Probability Notes
The Let’s Make a Deal Problem is as follows:
There is a car behind one of three doors, A, B or C, and a goat
behind the other two. A contestant picks one of the three doors.
The host then reveals one of the other two doors and shows a
goat. The contestant is then given a chance to “stay” or
“switch” doors. The question is, what is the probability of the
contestant winning if they stay? Also, what is the probability if
they switch?
Let’s consider the strategy of staying. Your chance of initially
picking the correct door (with the car) is 1/3. If the strategy is
staying, then you will only get the car in the cases when you
picked the door correctly to begin with. Thus, in this case, your
probability of winning is 1/3.
Let’s now consider the strategy of switching. There is a 2/3
chance of initially picking the incorrect door. When an
incorrect door is chosen, the host is
forced
to choose the other
incorrect door to reveal. Thus, if you switch, you will get the
correct door. Thus, the probability of winning when you switch
is 2/3. (You only lose in this case if you initially pick the correct
door, which only happens 1/3 of the time.)
This problem illustrates that our intuition about probability is
often misleading, just as it is about counting. Most people feel
that your probability of winning “goes up” when you are
presented with the door revealing the goat. But in fact, this
only occurs if you switch doors.
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View Full DocumentIn general, the probability of an event occurring is the number
of successes divided by the total number of possible outcomes
(known as the sample space) – assuming that each outcome is
equally likely.
For example, given a sixsided die the probability of rolling a 2
or a 5 is 2/6 because there are six possibilities, of which, two
are “successes.”
However, when rolling two dice, the probability of rolling a
sum of 2 is NOT 1/11. (There are 11 outcomes for the sum, 2
through 12.) This is because each of these possible sums, 2
through 12, are not equally likely.
The real sample space is the 36 ordered pairs (x,y) where 1 ≤
x,y ≤ 6.
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
For example, the probability of rolling a 2 is 1/36, since only
one of these 36 possibilities adds up to 2. The probability of
rolling a 3 is 2/36, a 4 is 3/36, etc. (Seven is the most frequent
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 Spring '09
 Conditional Probability, Probability, Probability theory, Sue, conditional probabilities

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