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EXAMPLE 1 In solving this problem I chose the strategy of starting with a simpler problem and looking for a pattern as the number of questions increased. Through this method I discovered that the pattern was Pascal's Triangle!
1--question quiz: Only 1 way to answer all correctly; 1 way to answer all wrong.
2--question quiz: 1 way to answer both correctly; 2 ways to answer only 1 question correctly (C/W, W/C); 1 way to answer both wrong.
3--question quiz: 1 way to answer all correctly; 3 ways to answer only 2 questions correctly; 3 ways to answer only 1 question correctly; and 1 way to answer no questions correctly.
After examing the possible outcomes of the first 3 smallest quizzes, I discovered the following pattern: 1 1 1 2 1 1 3 3 1 I discovered that I could continue the pattern by adding rows in which the entries reflect the sum of the two numbers above: 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 From this triangle pattern I could easily find the ways to answer a 5--question T/F quiz by adding the numbers in the 5th row. Since the sum of 1+5+10+10+5+1 is 32, I knew that was the correct number for the denominator of the mathematical probability. To find the numerator, the number of ways to pass the test, I had to find the numbers in the triangle which showed the ways to earn 100% (no wrong) or 80% (only 1 wrong). Based on the analysis of shorter quizzes, I realized that the 1 on the left side of the triangle always represents the only way to earn 100%. The number to the left of that represents not only the number of questions on the quiz, but also the ways to miss only 1 question, thereby earning 80%. On a 5--question T/F quiz there are only 2 possible passing grades, so to find the numerator, I only had to add the 1 + 5 to get 6. The mathematical probability I found is 6/32 or .19 or 19%.
You could also take those numbers from the right, instead of the left, because the pattern in this triangle is symmetric. When you have 2 possible outcomes from each independent event, you have a choice of which outcome you consider first. You could either consider the 1 on the left to mean 1 way to get all answers correct, or you could consider it to mean the 1 way to get all answers wrong. If you choose the latter, then the one on the right side becomes the 1 way to get all answers correct.
I started to wonder: Is there a pattern in the probabilities of passing the shorter quizzes? I found the total of the rows representing 1, 2, 3, 4, and 5--question quizzes, to be 2, 4, 8, 16, 32. These numbers would be the denominators showing the total ways to pass each quiz. The pattern here is that each number is a power of 10 in which the exponent is equal to the number of questions on the quiz. To find the numerator you have to think about the possible passing percentages for each quiz. For the 1--question and 2--question quizzes, there is only 1 possible passing grade; 100%. You could also pass the 3--question quiz with a grade of 66%. On the 4--question quiz, you could miss 1 question and still earn 3/4 or 75%. Therefore, in calculating the ways to pass the 3--question and 4--question quizzes, like the 5--question quiz, you must add both number on the left of the corresponding row from the triangle. Based on these calculations I found the following mathematical probabilities of passing the quiz: 1 Question: 1/2 or .5 or 50% 2 Question: 1/4 or .25 or 25% 3 Question: 4/8 or .5 or 50% 4 Question: 5/16 or .31 or 31% 5 Question 6/32 or .19 or 19%
I discovered, to my surprise, that the easiest quiz to pass by guessing is a 1--question or 3--question quiz, followed by a 4--question quiz, and then a 2--question quiz. It appears that the longer the quiz, the less the probability of earning a passing grade when guessing. None of the probabilities for guessing are favorable, so I guess there's a lesson here that studying is worth the effort. |
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