CSE 312 Homework 2, Due Wednesday, April 13, before class
Instructions:
• When asked for a short answer (such as a single number), also show and
... [Show More] explain your work briefly.
Unless you are asked to, leave your answer in terms of factorials, combinations, etc., for instance 267
or 26!/7! or 26 ·
26
7
.
• Please write down on your homework the names of all people you discussed the homework with.
• Turn-in procedure:
– You will turn in your homework on paper, in 3 separate bundles - you may hand-write your
homework legibly, or type it and print it out. If we can’t read your homework, we won’t grade it.
– Each bundle needs to have your name at the top, and the bundle letter A, B, or C.
– You need to staple the papers in each bundle - homework that is not stapled is not accepted -
turn in your favorite sheet for that bundle if you have multiple unstapled sheets.
This weeks bundles: (A) problems 1-2; (B) problems 3-6; (C) problems 7-8
Problems
A.1 For each of the following scenarios first answer the following two questions and then answer the question
stated. (a) What is the sample space and how big is it? (b) What is the probability of each outcome
in the sample space?
(a) You flip a fair coin 100 times. What is the probability that all 100 tosses are the same? What is
the probability of exactly 50 heads?
S = {(X1, X2, ...X100) : Xi ∈ {H, T}} and Xi represents the outcome of the i
th flip. Since
there are two possible outcomes for each flip, |S| = 2100. Each outcome occurs with probability
1
|S| =
1
2
100 . Since the cases where all 100 tosses are the same are when they are all heads or all
tails, the probability is: 2
2
100 . Since there are
100
50
ways for exactly 50 heads to appear in the
tosses, the probability that there are exactly 50 heads is: (
100
50 )
2
100 .
(b) You roll 2 fair dice. What is the probability that the sum of the two values showing is 5?
S = {(X, Y ) : X, Y ∈ {1, 2, 3, 4, 5, 6}}. X represents the outcome of the first die and Y represents
the outcome of the second die. Since there are 6 options for X and 6 options for Y , |S| = 62
.
Each outcome is equally likely so the probability of each outcome is 1
6
2 . There are 4 instances
when the sum of the values equal 5: (1, 4),(2, 3),(3, 2),(4, 1), so the probability is 4
6
2 .
(c) You are given a random 5 card poker hand (selected from a single deck). What is the probability
you have a flush (all cards have the same suit)?
The sample space is the set of all (ordered) five-card selections. Xi
is the i
th card in the hand.
S = {(X1, X2, X3, X4, X5) : Xi ∈ {52 possible cards}, Xi 6= Xj when i 6= j}. There are 52, 51,
50, 49, and 48 possible cards for the first, second, third, fourth, and fifth draws, respectively, so
|S| = 52×51×50×49×48. Since each outcome is equally likely, the probability of each outcome
is 1
52×51×50×49×48 . For any suit, there are 13×12×11×10×9 ways to get 5 cards from that suit.
Since there are 4 suits, there are 4 × 13 × 12 × 11 × 10 × 9 ways to draw a flush. The probability
of a flush is then: 4×13×12×11×10×9
52×51×50×49×48 .
(d) 20 labeled balls are placed into 10 labeled bins (with each placement equally likely). What is the
probability that bin 1 contains at least one ball?
Xi denotes the bin number that the i
th ball is placed in. S = {(X1, X2, ..., X20) : Xi ∈ Z, 0 ≤
Xi ≤ 10}. Since there are 10 options for each of the 20 balls, |S| = 1020. Each outcome has
probability 1
1020 . In order for bin 1 to have at least one ball, we look at |S| − |E| where E is t [Show Less]