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A fair 2 sided coin is flipped 6 times. What is the [#permalink ]
06 Dec 2009, 17:50
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A fair 2 sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?

A. 5/8

B. 3/4

C. 7/8

D. 57/64

E. 15/16

Hi all,

First post, I have to say is this site is an amazing resource. Thanks to everyone who contributes!

I understand how to get the denominator just fine, but I am missing something on the numerator. I read the answer, but something just isn't clicking.

Thanks!

Last edited by

Bunuel on 28 Nov 2013, 05:51, edited 2 times in total.

Edited the question and added the OA

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Re: Combination problem - Princenten Review 2009 Bin 4 Q2 [#permalink ]
06 Dec 2009, 18:14
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brentbrent wrote:

Hi all,

First post, I have to say is this site is an amazing resource. Thanks to everyone who contributes!

Question:

A fair 2 sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?

a) 5/8

b) 3/4

c) 7/8

d) 57/64

e) 15/16

I understand how to get the denominator just fine, but I am missing something on the numerator. I read the answer, but something just isn't clicking.

Thanks!

Welcome to Gmat Club forum.

It would be easier to calculate the probability of opposite event and subtract it from 1.

Opposite event: 0 tail, 1 tail, 6 tails.

Probability of getting no tails:

\frac{1}{2^6}=\frac{1}{64} ;

Probability of getting 1 tail:

6C1*\frac{1}{2^6}=\frac{6}{64} , we must multiply by 6C1 or by 6 as tail can occur for any flip from 6, hence in 6 ways;

Probability of getting 6 tails:

\frac{1}{2^6}=\frac{1}{64} P=1-(\frac{1}{64}+\frac{6}{64}+\frac{1}{64})=\frac{56}{64}=\frac{7}{8} Answer: C.

Please ask if any question remains.

For more on probability and combinatorics please refer to the link:

GMAT MATH BOOK _________________

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Re: Combination problem - Princenten Review 2009 Bin 4 Q2 [#permalink ]
06 Dec 2009, 19:14
1

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brentbrent wrote:

Hi all,

First post, I have to say is this site is an amazing resource. Thanks to everyone who contributes!

Question:

A fair 2 sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?

a) 5/8

b) 3/4

c) 7/8

d) 57/64

e) 15/16

I understand how to get the denominator just fine, but I am missing something on the numerator. I read the answer, but something just isn't clicking.

Thanks!

the long way:

You have 4 options, tails twice, tails three times, tails 4 times, and tails 5 times

let's start with tails twice( this means 2 tails 4 heads)

6!/2!4! = 15

tails three times

6!/3!3! = 20

tails four times

6!/4!2! = 15

tails 5 times

6!/5!1! = 6

sum all and you get 56

56/64 = 7/8

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Re: Combination problem - Princenten Review 2009 Bin 4 Q2 [#permalink ]
06 Dec 2009, 19:37

Bunnel and Iagomez, thanks for the timely responses! Bunnel: I was getting hung up on why 6C1 had to be multiplied by 6. Thanks again to both of you.

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Re: Combination problem - Princenten Review 2009 Bin 4 Q2 [#permalink ]
07 Dec 2009, 02:42

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A fair coin is flipped 6 times. what is the probability that heads will be the result at least twice but not more than 5 times

15/16

57/64

7/8

3/4

5/8

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Re: A fair 2 sided coin is flipped 6 times. What is the [#permalink ]
11 Mar 2013, 12:50

"at least twice, but not more than 5 times" means exactly 2 times, 3 times, 4 times and 5 times

The probability of getting exactly k results out of n flips is nCk/2^n

6C2/2^6+6C3/2^6+6C4/2^6+6C5/2^6=(20+15+15+6)/2^6=56/64=(7*8)/(8*8)=7/8

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Re: Combination problem - Princenten Review 2009 Bin 4 Q2 [#permalink ]
27 Nov 2013, 18:56

Bunuel wrote:

brentbrent wrote:

Hi all,

First post, I have to say is this site is an amazing resource. Thanks to everyone who contributes!

Question:

A fair 2 sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?

a) 5/8

b) 3/4

c) 7/8

d) 57/64

e) 15/16

I understand how to get the denominator just fine, but I am missing something on the numerator. I read the answer, but something just isn't clicking.

Thanks!

Welcome to Gmat Club forum.

It would be easier to calculate the probability of opposite event and subtract it from 1.

Opposite event: 0 tail, 1 tail, 6 tails.

