Question

Cathy is planning to take the Certified Public Accountant Examination (CPA exam). Records kept by the college of business from which she graduated indicate that 89% of the students who graduated pass the CPA exam. Assume that the exam is changed each time it is given. Let n = 1, 2, 3, ... represent the number of times a person takes the CPA exam until the first pass. (Assume the trials are independent.)

(a) What is the probability that Cathy passes the CPA exam on the first try? (Use 2 decimal places.)
P(1)

(b) What is the probability that Cathy passes the CPA exam on the second or third try? (Use 4 decimal places.)
P(2 or 3)

Answer 1

Answer:

a) $P(X=1)=(1-0.89)^{1-1} 0.89 = 0.89$

b) $P(X=2)=(1-0.89)^{2-1} 0.89 = 0.098$

$P(X=3)=(1-0.89)^{3-1} 0.89 = 0.0108$

And adding we got:

$P(2 \leq X\leq 3)=P(X=2)+P(X=3)= 0.098+0.0108= 0.1088$

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

Part a

For this case we want this probability

$P(X=1)=(1-0.89)^{1-1} 0.89 = 0.89$

Part b

For this case we want this probability:

$P(2 \leq X\leq 3)=P(X=2)+P(X=3)$

If we find the individual probabilities we got:

$P(X=2)=(1-0.89)^{2-1} 0.89 = 0.098$

$P(X=3)=(1-0.89)^{3-1} 0.89 = 0.0108$

And adding we got:

$P(2 \leq X\leq 3)=P(X=2)+P(X=3)= 0.098+0.0108= 0.1088$