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Stat 20 Study Group
Faculty: Professor Hank Ibser
Study Group Leader: Larry Wang, [email protected]
Location: MW 1-2 201A Chavez , http://www.csrjjsmp.com/stat20.html
Community through Academics and Leadership
Worksheet #9: Probability
1. Consider two events, A and B. Suppose P(A) = 99% and P(B) = 44%.
a. What is the chance that either A or B occurs?
At least 99% and at most 100%
b. What is the chance that they both occur?
At least 43% and at least 44%
c. How do your answers change if you know the events are independent?
We can calculate them exactly: P(A and B)=(.99)(.44) and P(A or B)=.99+.44-(.99)(.44)
2. In a class of 30, there are 18 male students and 12 female students. 10 each have A’s, B’s, and C’s.
A student’s grade and sex are independent.
a. How many female students have B’s?
4
b. How many students are female or have B’s?
12+10-4=18
c. Choose 2 students at random from the class without replacement. What is the probability
that they are both male and have different grades?
6*(18/30)(17/30)*(6/18)(6/17)= (18/30)(17/30)-3 (18/30)(17/30)*(6/18)(5/17)
d. Choose 10 students at random from the class without replacement. What is the probability
that they all have B’s? How can we write this concisely?
(10/30)*(9/29)*…*(1/21)=(10!20!/30!)
e. Choose 10 students at random from the class, with replacement. What is the probability
that 5 are male and 5 are female?
(18/30)(17/29)(16/28)(15/27)(14/26)*(12/25)(11/24)(10/23)(9/22)(8/21)(10!/(5!5!))
f. Choose 10 students at random from the class, with replacement. What is the probability
that 3 have A’s, 3 have B’s, and 4 have C’s?
(10/30)(9/29)(8/28)(10/27)(9/26)(8/25)(10/24)(9/23)(8/22)(7/21)(10!/(3!3!4!))
3. You flip four fair independent coins, labeled 1 through 4. Find the probability that:
Hint: List out the possible numbers of heads/tails that might come up, and find those
probabilities
a. You get more heads than tails.
(1/2)4+4(1/2)4
b. You get at least one head
1-(1/2)4
c. You get at least one head and at least one tail.
1-2(1/2)4
d. You get more heads than tails and (you get at least one head and at least one tail).
4(1/2)4
e. You get more heads than tails or (you get at least one head and at least one tail).
1-2(1/2)4+(1/2)4+4(1/2)4-4(1/2)4=1-(1/2)4=P(not all tails)
f. You get more heads than tails given (you get at least one head and at least one tail).
2/7
g. You get 1 head in the first two flips and you get 1 head in the last two flips
Stat 20 Study Group
Faculty: Professor Hank Ibser
Study Group Leader: Larry Wang, [email protected]
Location: MW 1-2 201A Chavez , http://www.csrjjsmp.com/stat20.html
Community through Academics and Leadership
(1/2)(1/2)
4. Gauss and Euler are playing poker. 5 cards are dealt to each player. A “Full House” is a specific
poker hand that consists of a “two-of-a-kind” and a “three-of-a-kind”. One example would be
KKKQQ, in any order. A “Two Pair” consists of 2 different pairs and a fifth non-matching card.
An example of this would be “AAQQ3”, also in any order.
Most of these are well-documented online.
a. What is the probability that Gauss gets a full house with 444JJ in that order?
(47!/52!)(4*3*2*4*3)
b. What is the probability that Gauss gets a full house with 444JJ in any order?
c. What is the probability that Euler gets a full house with JJJ44 in any order?
d. What is the probability that Euler gets a full house by first drawing 3 Jacks, then 2 of some
other card?
e. What is the probability that Euler gets a full house by first drawing 3 of any card, then 2 of
some other card?
f. What is the probability that Euler gets a full house?
g. What is the probability that Euler gets a full house with JJJ44 if Gauss has…
i. Three 4s and two Jacks?
0
ii. Three Queens and two 3s?
(5!/(2!3!))(4/47)(3/46)(2/45)(4/44)(3/43)
iii. Three Queens and two 4s?
(5!/(2!3!))(4/47)(3/46)(2/45)(2/44)(1/43)
h. What is the probability that both players get full houses?
(3,744/2,598,960)((11*4*61)/(47!/(5!42!)))
i. What is the probability that Gauss has two Jacks, two Queens, and a 7 in any order?
j. What is the probability that Gauss has two Jacks, two Queens, and any other card?
k. What is the probability that Gauss gets a two pair?
5. You are playing a lottery where each player chooses a number from 0-99999, and 5 numbers 0-9
are drawn. If any rearrangement of drawn numbers matches a player’s number, he wins. What is
the probability that a player will win if he chooses the number:
a. 99999?
c. 94704?
(1/10)5
(5!/2!) (1/10)5
b. 12345?
d. 66778?
5! (1/10)5
(5!/(2!2!)) (1/10)5
e. What kind of number should you choose to maximize your chance to win?
All digits different
6. Consider the sentence "the simians are jubilant." Neglecting the period at the end, but including the
spaces, there are 24 characters in this sentence. Suppose we put 1,000 monkeys in front of special
typewriters that have only lowercase letters and a spacebar—no numbers, punctuation marks, or
special characters, so there are 27 keys in all. Every time a monkey types 24 characters, we change
the paper in the typewriter. Assume that monkeys type independently of each other, and that they
pick which character to type independently each time and with equal probability of striking each of
Stat 20 Study Group
Faculty: Professor Hank Ibser
Study Group Leader: Larry Wang, [email protected]
Location: MW 1-2 201A Chavez , http://www.csrjjsmp.com/stat20.html
Community through Academics and Leadership
the 27 keys. Each monkey types 24 characters per minute for 8 hours a day, 365.25 days a year, for
20 years. Assume that it does not take any time to change the paper in the typewriter. What is the
chance that at least one of the monkeys types the sentence? You can use the approximation 1 −
(1−x)n is approximately n×x. (This is valid for very small x and essentially means we ignore the
probability the sentence is typed multiple times)
There are n=3,506,400,000 tries for all the monkeys. 1 − (1 − (1/27)24)3,506,400,000≈
3,506,400,000×(1/27)24 = 1.556e-25.
7. Cards are dealt without replacement from a standard deck of cards until a heart is dealt.
What must the sequence of cards look like?
a. What is the longest you might have to wait?
It might take as many as 40 cards
b. What is the probability that exactly 5 cards are required?
(39/52)(38/51)(37/50)(36/49)(13/48)
c. What is the probability that 5 or fewer cards are required?
1-(39/52)(38/51)(37/50)(36/49)(35/48)
d. Given that exactly 5 cards are required, what is the probability that 3 spades were dealt?
4(13/39)(12/38)(11/37)(26/36)
8. Keith sets up a maze for his mouse Ralph. Whenever Ralph reaches an intersection, he randomly
chooses a path to go down other than the one he just came out of. If he reaches a dead end, he turns
back the way he came. Given each maze, what is the probability Ralph reaches the food? What is
the probability he comes back out the entrance?
P(food)= (1/2)+(1/2)3+ (1/2)5+ (1/2)7+…=2/3
P(start)= (1/2)2+(1/2)4+ (1/2)6+ (1/2)8+…=1/3
P(food)=.6
P(start)=.4