Breaking Down A GMATPrep Probability Problem
I chose this weeks problem for two reasons: first, most people hate probability, so the more help we can get, the better. Second, aspects of this problem reminded me of one of the next generation GMAT problem examples published by GMAC last year. (Unfortunately, it looks like theyve taken the page down, so I cant show you the specific problem that made me think of the GMATPrep problem below.)
Set your timer for 2 minutes. and GO!
For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?(A) [pmath](0.6)^5[/pmath]
(B) [pmath]2(0.6)^4[/pmath]
(C) [pmath]3(0.6)^4(0.4)[/pmath]
(D) [pmath]4(0.6)^4(0.4) + (0.6)^5[/pmath]
(E) [pmath]5(0.6)^4(0.4) + (0.6)^5[/pmath]
Bonus question: figure out how someone would mistakenly arrive at any (or all!) of the four wrong answer choices.
So, weve got a coin, but we dont actually have 50/50 odds of getting heads or tails. In fact, they dont even tell us that the other side is tails or even that there are only two sides to the coin! Turns out that this doesnt matter - we can think of all possible outcomes simple as heads or not heads. So, lets see, the coin has a 0.6 chance of landing on heads, so what are the chances of the coin not landing on heads?
All possible outcomes add up to 1, so if we have a 0.6 chance of getting heads, then we have a 0.4 chance of not getting heads.
Okay. Were going to toss the coin 5 times. Each time, we can get heads or not-heads. They want to know the probability of getting heads at least 4 times. What possibilities are there? In one set of 5 tosses, I could get heads 4 times and not-heads 1 time. I could also get heads 5 times (and zero not-heads). The question is really asking: whats the probability of getting heads 4 times OR 5 times?
Ordinarily, when I see language similar to at least or at most on probability questions, I think about using the 1-x shortcut, because usually at least or at most language indicates that they are asking me to calculate the probability for all possible outcomes except for one. Is that what theyre asking me to calculate here?
No. Here, there are six possible outcomes (0 heads, 1 head, 2 heads, 3 heads, 4 heads, or 5 heads) and we need to calculate the probability of two of them. Hmm. So I need something other than my usual 1-x shortcut.
We still have one more thing to do before we can start to solve a PS problem: look at the answers. Interesting. They look complicated, but at least we dont actually have to solve all the way to a number. The test writers are interested in knowing whether we can construct the answer and this is what reminded me of the next-generation GMAT problem I referenced earlier. The next-gen questions seem to be much more focused on making us prove that we know how to set something up to solve. Theyre less focused on making us actually slog through the calculations (presumably because people are mostly going to use calculators or computers to do the grunt work in the real world).
So it actually makes sense that we cant use the 1-x shortcut here this problem is not about solving for a number. Its about knowing how to solve for the number. Our real task is just setting up the proper equation to solve.
Okay, back to our rephrased question: what is the probability of getting heads 4 times OR 5 times?
In probability, what does OR mean? It means addition. So we need to find the probabilities of each of these two events and then add them together. (Glance at the answers again. If you had to guess right now, what would you not pick?)
Lets start with the easier one: whats the probability of tossing this coin 5 times and getting heads all 5 times? In other words, we want H and H and H and H and H. What does AND mean, in probability terms?
Right, multiplication. So the probability of getting H all five times is
0.6 0.6 0.6 0.6 0.6 or
[pmath](0.6)^5[/pmath].
Glance at your answer choices again. If you had to guess right now, what would you not pick?
Not choice A, because A represents only the probability of getting heads all five times, and we know we also need to calculate the probability of getting heads four times (and not-heads once). Not B or C, either, because those dont contain the probability for getting heads all five times. Interesting. Whats the difference between D and E?
Make the answer choices work for you. D and E differ in only one way, so lets use that to help us figure out the rest of the correct set-up.
So, apparently, the probability of getting heads four times (and not-heads once) must include [pmath](0.6)^4[/pmath] and (0.4), because both remaining answer choices have it. What do those figures represent? 0.6 is the probability of getting heads. 0.4 is the probability of getting not-heads. Okay, that makes sense I want heads 4 times and the (0.6) probability has an exponent of 4. I want not-heads once, and the (0.4) probability doesn't show an exponent (which is the same thing as saying it has an exponent of 1). Perfect.
What about that third number in the term? Im going to multiply this probability either by 4 or by 5. Hmm. Why those two numbers? There are no coefficients in front of the all 5 heads figure.
Ah, right. Theres only one way to get all 5 heads. But thats not true for heads and heads and heads and heads and not-heads. Those coefficients must have to do with the number of different ways that I can get 4 heads and 1 not-heads.
When trying to determine the number of different ways that the combination can be constructed, concentrate on the unique option (if there is one). In this case, not-heads is our unique option. We could roll NHHHH. Or we could roll HNHHH. Or HHNHH. Keep going. How many options do we have for where to place that N (for not-heads)?
The not-heads has 5 different options: it can be placed 1st, 2nd, 3rd, 4th, or 5th (that is, NHHHH, HNHHH, HHNHH, HHHNH, HHHHN) So we want to multiply the probability, [pmath](0.6)^4(0.4)[/pmath], by the number of combinations that have that outcome: 5.
The correct answer is E.
Bonus question: how are the four wrong answers constructed?
Answer A reflects the probability of getting five heads but does not include the probability of getting four heads and one not-heads.
Answer B reflects part of the probability of getting four heads; perhaps the 2 coefficient comes from thinking that there are 2 ways to have at least 4 heads: (1) 4 heads + 1 not-heads or (2) 5 heads.
Answer C again reflects part of the probability of getting four heads, and it does include the not-heads part of the outcome, but it still omits the all five heads outcome and it mistakenly uses 3 as the coefficient.
Answer D is the closest to the real thing. Everything is right except for that coefficient at the beginning. Here, someone might think that there are 4 different ways to build this because they are focusing on the 4 heads part of the outcome without checking to see that there are actually five different possible combinations for the outcome.
Key Takeaways for Set-up Probability Problems:
- Look at your answer choices! Use the construction of the choices to help you figure out what steps to take. When they want a set-up, as opposed to a more normal solution (such as a value), you know that you need to work your way through that set-up very carefully. (And you also know you have the time to do so, because you arent going to have to solve for a value!)
- Make the situation as real and concrete as you can. Visualize the coin. Write out the possible outcomes and figure out what will actually happen each time you flip the coin. You can even draw out horizontal slots to help keep track of the different outcomes.
- If there are multiple possible combos that give the same probability, as we saw on the 4 heads / 1 not-heads outcome on this problem, make sure you multiply the probability of a certain outcome by the number of possible combos for that outcome.
* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.
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