Driver's Test and Probability

Today at lunch a coworker mentioned this story about a woman who took 950 tries to pass the written part of the driver's test and started asking about the probability of passing by guessing each answer.

Our guess is that the test contains 20 questions and that each one has 4 or 5 choices. Let's go with 4 choices for now. The article says she only needed 60% to pass. Well, to figure out the chances of getting 12 or more right by chance just sum the probability of getting 12 to 20 right answers. For a set number of right answers n the probability is (1/4)^n*(3/4)^(20-n)*(20 choose n) since you need n right answers, 20-n wrong answers and you have to account for all the ways of arranging them (they aren't unique so it is choose rather than permutations). So then the probability of getting 12 or more right in one sitting is sum (1/4)^(n)*(3/4)^(20-n)*(20 choose n) from 12 to 20 = 0.000935392 or 1 in 1069. Not very likely that someone would pass by blind guessing once.

However, what are the odds of passing at least once in 950 tries? This could be solved in basically the same way as the last question, but there is a faster way. This is the same as asking what is 1 minus the chances of failing 950 times which is simply 1-(1-0.000935392)^950 = 0.588949 which is approximately 59%. So even if she was purely blind guessing she was actually slightly unlucky it took that many tries to pass.

However, if there were actually 5 choices per question then she actually only had a 9% chance of passing in 950 tries. If that was the case she was lucky, but still not terribly far from the realm of blind guessing.

Remember how excited I was about wolfram alpha? This may be the first time I've used it since a few weeks after posting about it. I guess a combo of google calculator and matlab tend to keep me satisfied. Actually wolfram alpha is in a weird in between spot. It is not as immediately available as google calculator due to the firefox search box. On the other side it doesn't allow for scripting (at least it is not immediately obvious to me how to do it) or even have an obvious way to feed one answer into the next calculation so it's not that great for anything serious. I guess it fills in when I need to do some math at home.