Unit 1: Practice Problems
1. A car repair can be performed either on time or late and either satisfactorily or unsatisfactorily.
The probability of a repair being on time and satisfactory is 0.26.
The probability of a repair being on time is 0.74.
The probability of a repair being satisfactory is 0.41.
What is the probability of a repair being late and unsatisfactory?
On Time
0.48
Satisfactory
P(T
S)  0.26
0.15
1. P(T S)  0.26
2. P(T )  0.74 . The entire T circle is 0.74. Since we already have 0.26 in there, the left
over piece must be 0.48
3. P(S)  0.41 . The entire S circle is 0.41. 0.41 – 0.26 = 0.15.
4. Now we have 0.48 + 0.26 + 0.15 = 0.89 accounted for in our circles. That leaves 0.11
outside of the circles (representing events that are not on time and not satisfactory)
2. The length, width, and height of a manufactured part are classified as being either within or outside specified
tolerance limits. In a quality inspection:
86% of the parts were found to be within the tolerance limits for width, but only
80% of the parts are within the specified tolerance limits for all three dimensions.
However, 2% of parts are within the tolerance limits for width and length but not for height,
and 3% of parts are within the tolerance limits for width and height but not length.
Moreover, 92% of parts are within the tolerance limits for either width or height or both width & height.
If a part is within the tolerance limits for height, what is the probability that it will also be within the tolerance limits
for width?
If a part is within the limits for length and width, what is the probability that it will be within the tolerance limits for
all three dimensions?
Length
Width
0.02
0.01
0.80
0.03
0.06
Height
1.
2.
3.
4.
5.
P(L W H )  0.80
P(L W H ')  0.02
P(L ' W H )  0.03
P(W )  0.86 . Since 0.80+0.02+0.03 = 0.85 is already in the W circle, 0.01 is needed.
P(W H)  0.92 . This means we need a total of 0.92 in both the H and W circles.
We already have 0.86 in the W circle, so 0.06 is left for the H circle. We still do not
know exactly how to split this between the two sections of the H circle.
P(W | H ) 
P(W H ) 0.83

 0.933
P(H )
0.89
P(H | L
W)
P(L W H ) 0.80

 0.976
P(L W )
0.82
3. Suppose you get a job inspecting toys imported from China. You decide to sample some toys from each shipment and
test them for lead-based paint. If at least two of the toys you sample test positive for lead-based paint, you will reject
the entire shipment.
Suppose you receive one shipment of 100 toys. Unbeknownst to you, 20 of those toys contain lead-based paint. If
you sample and test 10 toys from this shipment, what’s the probability that you will incorrectly fail to reject shipment.
(In other words, what’s the probability that you will find 0 or 1 toys with lead-based paint in your sample?)
(A hypergeometric approach will work here. You could also count the total number of ways in which you could
sample 10 toys from 100 and the total number of ways you could sample 0 or 1 of the 20 dangerous toys)
The number of ways to select 10 toys from 100:
100  100!
C




100 10
 10  10!90!
100 99 98 97 96 95 94 93 92 91 90!


10 9 8 7 6 5 4 3 2 90!
100 99 98 97 96 95 94 93 92 91


10 9 7 1 6 8 5 2 3 2 1
 10 11 14 97 2 19 47 31 46 91 
 17310309456440
The number of ways to select 10 non-defective toys from 80:
 80 
80!
C10    

10  10!70!
80 79 78 77 76 75 74 73 72 71 70!


10 9 8 7 6 5 4 3 2 70!
80 79 78 77 76 75 74 73 72 71


10 1 6 7 4 5 3 2 1 9 8 1
 8 79 13 11 19 5 37 73 1 71 
80
 1646492110120
The probability of choosing 0 defective toys from a sample of 10:
 80   20 
  
1646492110120
10   0 
P(X  0) 
 0.095 or P(X  0) 
 0.095
100 
17310309456440


 10 
The probability of choosing 1 defective toy from a sample of 10:
 80   20 
  
 9  1 
P(X  1) 
 .2679
100 


 10 
4. Suppose I often come to work late and I need to sneak into my office without any administrators seeing me. I know
that I can take two different routes to my office. If I take route A, I will walk by one administrator who has a 30%
chance of seeing me. If I take route B, I will walk by two administrators who each have a 20% chance of seeing me.
Assume the probabilities that these administrators see me are independent. I also know that since route A is shorter, I
take it 75% of the time and I take route B 25% of the time.
Given all of this information, what’s the probability that an administrator will see me when I come into work late
tomorrow?
(Find the probability of getting through route A; then find the probability of getting through route B. Finally, use the
law of total probability to find the probability of getting through on any randomly selected day)
Admin B2
P = 0.20
Admin A
P = 0.30
Admin B1
P = 0.20
Route A
P = 0.75
Route B
P = 0.25
Probability of being seen on Route A = 0.30; Probability of making it through Route A = 0.70
I will not make it through Route B if (1) Admin 1 sees me, (2) Admin 2 sees me, or (3) both see me.
In other words, I will not make it through Route B if Admin B1 OR B2 see me.
P(B1 B2)  P(B1)  P(B2)  P(B1 B2)  0.2  0.2  0.04  0.36
Since B1 and B2 are independent, P(B1 B2)  P(B1) P(B2)  0.2 * 0.2  0.04
So the probability that I make it through Route B is 1 0.36  0.64
Law of Total Probability:
Let’s define M = (make it through) and A = (Route A)
P(M | A)P(A)  P(M | A')P(A')
We’re interested in
(0.70)(0.75)  (0.64)(0.25)  0.525  0.16  0.685