COT3100Probability01 - Probability Notes The Let's Make a...

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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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In 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 six-sided 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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COT3100Probability01 - Probability Notes The Let's Make a...

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