Class Activity 3
Probability Problems B –answers
1. Pregnancy Test Results.
Positive test Results(Pregnancy is indicated / Negative test Results
(Pregnancy is not indicated / Total
Subject is pregnant / 80 / 5 / A=85
Subject is not pregnant / 3 / 11 / 14
Total / B= 83 / 16 / S=99
A: subject is pregnant, B: subject test positive, S: sample space
a) Positive test result – If two different subjects are randomly selected, find the probability that they both test positive for pregnancy.Answer:
b) Pregnant. If one of the subject is randomly selected, find the probability of getting someone who test negative () or someone who is not pregnant ()
Answer:
c) Pregnant – If two different subjects are randomly selected, find the probability that they both pregnant.
Answer:
d) Nagative test result – If three different people are randomly selected, find the probability that they all test negative.
Answer:
e) If one of the 99 subjects is randomly selected, find the probability that the person tested was pregnant given that she tested positive.
Answer:
f) If one of the 99 subjects is randomly selected, find the probability that the person tested positive given that she was pregnant.
Answer:
g) Find the probability that the pregnancy test will be in error.
2. Redundancy. Assuming that your alarm clock has 0.975 probability of working on any given morning.
Let F: alarm clock fails to work
a) What is the probability that your alarm clock will not work on the morning of an important exam?
P(F)= P( your alarm clock will not work) = 1 – P(your alarm clock will work)
= 1 – 0.975
= 0.025
b) If you had two such alarm clocks, what is the probability that they both fail on the morning of an important final exam?
c) With one alarm clock , we have a 0.975 probability of being awakened. What is the probability of being awakened if we are using two alarm clocks?
3. In which of the following cases are events A and B independent?
Answer
a) P(A) = 0.3, P(B) = 0.8 and P(A and B) = 0.24 / a)
b) P(A) = 0.42, P(B) = 0.35 and P(A and B) = 0.145 / b)
c) P(A) = 0.8, P(B) = 0.5 and P(A or B) = 0.9 / c)
d)
Event / B / Not B
A / 0.42 / 0.6
Not A
0.3 / 1
/ d)
4. Counting techniques
a) 6! = 6(5)(4)(3)(2)(1) = 720 b) 15! = (15)(14)(13)(12) . . .(3)(2)(1) = 1,307,674,368,000
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