.57)^{1-1}+0.57*(1-0.57)^{2-1}+0.57*(1-0.57)^{3-1}\\ &= 0.920493 \end{aligned}

(d) The probability that Bob needs more than three attempts to pass the bar exam is

 \begin{aligned} P(N> 3) &= 1- P(N\leq 3)\\ &=0.57*(1-0.57)^{1-1}+0.57*(1-0.57)^{2-1}+0.57*(1-0.57)^{3-1}\\ &= 0.079507 \end{aligned}

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

$$E(N) = 1/p = 1/0.57 = 1.754386 \approx 2$$.