MAT470, Assorted Probability Questions
…and their solutions!
1.
Three “puppy factories” have purebred dogs ready for delivery to pet stores. The breeds available at each location are given in the following table.
German
Shepherd
Golden
Retriever
Labrador
Retriever
Total:
Location A
Location B
8
5
14
10
10
5
Location C
Total:
7
20
11
35
10
25
Suppose one dog is selected at random. Determine the following probabilities
A). P (selected dog is a Labrador ) = 25/80
B). P (selected dog is from Location B and is a Labrador) = 5/80
C). P ( dog is a Labrador | dog is from Location B ) = 5/20
32
20
28
80
D). P ( dog is from Location B | dog is a Labrador ) = 5/25
E). P( dog is a Shepherd | dog is from Location C ) = 7/28
F). P( dog is from Location C | dog is a Shepherd ) = 7/20
G). P( dog is Labrador or Golden | dog is from Location C ) = (11 + 10)/28 = 21/28
H). P( dog is from Location B or C | dog is a Golden Retriever )
I). P( dog is a Labrador or dog is from Location B )
J). P( dog is a Labrador or Golden Retriever )
= (10 + 11)/35 = 21/35
= (25 + 20 – 5)/80 = 40/80
= (35 + 25)/80 = 60/80
K). P( dog is a Labrador and Golden Retriever )
L). For the pairs of events listed below, identify any pairs that are independent events. i). “dog is Shepherd” and “dog from Location A”
Independent: P(shep)P(from A) = P(shep and from A) = 0.1
= 0, not possible since it’s purebred . ii). “dog is Golden” and “dog from Location B”
Dependent: P(Golden)P(from B) doesn’t equal P(Golden and from B) iii). “dog is Labrador” and “dog from Location A”
Independent: P(Lab)P(from A) = P(Lab and from A) = 0.125
iv). “dog is Labrador” and “dog from Location C”
Dependent: P(Lab)P(from C) doesn’t equal P(Lab and from C)

2.
Suppose there are two boxes containing colored balls. The probability of selecting a particular color of ball is given for each box in the table below.
Blue Red White Totals
Box # 1 0.125 0.150 0.350 0.625
Box # 2 0.075 0.225 0.075 0.375
Totals: 0.200 0.375 0.425 1.0
Assume one ball is selected at random and answer the following questions.
a). P ( the ball is White and came from Box # 1 ) = 0.350
b). P ( ball came from Box # 1 | ball is White ) = 0.350/0.425 = 0.8235
c). For the pairs of events listed below, CIRCLE any pairs which are independent events.
2. "came from Box # 2" and "ball is Blue" are independent because…
P( box #2 and Blue ) = 0.075 and P(box #2)P(Blue) = (0.375)(0.200) = 0.075
3.
As items come to the end of a production line, an inspector chooses which items are to go through a complete inspection. Five percent of items coming off the line are indeed defective. Seventy percent of the defective items get selected for a complete inspection, and
10% of non-defective items get selected for a complete inspection.
(try drawing the tree that represents these events) a). What is the probability an item goes through a complete inspection?
P( I ) = P( I and D ) + P( I and not D ) = 0.13
b). What is the probability at item is not defective, given that is was selected for a complete inspection?
P( not D, given I ) = P( I and not D)/P( I ) = 0.73077
4.
Two events, A, B and C , satisfy the probabilities
( )

P B A

0.5, (

B )

P C A

B )

0.15.
a). Compute the probability (

B )

P B A P A

0 3
and by the addition rule,
( )

(

B )

P A

P A

B )
   
0. 0 b). Determine whether events A and B are independent and justify your answer.
Since we found P(B) = 0.5, we note that P( B | A ) = P( B ) and so they’re independent . c). Compute the probability
( B C ) = ( |

) (

B )

( 0.15 )( 0.30 ) = 0.045
d). Compute the probability (

| ) =
(
 
C )
= 0.045 / 0.6 = 0.075

5.
There are two urns containing balls. Urn A has 3 black balls and 2 white balls. Urn B has 1 black and 4 white. An experiment consists of selecting an urn and randomly selecting one ball from the urn. a). For this experiment, how many sample points are in the sample space.
By the multiplication principle, there are a total of (2)(5) = 10 outcomes from this sequence of decisions. You could even list them
S = {(A, b
1
), (A, b
2
), (A, b
3
), (A, w
1
), (A, w
2
), (B, b
4
), (B, w
3
), (B, w
4
), (B, w
5
), (B, w
6
)} or illustrate it using a tree diagram.
b). If the urn is chosen at random, so that all outcomes are equally likely, find the probability that a black ball is selected.
Of the 10 possible outcomes, 4 involve selecting a black ball, ( )

4 /10

0.40
.
More formally, we could write
( )

(
 black )

(
 black )

1
2
3
5

1 1
     

4 /10

0.40
c). If the urn is not chosen at random but Urn A is twice as likely to be chosen, then what is the probability the experiment results in a black ball being drawn from
Urn A?
Given that A is twice as likely, we know P( urn A ) = 2/3 and P( urn B ) = 1/3.
(
 black )

( | ) ( )

3 2

2 / 5

0.40
d). If the urn is not chosen at random but Urn A is twice as likely to be chosen, then what is the probability that the ball was selected from Urn A, given that a black ball was chosen?
| ) ( )
| )

)

0.4
0.8571
2 3
3 5

1
3
1
5
7