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Admissions Success Stories Coin flip questions made easy

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Coin flip questions made easy

Post Thu Sep 04, 2008 10:42 pm
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.

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snuman Junior | Next Rank: 30 Posts Default Avatar
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Post Fri Sep 05, 2008 11:10 am
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

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Post Fri Sep 05, 2008 11:45 am
snuman wrote:
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
Have you even tried understanding the most impressive coin flip strategy for most of the GMAT related questions.

It has nothing to do with KAPLAN, MGMAT, Princeton or any other prep courses or its folks!!!

Thanks a ton Stuart for such strategies.

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Post Fri Sep 05, 2008 11:50 am
snuman wrote:
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
My original post was in direct response to https://www.beatthegmat.com/probability-question-t16765.html#75446, a question posted by a user of this board.

Many test takers see coin flip (or pseudo-coin flip) questions on the GMAT. The better you're doing, the more complicated the probability questions are likely to be.

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Post Fri Sep 05, 2008 12:43 pm
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.

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Post Fri Sep 05, 2008 2:33 pm
Mpalmer22 wrote:
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.
You're correct: if you want exactly 3 or 4 flips, you just add up the 4th and 5th positions in the row.

Quote:
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).
4C3 indicates that I'm using the combinations formula, which we use when we want to count the number of unordered subgroups that can be made out of a group of entities.

Here's the generic formula:

nCk = n!/k!(n-k)!

n = total number of items
k = number of items that you're choosing

When we speak out loud, we usually translate nCk as "n choose k".

In coin flip questions, n is the total number of flips and k is the number of results you want to get. Going back to the first "4 flips, 3 heads" example, we have n=4 and k=3, which is why I solved for 4C3.

To illustrate the math at work:

4C3 = 4!/3!(4-3)! = 4!/3!1! = 4*3*2*1/3*2*1*1 = 4

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Post Fri Sep 05, 2008 3:47 pm
snuman wrote:
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
While I agree that some test prep books go a bit 'overboard' with their questions, Stuart certainly is not doing that here. You can definitely see a question like "If you flip a coin five times, what's the probability you get exactly three heads?" on the GMAT. The answer can be found in ten seconds- it's 5C3/2^5 = 5/16, so it isn't particularly laborious. All Stuart is pointing out is that numbers like 5C3, 5C4, and 5C5 are found in Pascal's triangle, a fundamental fact of combinatorics. There are many interesting connections that can be learned from Pascal's triangle- for example, that 6C3 + 6C4 = 7C4 (add two numbers from one row, you get a number in the next row), or that 4C0 + 4C1 + 4C2 + 4C3 + 4C4 = 2^4 (add all the numbers in row n, and you get 2^n). A long way of saying- nice post, Stuart!

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Post Fri Sep 05, 2008 5:36 pm
A great post stuart! you've made the lives of many gmat aspirants a bit easy! Thanks.

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Post Thu Sep 11, 2008 8:27 am
Stuart Kovinsky wrote:
Many test takers see coin flip (or pseudo-coin flip) questions on the GMAT. The better you're doing, the more complicated the probability questions are likely to be.
thanks for the wonderful strategy Smile

what do you mean by the "pseudo-coin flip" questions ?

Thanks Smile

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Post Thu Sep 11, 2008 9:40 am
amitdgr wrote:
Stuart Kovinsky wrote:
Many test takers see coin flip (or pseudo-coin flip) questions on the GMAT. The better you're doing, the more complicated the probability questions are likely to be.
what do you mean by the "pseudo-coin flip" questions ?

Thanks Smile
Any situation in which there are two possible outcomes, each one with a 50% chance.

For example:

"A certain volcano has a 50% chance to erupt in any given year. What's the probability that it erupts in exactly 2 years in a 5 year period?"

is really the same question as:

"A fair coin is flipped 5 times. What's the probability that it lands on heads exactly twice?"

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Post Thu Sep 11, 2008 9:42 am
I get it now Stuart Smile Thanks a lot .... Please share other such strategies also Smile

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Post Wed Sep 17, 2008 12:27 pm
Stuart Kovinsky wrote:
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
Stuart,

I posted this question elsewhere in the forum. But I'll post it here again, in hopes that you'll see it:

I don't understand your logic for the coin flip using combinations. In the numerator, don't you have to account for the fourth flip? Shouldn't the numerator be 4C3 (4 flips, 3 heads)* 4C1 (the fourth flip, tails)?

Here's a question I found on a combinatrics web site that proves my theory:

Three letters are randomly chosen from the word TUESDAY. How many possible selections are there? How many of these selections have:

(b) exactly 2 vowels?


No. of selections = 3C2* 4C1= 12

Here, the 4C1 represents the chances of NOT getting a vowel for the 3rd letter. But in your coin flip question, you never account for NOT getting a heads in one of the four flips.

Shouldn't the coin flip numerator be:

4C3 * 4C1? If not, please explain why it's different from the above posted question.

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Post Wed Sep 17, 2008 12:37 pm
Responded in the other thread!

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Post Fri Sep 19, 2008 2:28 am
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

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Post Fri Sep 19, 2008 8:17 am
amitdgr wrote:
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
If we want fewer (as a sentence correction aside, we use "fewer" when the noun is countable - less vs fewer is commonly tested on the GMAT Very Happy ) than 2 boys, we want 0 boys or 1 boy.

So, we go to the n=10 row (which is the 11th row down - remember that the apex of the triangle is the n=0 row) and add up the first 2 entries. The sum of the row will be 2^10.

It would actually be much quicker to do this using combinations, since we're looking at k values of 0 and 1:

10C0 = 1 (any nC0 = 1)
10C1 = 10 (any nC1 = n)

In fact, by the above principles (or just by looking at the triangle), we can see that the first two entries in every row are 1 and n (and, since the rows are symmetrical, the last two entries in every row are n and 1).

So, the answer would be:

1 - (1 + 10)/2^10

1 - 11/2^10

1024/1024 - 11/1024

(1024 - 11)/1024 = 1013/1024

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