Class Activity 3
Probability Problems B –answers
1. Pregnancy Test Results.
Positive test Results
(Pregnancy is indicated
Subject is pregnant
Negative test Results
(Pregnancy is not indicated
Total
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.
83 82
Answer: P(B1 and B 2 )  P(B1 )P(B 2 | B1 ) 

 0.7015
99 98
b) Pregnant. If one of the subject is randomly selected, find the probability of getting someone
who test negative ( B ) or someone who is not pregnant ( A )
16 14 11 19
Answer: P( B or A)  P( B )  P( A)  P( B and A) 



 0.1919
99 99 99 99
c) Pregnant – If two different subjects are randomly selected, find the probability that they both
pregnant.
85 84
Answer: P( A1 and A 2 )  P( A1 )P( A 2 | A1 ) 

 0.7359
99 98
d) Nagative test result – If three different people are randomly selected, find the probability that
they all test negative.
P( B1 and B 2 and B3 )  P( B1 )P( B 2 | B1 )P( B3 | B1 and B 2 )
Answer:
16 15 14

 
 0.00357
99 98 97
e) If one of the 99 subjects is randomly selected, find the probability that the person tested was
pregnant given that she tested positive.
80
Answer: P( A | B) 
 0.96386
83
f) If one of the 99 subjects is randomly selected, find the probability that the person tested
positive given that she was pregnant.
80
Answer: P(B | A) 
 0.94118
85
g) Find the probability that the pregnancy test will be in error.
3
5
8
P( Positive and not pregnant )  P(Pr egnant and negative test) 


 0.08  8%
99 99 99
80
h ) Which of the following i ) PPositive test result | subject is pregnant  
85
80
ii ) Psubject is pregnant | Positive test result  
measures the efficiency of the test ?
83
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?
P(F1 and F2 )  P(F1 ) P( F2 )  (0.025)(0.025)  0.000625
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?
P( F1 or F2 )  P( F1 )  P( F2 )  P( F1 and F2 )
 0.975  0.0925  (0.975)  (0.0925)
 0.999375
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)
d)
Event
A
Not A
B
0.42
Not B
0.6
0.3
1
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
25! 25(24)( 23! )
100!
c) 25 P2 

 24(23)  552
c) 100 P3 
 100(99)(98)  970200
23!
23!
97!
c) 25 C 2 
25! 25( 24)( 23! ) 24( 23) 552



 276
23!
2! 23!
2
2
c) 100 C 3 
100! 100(99)(98)

 161700
3! 97!
3( 2)(1)