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ECSE-2500 Engineering Probability HW#3 Solutions
Due 9/8/14
1. (8 points, 4 points each) As in class, we pick a number  1 at random from the interval [0,1]. We
then independently pick a 2nd number  2 at random from the interval [0,1].
1.a. Define the set A in the sample space which represents the event that two times the first number
drawn is less than the 2nd number drawn. Then draw a picture of the set A in the sample space,
and compute its probability.
Solution
ζ2
1
A    1,  2   S 2 1   2 
P  A  area of the triangle  12  12  1= 14 .
0
S
0
½
1 ζ1
1.b. Define the set A in the sample space which represents the event that the pair  1 ,  2  is inside
the unit circle. Then draw a picture of the set A in the sample space, and compute its
probability.
Solution
ζ2
1
P  A  area of the quarter circle of radius one
A
2
 14  1 = 4 .
0
0
½
S
1 ζ1
Page 1 of 3
ECSE-2500 Engineering Probability HW#3 Solutions
Due 9/8/14
2. (8 points) Recall Problems 1.a and 2.a from HW#1. (There are three balls in an urn numbered
1,2,3. You randomly select one ball from the urn.) You found that the smallest sample space for
this problem is S  1,2,3. Then you showed that the set of all possible events for this sample
space is 1,2,3,1,2,1,3,2,3,1, 2, 3, . Then in HW#2 you showed that, assuming
that the 3 balls were equally likely, the probability of these sets is P   0,
P 1  P 2  P 3  13 , P 1,2  P 1,3  P 2,3  32 , and P 1,2,3  1.
Now, for each of these sets, compute the conditional probability of that set given B  1,3. Use
the definition of conditional probability and show your work.
Solution
P  B 
P  1,3 P  0

  0,
2
2
P 1,3
3
3
P 1 B 
P 1 1,3 P 1


2
P 1,3
3
1
2
3
3
1
 ,
2
P 2 B 
P 2 1,3 P  0

  0,
2
2
P 1,3
3
3
P 3 B 
P 3 1,3 P 3


2
P 1,3
3
P 1,2 B 
P 1,3 B 
P 2,3 B 
1
2
3
3
1
 ,
2
P 1,2 1,3 P 1


2
P 1,3
3
1
2
3
3
P 1,3 1,3 P 1,3


2
P 1,3
3
P 2,3 1,3 P 3


2
P 1,3
3
P 1,2,3 B 
1
2
1
 ,
2
2
2
3
3
3
 1,
3
1
 ,
2
P 1,2,3 1,3 P 1,3


2
P 1,3
3
2
2
3
 1,
3
Page 2 of 3
ECSE-2500 Engineering Probability HW#3 Solutions
Due 9/8/14
3. (8 points, 4 points each) Consider a standard 52-card deck of cards. If you are unfamiliar with
this concept, there is a Wikipedia page: http://en.wikipedia.org/wiki/Standard_52-card_deck
3.a. You randomly select 7 cards from this deck. What is the probability your 7 cards do not
include any aces? Clearly show your work.
Solution
 
 
52
48
There are 7 distinct possible hands of 7 cards. Since there are 4 aces, there are 7
distinct possible hands of 7 cards with no aces. Since all hands are equally likely, the
probability of a 7 card hand with no aces is
48!
48


7
  48!  45!  45  44  43  42
P no aces 
 7! 41! 

 0.5504
52!
52
52!  41!  52  51  50  49
 7  7! 45!
3.b. Five cards are dealt face up. How many distinct hands could be dealt which form a “straight”
dealt in order? A “straight” is 5 cards that are consecutive in the ordering of cards. Here, suits
do not matter. Note that an ace can be the lowest card or the highest card. So the lowest
straight would consist of any ace being dealt first, then any two, then any three, then any four,
then any five. The highest straight would be any 10, then any jack, then any queen, then any
king, then any ace. Here we are not counting all possible 5-card straights, only those dealt in
increasing order. Clearly show your work.
Solution
The first card cannot be a “picture card” i.e. a jack, queen or king, since there could not be 4
consecutively higher cards to follow. So the first card can be any of the 4 aces or any of the 4
twos or … or any of the 4 tens. That is there are 10 times 4 choices for the first card, so
there are 40 possible first cards. Once that card has been dealt, there are exactly 4 choices for
the next card, and four for each of the subsequent 3 cards. So the total combinations is
40  4  4  4  4  10  45  40  256  10,240.
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