Coin flip questions made easy

Stuart Kovinsky · Posted: Thu Sep 04, 2008 10:42 pm Post subject: Coin flip questions made easy

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 Wink

). 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.

snuman · Posted: Fri Sep 05, 2008 11:10 am Post subject:

I think Kaplan going way overboard in posting such laborious questions. GMAT never wants you to do such labor and expect you to know some tricks to solve such laborious questions. GMAT's focus is basic concepts and presents short amount of labor, if need be, to test basic concepts.

I think test preparation companies get folks scared of GMAT by posting such questions and tricks to solve them.

Nauman

parallel_chase · Posted: Fri Sep 05, 2008 11:45 am Post subject:

Stuart Kovinsky · Posted: Fri Sep 05, 2008 11:50 am Post subject:

Mpalmer22 · Posted: Fri Sep 05, 2008 12:43 pm Post subject:

Thanks Stuart for that. Just to make sure, when they say 3 or 4, but NOT 5 times, and using the triangle you came up invented Wink

You only need to add the 4th and 5th position and just ignore the 6th position, not subtract the 5th position or anything like that.

This question may also be elementary too, but I really don't understand the coin flip formula. when you write 4C3, and it solves out to 4, what is the 3 and how does that affect anything. (If you know of a good probability guide or reference I can use, I can look for the answer).

Anywho, thanks for the help with probability q's they are the main thing giving me trouble.

Stuart Kovinsky · Posted: Fri Sep 05, 2008 2:33 pm Post subject:

Ian Stewart · Posted: Fri Sep 05, 2008 3:47 pm Post subject:

karthikeyan.sivanantham · Posted: Fri Sep 05, 2008 5:36 pm Post subject: great post

A great post stuart! you've made the lives of many gmat aspirants a bit easy! Thanks.

amitdgr · Posted: Thu Sep 11, 2008 8:27 am Post subject:

Stuart Kovinsky · Posted: Thu Sep 11, 2008 9:40 am Post subject:

amitdgr · Posted: Thu Sep 11, 2008 9:42 am Post subject:

I get it now Stuart Smile

Thanks a lot .... Please share other such strategies also Smile

Stockmoose16

Stuart Kovinsky · Posted: Wed Sep 17, 2008 12:37 pm Post subject:

Responded in the other thread!

amitdgr · Posted: Fri Sep 19, 2008 2:28 am Post subject:

Stuart I am not sure if you are still following this thread. If you are, I'd like to know how to solve this particular problem (This is a pseudo coin-flip question)

The probability of a new born baby is a boy is 50% and that of a new born baby is a girl is 50%. Ten mothers are going to give birth to ten babies this morning. What is the probability that at least two babies are boys?

One way of solving this would be, write the pascal's triangle up to the ten coins row and then solve for it. That may take us some time, so can we do it in this way instead

1-probability of less than 2 boys being born

Now my question is, how to find the value of probability of LESS THAN 2 boys being born using the Pascal's triangle.

Thanks

Stuart Kovinsky · Posted: Fri Sep 19, 2008 8:17 am Post subject: