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Coin flip questions made easy

This topic has 24 expert replies and 45 member replies
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jagdeep Junior | Next Rank: 30 Posts Default Avatar
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Post Sun Nov 22, 2009 6:29 pm
Can anyone answer this question please


a fair two sided coin is flipped 6 times. what is the probability that tails will be the result atleast twice but not more than 5 times

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Post Mon Nov 23, 2009 12:25 am
jagdeep wrote:
Can anyone answer this question please


a fair two sided coin is flipped 6 times. what is the probability that tails will be the result atleast twice but not more than 5 times
There are a few ways we can solve. Before we do, we need to understand exactly what the question is asking.

"At least twice but not more than 5 times" means that, out of 6 flips total, we want 2, 3, 4 or 5 tails; we don't want 0, 1 or 6 tails.

One way we could solve is using the coin flip formula:

Prob (k results out of n flips) = nCk/2^n

Since this is an alternative probability question (we want 2 tails OR 3 tails OR 4 tails OR 5 tails), we add the individual results to get:

(6C2 + 6C3 + 6C4 + 6C5)/2^6

We could also use the "one minus" approach and focus on what we don't want:

1 - (6C0 + 6C1 + 6C6)/2^6

This second approach is almost always quicker on at least/at most questions (the questions are designed to reward people who see creative solutions).

In this particular case, the second approach is super quick if we remember some combination shortcuts:

nC0 = 1
nC1 = n
nCn = 1

So, we get:

1 - (1 + 6 + 1)/64 = 1 - 8/64 = 1 - 1/8 = 7/8

Of course, the quickest way to solve is to use Pascal's Triangle. As noted in an earlier post in this thread, the 6 row of the triangle is:

1 6 15 20 15 6 1

Since we want 2, 3, 4 or 5 tails, we add up the 3rd, 4th, 5th and 6th entries in the row:

15 + 20 + 15 + 6 = 56

and divide by the sum of the entire row, 64,

to get:

56/64 = 7/8

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moatazyousif Junior | Next Rank: 30 Posts
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Post Sat Apr 03, 2010 11:36 am
Thanks a lot Stuart for this great tool and explaination.

mepinoargote Junior | Next Rank: 30 Posts
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Post Thu Jun 17, 2010 3:38 pm
Stuart Kovinsky wrote:
GMATCHPOINT wrote:
(question from someone terrible in math)

Does it make sense to apply this formula on below question from OG11? If so, how can I do it? I have tried with no success.

From OG11 page 241 ex 173

THE PROBABILITY IS 1/2 THAT A CERTAIN COIN WILL TURN UP HEADS ON ANY GIVEN TOSS. IF THE COIN IS TO BE TOSSED THREE TIMES, WHAT IS THE PROBABILITY THAT ON AT LEAST ONE OF THE TOSSES THE COIN WILL TURN UP TAILS?

A) 1/8
B) 1/2
C) 3/4
D) 7/8
E) 15/16
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.
Using the Coin flip formula, for the same question lets say now, we want that on at least two of the tosses the coin will turn up tails.

In other words, the probability of getting one head: P(getting one head)= 3C1/2^3 = 3/8

Can i conclude then, that for x number of tosses, the probablity of getting at least n tails (when n>1) equals to the probability of getting x-n heads???

sdilmanian Junior | Next Rank: 30 Posts Default Avatar
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Post Tue Jun 07, 2011 5:26 am
Great stuff Stuart - I really do need a refresher in these concepts from my math days.

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Post Wed Sep 21, 2011 1:27 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

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.
Regarding this approach:

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


How would you tackle the following question with the approach above?

The probability is 0.6 that an “unfair” coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails?

Thx a lot

kumadil2011 Junior | Next Rank: 30 Posts Default Avatar
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Post Fri Nov 18, 2011 8:41 pm
How can we solve the below problem using the same formula ?

http://www.beatthegmat.com/three-fair-coins-t53691.html

tx

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Post Mon Nov 21, 2011 3:50 pm
kumadil2011 wrote:
How can we solve the below problem using the same formula ?

http://www.beatthegmat.com/three-fair-coins-t53691.html

tx
That question asks you to solve for the standard deviation of a set; since the GMAT never asks you to solve for standard deviation, I have no idea why anyone would even want to solve that problem!

