Download Module 5-7 Questions and Answers

Q&A on Modules 5-7
Can you explain how the probability that at least two of the 60 people share the
same birthday is above 99% chance?
The probability that two people have the same birthday is 1/365. The
probability that they have different birthdays is 1-1/365 = 364/365. If a third
person comes along, the probability that she has a different birthday from the
first two is 363/365, because two days are already used. Thus the probability
that three people have different birthdays is (364/365)(363/365). Continuing
this argument, we get the probability that 60 people have different birthdays is
(364/365)(363/365)...(305/365) = 0.0049
Still not completely clear on how to compute getting at least one factor when
considering two factors like in the coin example of the likelihood of getting at least
one heads when tossing a coin a couple of times.
Still a little unclear about how to find the probability of an event when all
outcomes are equally likely.
If you toss a coin three times, for example, there are eight equally likely
outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Each has probability
1/8, so the probability of exactly one H is 3/8, since there are three outcomes
with exactly one H.
how do you calculate the exactly one, at least one of them, neither of them, by
using probability rules? it seems very confusing to me.
Using probability rules, the easiest way to calculate the probability of "at least
one" is to calculate the probability of "none" and take the complement. This
works because "at least one" is the negation of "none". If you toss three coins,
the probability of no heads is 1/8, so the probability of at least one head is 11/8 = 7/8.
when measuring probability on the scale of 0 through 1 are negative numbers
possible?
No. Probabilities are always decimal numbers between 0 and 1.
i'm still a little fuzzy on how to set up possible outcomes. And while i understand
the video about the prizes behind the doors, why would you split the three doors
into two uneven groups?
When you play "Let's Make a Deal" and you choose a door, you have divided the
doors into two unequal groups: the one you chose and the two you didn't
choose. The offer to switch your choice is really the offer to switch from the door
you choose to the two doors you didn't initially choose.
How do you construct a two way probability table?
Finding when to use the complement rule and figuring out when an event is
independent or disjointed or not.
the concept of disjointed probabilities.
The disjoint is kind confusing.
First, the term is "disjoint", not "disjointed". There are no such things as
disjoint probabilities, but there are disjoint events. Two events are disjoint if
they both cannot happen at once. If you toss three coins, and if A is the event of
getting 2 heads and B is the event of getting 2 tails, then A and B are disjoint.
They cannot both happen on one toss with just three coins. When two events
are disjoint we have the simple rule P(A or B) = P(A) + P(B).
Interpreting the data in a two way probability table. What goes in it and how to
answer questions about it.
pl explain in details the two way probability table.
the Complement Rules for the 2- way Preobability Table
Suppose you have two events about the same general context, for example (to
give away a quiz question) event A is "a dalmatian is blue-eyed" and event B is
"a dalmatian is deaf". We can make a table with 9 squares, each one
representing part of the population of dalmations:
B
not B
Total
A
P(A and B)
P(A and not B)
P(A)
not A
P(A and not B)
P(not A and not B)
P(not A)
Total
P(B)
P(not B)
1
Since the first two probabilities in each row and column add to the total
probability, and the total row and column probabilities add to 1, you can fill in
the entire table if you are given some of the data.
For example, suppose you know that 31% of dalmations are blue-eyed, 38% are
deaf and 42% of blue-eyed dalmations are deaf. If B is "blue-eyed" and D is
"deaf", then we know P(B) = 0.31, P(D) = 0.38 and P(D | B) = 0.42. Thus
P(D and B) = P(D | B) P(B) = 0.42 x 0.31 = 0.13.
Now we can fill in part of the table:
blue-eyed
not blue-eyed
Total
deaf
0.13
not deaf
Total
0.31
0.38
1
Using addition, we get:
blue-eyed
not blue-eyed
Total
deaf
0.13
0.25
0.38
not deaf
0.18
0.62
Total
0.31
0.69
1
deaf
0.13
0.25
0.38
not deaf
0.18
0.44
0.62
Total
0.31
0.69
1
and finally:
blue-eyed
not blue-eyed
Total
From this table you can answer any question about dalmations, being blueeyed and being deaf. For example, the probability that a dalmation is blue-eyed
or deaf is 0.13+0.25+0.18 = 0.56. The probability that a deaf dalmation is blueeyed is 0.13/0.38 = 0.34.
I'm not sure what the complement rule is.
Summing up and multiplying probabilities.
Probabilities are not summed or multiplied except in special cases. When two
events A and B are independent, the P(A and B) = P(A)P(B). When two events are
disjoint, then P(A or B) = P(A) + P(B). More frequently you have to use the more
complicated rules:
P(A and B) + P(A or B) = P(A) + P(B)
P(A | B) P(B) = P(A and B)
How to determine if two events are independent or not more efficiently.
There is no "efficient" way to determine independence. If you know P(A) and P(B)
and P(A and B) , then you can determine if A and B are independent by
checking if P(A and B) = P(A)P(B). Also if you know P(A) and P(A | B), then A and
B are independent if P(A) = P(A | B).
More often you use common sense to determine if A and B are independent, if
the occurrence of one has no effect on the other, and then you use their
independence to compute P(A and B) by multiplying P(A) times P(B). If A and B
are independent, then you can set P(A | B) = P(A).
I'm still not sure how to tell when i should or shouldn't use probablity trees.
How to start to construct and interpret probability trees.
There is one question on the quiz for Module 7 about probability trees. There
will be no questions on the final about probability trees. Any problem that can
be solved with probability trees can be solved with tables.