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