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tomatos Newbie | Next Rank: 10 Posts Default Avatar
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Post Tue Mar 27, 2012 9:04 am
How do you solve
Stuart Kovinsky wrote:
"A fair coin is flipped 3 times. What's the probability of getting exactly 2 heads?"
with nCk/2^n

gmathater2 Newbie | Next Rank: 10 Posts Default Avatar
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Post Thu Apr 19, 2012 6:41 am
Hi Stuart, i hope you are still checking this forum. First thank you for the formula, i'm solving all the coins exercises and more! I was just wondering, using you formula is it possible to use the following for die (dice) questions nCk/6^n? if not, is there a formula to make the die problems easier?

Best Regards

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Stuart Kovinsky GMAT Instructor
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Post Thu Apr 19, 2012 8:34 pm
gmathater2 wrote:
Hi Stuart, i hope you are still checking this forum. First thank you for the formula, i'm solving all the coins exercises and more! I was just wondering, using you formula is it possible to use the following for die (dice) questions nCk/6^n? if not, is there a formula to make the die problems easier?

Best Regards
Hi! I was away for a while, but I am indeed back!

You can indeed use that version of the formula for die roll questions.

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gmathater2 Newbie | Next Rank: 10 Posts Default Avatar
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Post Sat May 19, 2012 11:56 pm
Hi Stuart,

Sorry i'm bothering you again with the same issue.

the formula for the die problems is not really working for me. Let's take the following example:
A fair 6 sided die is rolled 4 times. what is the probability that there will be a "3" at least 3 times?

according to the formula the answer is: 4C3/6^4 + 4C4/6^4

but the correct answer is: 5 * 4C3/6^4 + 4C4/6^4 OR 4 * (1/6)^3 * (5/6) + (1/6)^4

so should we add to the formula something like (number of non chosen numbers to show)^(n-k) and the formula becomes (number of non chosen numbers to show)^(n-k) * nCk/6^n

It could be possible because for the coin it's either this or that so this added factor is always 1, but for the die there are 5 other possibilities.

cheers

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Post Sun May 20, 2012 12:45 am
gmathater2 wrote:
Hi Stuart,

Sorry i'm bothering you again with the same issue.

the formula for the die problems is not really working for me. Let's take the following example:
A fair 6 sided die is rolled 4 times. what is the probability that there will be a "3" at least 3 times?

according to the formula the answer is: 4C3/6^4 + 4C4/6^4

but the correct answer is: 5 * 4C3/6^4 + 4C4/6^4 OR 4 * (1/6)^3 * (5/6) + (1/6)^4

so should we add to the formula something like (number of non chosen numbers to show)^(n-k) and the formula becomes (number of non chosen numbers to show)^(n-k) * nCk/6^n

It could be possible because for the coin it's either this or that so this added factor is always 1, but for the die there are 5 other possibilities.

cheers
Hi!

The formula is more complicated for die rolls because there are more than 2 possible outcomes, so "what you want" doesn't equal "what you don't want".

If you're looking for one specific result, there's a 1/6 chance of what you want and a 5/6 chance of what you don't want (unlike coin flips, for which it's 1/2 and 1/2).

Let's look at an example before we derive the formula:

A fair six-sided die is rolled 5 times. What's the probability of getting exactly two 4s?

There are 5C2 ways to get two 4s on 5 dice, so we need to multiply by that number.

For our two 4s, there's a 1/6 chance for each one, so we need to multiply by (1/6)^2.

For our three non-4s, there's a 5/6 chance for each one, so we need to multiply by (5/6)^3.

Accordingly, the final answer would be:

5C2 * (1/6)^2 * (5/6)^3.

Now to genericize the formula:

If you want exactly k results of a specific number out of a total of n rolls, the probability is:

nCk * (1/6)^k * (5/6)^(n-k)

or, if you prefer:

(nCk * 5^(n-k))/(6^n)

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dj_vinayak Senior | Next Rank: 100 Posts
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Post Wed Aug 01, 2012 6:58 pm
For any Coin Flip Question I prefer to go with the Anagram Method.

In the originally asked question:
A fair coin is flipped three times. What is the probability that the coin lands on heads exactly twice?
(A) 1/8 (B) 3/8 (C) 1/2 (D) 5/8 (E) 7/8

Total combination per coin flip =2
For 3 Coin Flips , Total combination = 8 (2*2*2)

We want number of combination of anagram HHT
3!/2! = 3

Thus,Prob = 3/8.

This solution I believe is more easily scalable to 5 flips etc.

_________________
Stay Hungry Stay Foolish

Post Thu Oct 04, 2012 11:43 am
Just wanted to expand the universe of coin-flip questions by adding the following scenario: "Find the probability that on five coin tosses we would get three heads and two tails" (Source: MGMAT).

P(3H and 2T) = [5!/3!*2!] / 2^5 = 10/32 = 5/16

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