Suppose you wanted to evaluate the performance of the three judges in Smallville, Texas: Judge Adams, Judge Brown, and Judge Carter. Over a three-year period in Smallville, Judge Adams saw 27% of the cases, Judge Brown saw 31% of the cases, and Judge Carter saw the remainder of the cases. 5% of Judge Adams' cases were appealed, 7% of Judge Brown's cases were appealed, and 9% of Judge Carter's cases were appealed. (See the case problem on pages 216-218 of your textbook for a similar problem.) Given a case from this three-year period was not appealed, what is the probability the judge in the case was not Judge Brown? Question 1 Of 46 bank accounts at a small bank, 25 accounts have values of less than $1,000 and the rest have values of at least $1,000. Suppose 4 accounts are randomly sampled (See exercise 9 on page 184 of your textbook for a similar problem.) What is the probability that exactly 3 of the 4 accounts have values of at least $1,000?

Answer 1: Given that the case from this three-year period was not appealed, the probability that the judge was not Judge Brown is 0.689.

Answer 2: Probability that exactly 3 of the 4 accounts have values of at least $1000 is 0.2068.

Step-by-step explanation:

Answer 1:

Probabilities that each of the judges saw the cases are:

P(Judge Adams) = 0.27

P(Judge Brown) = 0.31

P(Judge Carter) = 1 - (0.27+ 0.31)

P(Judge Carter) = 0.42

Probabilities that the cases of each of the judges were appealed(A) can be represented as a conditional probability:

P(A | Judge Adams) = 0.05

P(A | Judge Brown) = 0.07

P(A | Judge Carter) = 0.09

Hence we can find the probabilities that the cases of each of the judges were not appealed:

P(Not A | Judge Adams) = 1 - 0.05 = 0.95

P(Not A | Judge Brown) = 1 - 0.07 = 0.93

P(Not A | Judge Carter) = 1 - 0.09 = 0.91

We need to find the probability P(Judge Brown | Not A). For this we will use the Baye's Theorem:

P(Ai|B) = [P(Ai)P(B|Ai)]/[P(A1)P(B|A1) + P(A2)P(B|A2) + P(A3)P(B|A3)]

P(Judge Brown | Not A) = [P(Judge Brown)*P(Not A | Judge Brown)]/[P(Judge Adams)*P(Not A | Judge Adams) + P(Judge Brown)*P(Not A | Judge Brown) + P(Judge Carter)*P(Not A | Judge Carter)]

P(Judge Brown | Not A) = [0.31*0.93]/[(0.27*0.95) + (0.31*0.93) + (0.42*0.91)]

                                       = 0.2883/(0.2565 + 0.2883 + 0.3822)

                                       = 0.2883/0.927

P(Judge Brown | Not A) = 0.311

P(Not Judge Brown | Not A) = 1 - 0.311 = 0.689

Given a case from this three-year period was not appealed, the probability the judge in the case was not Judge Brown is 0.689

Answer 2:

We will use the binomial distribution formula to find out the probability that exactly 3 of the 4 accounts have values of at least $1000. The binomial distribution formula is:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

Where n = no. of trials

           x = no. of successful trials

           p = probability of success

           q = probability of failure = 1-p

We have,

n = 46

x = 4

p = (46-25)/46 = 21/46

q = 25/46

P(X=3) = ⁴C₃ (21/46)³ (25/46)⁴⁻³

           = 4 * (21/46)³ (25/46)¹

P(X=3) = 0.2068

step-by-step explanation:

where is the questions

translate k to n and reflect across the line containing jk.