Transcript Sushi Roulette - Math For Life

Sushi Roulette
• In the Japanese game show Sushi Roulette, the
contestant spins a large wheel that’s divided
into 12 equal sections. Nine of the sections
have a sushi roll, and three have a “wasabi
bomb.” When the wheel stops, the contestant
must eat whatever food is on that section. To
win the game, the contestant must eat one
wasabi bomb. Find the probability that it
takes 3 or more spins for the contestant to get
a wasabi bomb. Show your method clearly.
Knee Surgery
• Patients receiving artificial knees often
experience pain after surgery. The pain is
measured on a subjective scale with possible
values of 1(low) to 5 (high). Let X be the pain
score for a randomly selected patient. The
following table gives part of the probability
distribution for X.
Value
Probability
1
2
3
4
5
0.1
0.2
0.3
0.3
??
a) Find P(X = 5)
b) If two patients who received artificial knees
are chosen at random, what’s the probability
that both of them report pain scores of 1 or
2?
c) Compute the mean and standard deviation of
X.
Keno
• In a game of 4-spot Keno, the player picks 4 numbers
from 1 to 80. The casino randomly selects 20
winning numbers from 1 to 80. The table below
shows the possible outcomes of the game and their
probabilities, along with the amount of money
(Payout) that the player wins for a $1 bet. If X = the
payout for a single $1 bet, you can check that 𝜇𝑋 =
$0.70 and 𝜎𝑋 = $6.58.
Matches
Probability
Payout
0
1
2
3
4
0.308
0.433
0.213
0.043
0.003
$0
$0
$1
$3
$120
a) Interpret the values of 𝜇𝑋 and 𝜎𝑋 .
b) Jerry places a single $5 bet on 4-spot Keno. Find
the expected value and the standard deviation
of his winnings.
c) Marla plays five games of 4-spot Keno, betting
$1 each time. Find the expected value and the
standard deviation of her total winnings.
d) Based on your answers to (b) and (c), which
player would the casino prefer? Justify your
answer.
Winner, Winner, Chicken Dinner
• As a special promotion for its 20 –ounce bottles
of soda, a soft drink company printed a message
on the inside of each cap. Some of the caps said,
“Please try again,” while others said, “You’re a
winner!” The company advertised the promotion
with the slogan “1 in 6 wins a prize.” Suppose the
company is telling the truth and that every 20ounce bottle of soda it fills has a 1-in-6 chance of
being a winner. Seven friends each buy one 20ounce bottle of the soda at a local convenience
store. Let X = the number who win a prize.
a) Explain why X is a binomial random variable.
b) Find the mean and standard deviation of X.
Interpret each value in context.
c) The store clerk is surprised when three of the
friends win a prize. Is this group of friends
just lucky, or is the company’s 1-in-6 claim
inaccurate? Compute P(X > 3) and use the
result to justify your answer.
Cranky Mower
• To start her old mower, Rita has to pull a cord
and hope for some luck. On any particular
pull, the mower has a 20% chance of starting.
a) Find the probability that it takes her exactly 3
pulls to start the mower.
b) Find the probability that it takes her more
than 10 pulls to start the mower.
12,000 Miles & Running
• A study of 12,000 able-bodied male students
at the University of Illinois found that their
times for the mile run were approximately
Normal with mean 7.11 minutes and standard
deviation 0.74 minutes. Choose a student at
random from this group and call his time for
the mile Y. Find P(Y < 6) and interpret the
result.
Too much juice?
• A company has developed a drug test to
detect steroid use by athletes. The test is
accurate 95% of the time when an athlete has
taken steroids. It is 97% accurate when an
athlete hasn’t taken steroids. Suppose that
the drug test will be used in a population of
athletes in which 10% have actually taken
steroids. Let’s choose an athlete at random
and administer the drug test.
a) Make a tree diagram showing the sample
space of this chance process.
b) What’s the probability that the randomly
selected athlete tests positive?
c) Suppose that the chosen athlete tests
positive. What’s the probability that he or
she actually used steroids?
Johnny’s Pizza
• You work at Johnny’s pizza shop. You have the
following information about the 7 pizzas in the
oven: 3 of the 7 have thick crust, and of these
1 has only sausage and 2 have only
mushrooms. The remaining 4 pizzas have
regular crust, and of these 2 have only
sausage and 2 have only mushrooms. Choose
a pizza at random from the oven.
a) Are the events {getting a thick-crust pizza)
and {getting a pizza with mushrooms}
independent? Explain.
b) You add an eighth pizza to the oven. This
pizza has thick crust with only cheese. Now
are the events {getting a thick-crust pizza)
and {getting a pizza with mushrooms}
independent? Explain.
Too cool for the cabin?
• During the winter months, the temperatures at the
Young’s Colorado cabin can stay well below freezing
(32℉ or 0℃) for weeks at a time. To prevent the pipes
from freezing, Mrs. Young sets the thermostat at 50 ℉.
She also buys a digital thermometer that records the
indoor temperature each night at midnight.
