Question

Susan is taking Western Civilization this semester on a pass/fail basis. The department teaching the course has a history of passing 85% of the students in Western Civilization each term.
Let n = 1, 2, 3, ... represent the number of times a student takes Western Civilization until the first passing grade is received. (Assume the trials are independent.)
(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 Susan passes on the first try (n = 1)? (Use 2 decimal places.)
(c) What is the probability that Susan first passes on the second try (n = 2)? (Use 3 decimal places.)
(d) What is the probability that Susan needs three or more tries to pass Western Civilization? (Use 3 decimal places.)
(e) What is the expected number of attempts at Western Civilization Susan must make to have her (first) pass? Hint: Use μ for the geometric distribution and round.

Answer 1

Answer:

a. P(n) = 0.85 * (0.15)^(n-1)

b. P(n=1) = 0.85

c. P(n= 2) = 0.1275

d. P(n≥3) = 0.0225

e. Expected number of attempts is 1.176

Step-by-step explanation:

a.

Given

p = success = 85% = 0.85

q = failure = 1 - q = 1 - 0.85 = 0.15

The results of passing/failing takes a Bernoulli distribution

Since, there are independent trials

The number of trials until the first successful event occurs is given by

P(n = k) = p . (1 - p)^(k-1)

P(n = k) = p.q^(k-1)

This is so because it is a Bernoulli distribution and it is modeled by a geometric distribution.

Substitute 0.85 for p

P(n) = 0.85 * (0.15)^(n-1)

b.

Given

n = 1

Using P(n=1) = 0.85 * (0.15)^(n-1)

P(1) = 0.85 * 0.15^(1-1)

P(1) = 0.85 * 0.15°

P(1) = 0.85 * 1

P(1) = 0.85

Therefore, the probability that Susan passes on the first try is 0.85.

c.

n = 2

Using P(n=2) = 0.85 * (0.15)^(2-1)

P(2) = 0.85 * 0.15^(2-1)

P(2) = 0.85 * 0.15¹

P(2) = 0.85 * 0.15

P(2) = 0.1275

Therefore, the probability that Susan passes on the first try is 0.1275

d.

We'll make use of the probability of Susan passing the course after an infinite number of trials is 1.

i.e.

P(n=1) + P(n=2) + P(n=3) + P(n=4) + ......... = 1 --- This is then simplified to

P(n=1) + P(n=2) + P(n≥3) = 1

P(n≥3) = 1 - P(n=1) - P(n=2)

P(n≥3) = 1 - 0.85 - 0.1275

P(n≥3) = 0.0225

Therefore, the probability that Susan needs at least 3 attempts to pass is 0.0225

e.

In (a) above, we explained that the distribution is modeled by an exponential distribution.

The Expected Value for this is inverse of p, where p = 0.85

So, E(n) = 1/p

E(n) = 1/0.85

E(n) = 1.176470588235294

E(n) = 1.176 --- Approximated

Hence the Expected number of attempts is 1.176