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The formula works very well for this question. First, let's simplify.

If we want AT LEAST 1 tails, then the only thing that we DO NOT want is all 3 heads.

Remember this general formula:

Prob (want) = 1 - Prob (don't want)

Applying that to this question:

Prob (at least 1 tails) = 1 - Prob (all 3 heads)

Prob(all 3 heads) = 3C3/2^3 = 1/8

Prob (at least 1 tails) = 1 - 1/8 = 7/8

Of course, you could also answer this question without the forumla, using a bit of logic and common sense.

The only result we don't want is HHH. For 3 flips, there are 2^3 = 8 total possible outcomes. If there's 1 outcome we don't want, there must be 7 outcomes that we do want, so the chance of getting what we want is 7/8.

You could also look at the n=3 row of Pascal's Triangle. The only thing we don't want is 0 tails, so if we add up 1 tails + 2 tails + 3 tails we get 3 + 3 + 1 = 7. The whole row totals to 8, so again we end up with an answer of 7/8.

I just typed up this detailed explanation to address a specific question and figured that I might as well share it with everyone!

Coin flip strategies

There are numerous ways to solve coin flip questions. One of the quickest is to apply the coin flip formula.

The probability of getting exactly k results out of n flips is:

nCk/2^n

For example , if one wanted to know the probability of getting exactly 3 heads out of 4 flips:

4C3/2^4 = 4/16 = 1/4

As quick as it was to apply the formula, there's an even BETTER way to solve coin flip questions, involving memorizing a few numbers.

Here are the numbers to remember:

1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

Some of you may recognize those patterns as rows of numbers from Pascal's Triangle (I swear, I came up with them first ). The Triangle has a number of uses, but for GMAT purposes one of its most useful applications is to coin flip questions.

The first row applies to 3 flip questions, the second to 4 flip questions and the third to 5 flip questions.

Let's start with the first row, 1 3 3 1, and see how it helps.

"A fair coin is flipped 3 times. What's the probability of getting exactly 2 heads?"

The entries in the row represent the different ways to get 0, 1, 2 and 3 results, respectively. In our question, we want 2 heads, so we go to the 3rd entry in the row, "3".

To find the total number of possibilities, add up the row... 1+3+3+1 = 8

So, our answer is 3/8.

Going back to our original question (exactly 3 heads out of 4 flips):

4 row is 1 4 6 4 1

For 3 heads, we use the 4th entry: 4
Sum of the row is 16
Answer: 4/16 = 1/4

Let's look at a much more complicated question:

"A fair coin is flipped 5 times. What's the probability of getting at least 2 heads?"

If we want at least 2 heads, we want 2 heads, 3 heads, 4 heads OR 5 heads. Pretty much whenever we see "OR" in probability, we add the individual probabilities.

Looking at the 5 flip row, we have 1 5 10 10 5 1. For 2H, 3H, 4H and 5H we add up the 3rd, 4th, 5th and 6th entries: 10+10+5+1=26.

Summing the whole row, we get 32.

So, the chance of getting at least 2 heads out of 5 flips is 26/32 = 13/16.

"A fair coin is flipped 3 times. What's the probability of getting exactly 2 heads?"