Unfortunately, the thermometer is programmed to
measure the temperature in degrees Celsius. Based on
several years’ worth of data, the temperature T in the
cabin at midnight on a randomly selected night follows
a Normal distribution with mean 8.5 ℃ and standard
deviation 2.25 ℃.
a) Let Y = the temperature in the cabin at
midnight on a randomly selected night in
degrees Fahrenheit (recall that F = 9/5 C +32).
Find the mean and standard deviation of Y.
b) Find the probability that the midnight
temperature in the cabin is below 40 ℉.
Solutions
• Sushi Roulette
– Success = landing on wasabi bomb
– 2 outcomes, independent trials, probability of success
is the same from spin to spin (p = .25)
– We do not know the number of spins the contestant
will take, making this variable a geometric random
variable
– Let X = # of spins until contestant lands on wasabi
bomb
– P(X > 3) = P(1st 2 spins are sushi rolls)
= (1 - .25)2 = .5625
Knee Surgery
a) P(X = 5) = 1 – P(X < 4) = 1 – 0.9 = 0.1
b) P(X = 1 or X = 2)
= P(one patient reports 1 or 2)
= .1 +.2 = .3
P(two patients report 1 or 2)
= 2 (.3)
= 0.6
c)   1(.1)   5(.1)  3.1
  (1  3.1) (.1) 
2
X
X
 1.136
 (5  3.1) (.1)
2
Keno
a) In the long run, a typical player can expect a payout of
$0.70, which may vary as much as $6.58.
b) Jerry’s mean payout is 5 times the original
mean…$3.50. The standard deviation of his winnings
is 5 times the original standard deviation…$32.90
c) Marla’s mean payout is .7 +.7 +.7+.7+.7 = $3.50. The
standard deviations of her winnings is 6.582 ∗ 5
= $14.71
d) The casino would prefer Marla. The standard
deviations of her winnings is lower than Jerry’s. The
casino would payout lower winnings more
consistently.
Winner, Winner, Chicken Dinner
a)
There are two outcomes…winner or not. Whether or not one
friend is a winner does not change the probability of another
friends winning. The probability of success is the same from bottle
to bottle (p = 1/6). There are seven friends who will purchase
soda (known number of trials).
b) 𝜇𝑋 = 𝑛𝑝 = 7
1
6
7
5
= 1.17; 𝜎𝑋 = 𝑛𝑝(1 − 𝑝) =
× =0.986; If
6
6
seven friends purchased bottles of soda over and over again, we
would expect 1.17 of them to win, on average. This numbers of
wins would vary by as much as 0.986 on average.
c)
2 7
1 𝑘
1 𝑘 5 7−𝑘
( )
=0.7656
6
6
P(X > 3) = 1 – P(X < 2) = 1 –
This probability suggests that we could expect 3 or more friends in
a group of 7 to win about 77% of the time. The company’s claim
appears to be inaccurate.
Cranky Mower
• This is a geometric setting…
– Success = the mower cranks
– Attempts are independent.
– Probability of success does not change from attempt
to attempt (p = 0 .2)
– The number of attempts to be made is unknown
a) Let X = # of attempts until mower cranks. Then
P(X = 3) = (1 – 0.2)2(0.2) = 0.128
b) P(X > 10) = P(mower did not crank the 1st nine
times) = (1 – 0.2)9 = 0.818
12,000 Miles & Running
• P(Y < 6) = P(z <
6−7.11
)
.74
= 0.6681
• The probability that a randomly selected male
student at the University of Illinois will run
the mile in less than 6 minutes is 0.6681.
Too much juice?
Postive Result
= 0.95
Yes = 0.1
Negative Result
= 0.05
Steroids?
Postive Result
= 0.03
No = 0.9
Negative Result
= 0.97
b) P(tests positive)
= P(positive and has used OR positive and has not used)
= 0.95(0.1) + (0.03)(0.9) = 0.122
c) P(has used|tests positive) =
P(positive and has used) .95(.1)
= .122 = .7787
P(positive)
Johnny’s Pizza
• Let A = getting a thick-crust pizza
• Let B = getting a pizza with mushrooms
a) P(B|A)=P(B) iff A & B are independent
2
7
P(B|A) = 4 =
7
2
;
4
4
P(B)= .
7
Because the probability of getting a pizza with mushrooms
changes when we know that the pizza has thick crust,
these events are dependent.
2
8
b) P(B|A) = 4 =
8
2
;
4
4
P(B)= .
8
Because the probability of getting a pizza with mushrooms
does not change when we know that the pizza has thick
crust, these events are now independent.
Too cool for the cabin?
5
9
𝜇𝐶
5
𝜎𝑌 = 𝜎9𝐶+32 =
9
𝜎𝐶
5
a) 𝜇𝑌 = 𝜇9𝐶+32 =
+ 32 =
9
5
8.5 + 32
= 47.3℉
5
b) P(Y < 40) = P(z <
=
9
5
40−47.3
)
4.05
2.25 = 4.05 ℉
= 0.0357