Wednesday 1/19
7.1: 3, 6,7,10 (do straight flush, flush and full house),
12
Plus the following additional problems:
1. Powerball. Balls numbered 1-49 are placed in a container and
5 of them are drawn out of the container without replacement.
Then a separate "power ball" is drawn from another contained
containing balls numbered 1-42. Meanwhile, you write down 5 numbers
from 1-49 and one number from 1-42 on a card. To win the jackpot,
your first 5 numbers must match the numbers of the first 5 balls
(not necessarily in order) and your 6th number must match the
number of the 6th ball (the "powerball"). What is the
probability that you will win ?
2. You can view a birthday as a random choice of a number from
1-365. Let E be the event that in a room of 23 people, at least
two have the same birthday.
a. Describe the complementary event Ec.
b. Compute the probability P(Ec) and use it to find P(E).
HW 7.1: 8,9,10 (do four of a kind)
Monday 1/24
7.1: 15, 16
7.2: 3 (see below), 4ab, 5, 8 (see below)
7.3: 1 ii,iv,vi
More about 7.2.3. The probabilities for the loaded dice are:
P(1)=1/21 P(2)=2/21 P(3)=3/21 P(4)=4/21 P(5)=5/21 P(6)=6/21
We get the 21 from 21 = 1+2+3+4+5+6. For the sample points for
a pair of dice, just multiply the probabilites for the individual
dice. For example the probability that the first die is a 2 and
the second is a 5 would be
P(2,5) = 2/21 x 5/21 = 10/441
More about 7.2.8: let E and F be the events
E = "first die is a 3" F = "second die is a 3"
Describe the events E»F and E«F both in words and by listing the
elements of each event. Then verify formula 7.1 by separately
computing all four of the probabilities in it.
HW: 7.2.4cd and 7.3.1 iii
Wednesday 1/26
7.3: 4, 5, 6, 9, 11
7.6: 3, 4
HW 7.3: 10 , 7.6: 5
Monday 1/31
7.6: 6, 7i,ii, 11, 15, 17, 16 (see below)
HW 7.6: 7 iii,iv + Study for test
Here is one stategy for solving problem 7.6.16, the notorious "Monty Hall Problem"
You, the contestant, choose door number 2. Now identify several
events.
C1 = Car behind door number 1
C2 = Car behind door number 2
C3 = Car behind door number 3
It is reasonable to assign probability 1/3 to each of these.
M1 = Monty opens door number 1
M3 = Monty opens door number 2
Of course Monty won't open door number 2. Also (and this is the
source of the difficulty) Monty will not open the door with the
car behind it (and he knows which door that is). The information
about Monty's behavior can be summarized using conditional probabilities
as follows:
P(M1 | C3) = 1 P(M1 | C1) = 0
P(M3 | C1) = 1 P(M3 | C3) = 0
In other words given that the car is behind C3 Monty will definitely
open door number 1 and vice versa. If the car is really behind
door 2 Monty chooses which of doors 1 or 3 to open at random so
P(M1 | C2) = 1/2 P(M3 | C2) = 1/2
Now suppose Monty opens door number 1. Should you stick with door
2 or switch to door 3 or doesn't it matter? To find out, compute
the conditional probabilities:
P(C2 | M1) and P(C3 | M1)
to see which is larger.
Wednesday 2/2 ---- Chapter 7 Test