## The SBC GMAT Files

**Three Approaches To Probability Questions**

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GMAT probability questions are often difficult. They’re even more difficult if you fail to recognize that you often have three possible approaches at your disposal:

1. Apply probability rules (e.g., “and” probabilities, “or” probabilities, the complement, etc.)

2. Apply counting techniques (e.g., Fundamental Counting Principle, Mississippi Rule, combinations, etc.)

3. List outcomes

In many cases, choosing the fastest approach can save you considerable time. So before you begin pursuing the first approach that comes to mind, take a moment to consider your options.

Consider the following question:

The probability is ½ that a certain coin will turn up heads when tossed. If a coin is tossed three times, what is the probability that all 3 tosses turn up heads?

Let’s solve this question using all three approaches.

Probability rules:

P(all 3 heads) = P(heads on toss #1 AND heads on toss #2 AND heads on toss #3)

From here, we can apply the “and” probability rule to get:

P(all 3 heads) = P(heads on toss #1) Ã— P(heads on toss #2) Ã— P(heads on toss #3)

So, P(all 3 heads) = (½) Ã— (½) Ã— (½) = 1/8

Counting techniques:

Here, we’ll apply the basic probability formula to get: P(all 3 heads) = [number of outcomes with all 3 heads]/[total number of possible outcomes]. At this point, we’ll apply some counting techniques to find the numerator and denominator.

Let’s begin with the denominator (always begin with the denominator!). To determine the number of possible outcomes when a coin is tossed 3 times, we can take the task of tossing a coin 3 times and break it into stages:

Stage 1: toss the coin the first time

Stage 2: toss the coin a second time

Stage 3: toss the coin a third time

At this point, we’ll determine the number of ways to accomplish each stage.

Stage 1: the toss can be head or tails, so there are 2 ways to accomplish stage 1.

Stage 2: there are 2 ways to accomplish this stage.

Stage 3: there are 2 ways to accomplish this stage.

At this point, we can apply the Fundamental Counting Principle, which says the total number of ways to accomplish all 3 stages will equal the product 2Ã—2Ã—2 (which equals 8). So, there are 8 possible outcomes when a coin is flipped 3 times.

Of these 8 outcomes, how many are such that all three tosses turn up heads? There’s only 1 such outcome. So, P(all 3 heads) = 1/8

List outcomes:

This is a slight variation of the last approach. We’ll begin with P(all 3 heads) = [number of outcomes with all 3 heads]/[total number of possible outcomes]. However, rather than use counting techniques to find the numerator and denominator, we’ll just list the possible outcomes.

The possible outcomes are: (HHH), (THH), (HTH), (TTH), (HHT), (THT), (HTT) and (TTT)

From the list, we can see that there are 8 possible outcomes, and 1 of them is such that all 3 tosses are head. So, P(all 3 heads) = 1/8

So, we were able to solve the question using 3 different approaches. Which of them was the best in this situation? It’s hard to say. With some questions, all three approaches are equally matched. With other questions, one approach will be much more efficient than the others. In fact, in some cases, it will be impossible to apply all three approaches.

The big takeaway here is that, whenever you encounter a probability question, be sure to consider your options before you pursue the first approach that comes to mind.

For more information on the Fundamental Counting Principle, you might wish to watch the following video: http://www.gmatprepnow.com/module/gmat-counting?id=775

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