1. Problem 2-26 - Northside Rifle Team
Jan White
BUS 340
October 31, 2006
Slide #1

2. Dick hits the bull’s-eye 90% of the time
Dick and Sally are markspersons on the Northside Rifle Team – their statistics are:
Dick hits the bull’s-eye 90% of the time
Sally hits the bull’s-eye 95% of the time
Jan White
BUS340
Slide #2

3. Dick = Event A, Sally = Event B P(A) + P(B) - P(A and B) = P(A or B)
Part A. What is the probability that either Dick or Sally or both will hit the bull’s-eye if each takes one shot?
Dick = Event A, Sally = Event B
P(A) + P(B) - P(A and B) = P(A or B)
Not mutually exclusive (there are other outcomes)
= .995, or 99.5% of the time either Dick or Sally or both will hit the bull’s-eye.
Can also be shown in terms of “misses”, as
1 – {P(NA) x P(NB)} = 1 – (.1 x .05) =
= .995, or 99.5% chance of not missing
Jan White
BUS340
Slide #3

4. Part B. What is the probability that Dick and Sally will both hit the bull’s-eye?
P(A) x P(B) = .9 x .95 = .855, or 85.5% of the time Dick and Sally will both hit the bull’s-eye
Joint probability of multiple, independent events occurring at the same time.
Jan White
BUS 340
Slide #4

5. Yes, I had the assumption that the events were completely independent.
Part C. Did I have any assumptions in answering the questions in parts A and B? If so, do I think I was justified?
Yes, I had the assumption that the events were completely independent.
My justification is that Dick’s marksmanship would not affect Sally’s, and vice-versa, as they each had a record of performance.
Jan White
BUS 340
Slide #5