Translating a Tough Probability Word Problem
A few months ago, we discussed various strategies for translating word problems into math. Lets put that knowledge to the test on this challenging GMATPrep problem.
Set your timer for 2 minutes and go!
A contest will consist of n questions, each of which is to be answered either True or False. Anyone who answers all n questions correctly will be a winner. What is the least value of n for which the probability is less than 1/1000 that a person who randomly guesses the answer to each question will be a winner?(A) 5
(B) 10
(C) 50
(D) 100
(E) 1,000
One of the strategies we discussed in the translation article was make the situation real. Put yourself into the situation and imagine youre the one doing whatever the problem is describing. That will help you to set things up cleanly and correctly.
So whats going on in this particular situation? Youre the contestant. You have to guess randomly on a bunch of questions to which the answer is either True or False. What is the chance that you guess correctly? The probability is one-half, because there are two possible options and only one is right.
Figure out the next step dont try to go straight for that annoying 1/1000 number. Understand how the moving parts work first. Lets say you get the first one right. What happens next? You get another question. Whats the probability of getting the first one right and getting the second one right?
For each question, the probability of answering correctly is . If you want to answer the first and the second correctly, then we multiply the two probabilities:
(1/2)*(1/2) = 1/4
(Remember that and means multiply when were dealing with probability.)
And what about if I want to answer three questions in a row correctly? That would be:
(1/2)*(1/2)*(1/2) = 1/8
Ah, okay, Im starting to see the pattern. For each additional question answered, how does the number change? Each time, I multiply by 1/2. The result is that, each time, the denominator of the fraction goes up by a factor of 2.
Okay, I think I understand generally how things are working for this situation now. What else do I need to figure out? Hmm. They named a variable, n, for the number of questions answered. How does n fit into the scenario that I figured out, above?
For n = 1 question, probability = 1/2
For n = 2 questions, probability = 1/4
For n = 3 questions, probability = 1/8
Can I rewrite that in some way to show the variable n as part of the number? I can! Im actually raising the denominator to the power of n, the number of questions: [pmath](1/2)^n[/pmath]
For n = 1 question, probability = 1/2 = [pmath](1/2)^1[/pmath]
For n = 2 questions, probability = [pmath]1/4[/pmath] = [pmath](1/2)^2[/pmath]
For n = 3 questions, probability = [pmath]1/8[/pmath] = [pmath](1/2)^3[/pmath]
Check it out! So the probability of getting n questions in a row right is [pmath](1/2)^n[/pmath]. What can we do with that knowledge?
Heres where we get to the actual question and its a doozy.
What is the least value of n for which the probability is less than 1/1000 that a person who randomly guesses the answer to each question will be a winner?
Wow. Okay, were looking to win (of course!). Somehow, the probability 1/1000 is going to come into play. And then they ask me for the least value of n the least value for which the probability is less than 1/1000. What does that mean?
Hmm. Lets start with a probability less than 1/1000. Whats an example of such a probability? Is the probability 1/900 less than the probability 1/1000? Or is the probability 1/1100 less than the probability 1/1000?
Lets use the same kind of trick we used earlier. Remember, when we started, we figured out the probabilities for just the first few guesses, rather than jumping straight to the annoying 1/1000 fraction? Lets look at those smaller fractions from the beginning to figure out what it means for one probability to be less than another.
So, the probability of guessing one question correctly was 1/2. And the probability of guessing two questions correctly was 1/4. Which one of those is more likely to happen and which one is less likely to happen?
The probability 1/4 is less than the probability 1/2. Is the lesser probability a larger or smaller fraction? Its a smaller fraction. How do we know its smaller, from a mathematical perspective? The number in the denominator is larger, so that means the fraction itself is smaller.
Apply that to our 1/1000 situation. What would be a smaller fraction? Something with a larger denominator, such as 1/1100. Okay, so were looking for something that would make the probability less than 1/1000], or push the denominator above 1,000.
For what value of n would our probability push past 1,000? Lets start doing powers of 2! 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024 bingo! That last number, 1024, would give us this fraction: 1/1024. Thats past or smaller than 1/1000.
How many ns do we have? There are 10 powers of 2 to get to 1024, so the smallest value of n that will push us past the 1/1000 probability is 10.
The correct answer is B.
For me, the biggest trap on this one is answer A. If we misread the question (and thats very easy to do with the tricky wording on this one), we might decide that we want the denominator to be less than 1,000. If thats the case, then the smallest possible value of n would be 9. The number 9 isnt in the answers, but of the answers, 5 is the smallest possible value that would fit this (mistaken) criterion so we might end up picking 5. (Theres one big clue to tell me I should think this through very carefully if I want to pick 5: they ask for the least value of n and its very unlikely that the smallest value given in the answers will be the right answer. Thats a trap because people who dont know how to answer the question may pick the smallest answer simply because the question asked for the least value.)
Key Takeaways for Wordy Probability Problems:
(1) For wordy story problems in general, we want to insert ourselves into the story and make it real. Understand the sequence of events and actually figure out each step that needs to take place, one by one.
(2) Whenever necessary, use smaller or simpler numbers or examples to understand how something works. In this case, it was much easier to figure out the whole least value for which the probability is less than mumbo jumbo using and as our test numbers.
(3) Study the traps! Did you already know that a question that asks you for the least value of something will almost never have the smallest available answer as the correct answer? Thats not the kind of thing you know unless somebody tells you or unless you notice the patterns across multiple problems. Its not something you just see in the moment without having to think about it you actually have to figure it out in advance of a real testing situation.
* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.