Probability of getting no tails:

\frac{1}{2^6}=\frac{1}{64} ;

Probability of getting 1 tail:

6C1*\frac{1}{2^6}=\frac{6}{64} , we must multiply by 6C1 or by 6 as tail can occur for any flip from 6, hence in 6 ways;

Probability of getting 6 tails:

\frac{1}{2^6}=\frac{1}{64} P=1-(\frac{1}{64}+\frac{6}{64}+\frac{1}{64})=\frac{56}{64}=\frac{7}{8} Answer: C.

Please ask if any question remains.

For more on probability and combinatorics please refer to the link:

GMAT MATH BOOK How do you calculate the prob of getting one tail in the any of the 6 flips as 6/64; can you let me know?

Thanks in advance

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Re: Combination problem - Princenten Review 2009 Bin 4 Q2 [#permalink ]
28 Nov 2013, 05:52
rango wrote:

Bunuel wrote:

brentbrent wrote:

Hi all,

First post, I have to say is this site is an amazing resource. Thanks to everyone who contributes!

Question:

A fair 2 sided coin is flipped 6 times. What is the probability that tails will be the result at least twice, but not more than 5 times?

a) 5/8

b) 3/4

c) 7/8

d) 57/64

e) 15/16

I understand how to get the denominator just fine, but I am missing something on the numerator. I read the answer, but something just isn't clicking.

Thanks!

Welcome to Gmat Club forum.

It would be easier to calculate the probability of opposite event and subtract it from 1.

Opposite event: 0 tail, 1 tail, 6 tails.

Probability of getting no tails:

\frac{1}{2^6}=\frac{1}{64} ;

Probability of getting 1 tail:

6C1*\frac{1}{2^6}=\frac{6}{64} , we must multiply by 6C1 or by 6 as tail can occur for any flip from 6, hence in 6 ways;

Probability of getting 6 tails:

\frac{1}{2^6}=\frac{1}{64} P=1-(\frac{1}{64}+\frac{6}{64}+\frac{1}{64})=\frac{56}{64}=\frac{7}{8} Answer: C.

Please ask if any question remains.

For more on probability and combinatorics please refer to the link:

GMAT MATH BOOK How do you calculate the prob of getting one tail in the any of the 6 flips as 6/64; can you let me know?

Thanks in advance

Explained here:

a-fair-2-sided-coin-is-flipped-6-times-what-is-the-87673.html#p659490 Hope it helps.

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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!! PLEASE READ AND FOLLOW: 11 Rules for Posting!!! RESOURCES: [GMAT MATH BOOK ]; 1. Triangles ; 2. Polygons ; 3. Coordinate Geometry ; 4. Factorials ; 5. Circles ; 6. Number Theory ; 7. Remainders ; 8. Overlapping Sets ; 9. PDF of Math Book ; 10. Remainders ; 11. GMAT Prep Software Analysis NEW!!! ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!! ; 12. Tricky questions from previous years. NEW!!! ;COLLECTION OF QUESTIONS: PS: 1. Tough and Tricky questions ; 2. Hard questions ; 3. Hard questions part 2 ; 4. Standard deviation ; 5. Tough Problem Solving Questions With Solutions ; 6. Probability and Combinations Questions With Solutions ; 7 Tough and tricky exponents and roots questions ; 8 12 Easy Pieces (or not?) ; 9 Bakers' Dozen ; 10 Algebra set. ,11 Mixed Questions , 12 Fresh Meat DS: 1. DS tough questions ; 2. DS tough questions part 2 ; 3. DS tough questions part 3 ; 4. DS Standard deviation ; 5. Inequalities ; 6. 700+ GMAT Data Sufficiency Questions With Explanations ; 7 Tough and tricky exponents and roots questions ; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!! ; 10 Number Properties set. , 11 New DS set. What are GMAT Club Tests ? 25 extra-hard Quant Tests Get the best GMAT Prep Resources with GMAT Club Premium Membership

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Re: Combination problem - Princenten Review 2009 Bin 4 Q2 [#permalink ]
28 Nov 2013, 06:25

Bunuel wrote:

brentbrent wrote:

Bunnel and Iagomez, thanks for the timely responses! Bunnel: I was getting hung up on why 6C1 had to be multiplied by 6. Thanks again to both of you.

What I meant was, when counting probability of getting 1 tail when flipped 6 times, 1 tail can occur in 6 different ways:

THHHHH

HTHHHH

HHTHHH

HHHTHH

HHHHTH

HHHHHT

Generally probability of occurring event k times in n-time sequence could be expressed as:

P = C^n_k*p^k*(1-p)^{n-k} In our case

k=1 and

n=6 , so we get:

P = C^6_1*\frac{1}{2}*\frac{1}{2^5}=6*\frac{1}{2^6}=\frac{6}{64} So

\frac{1}{64} should be multiplied by

C^6_1 , which is

6 .

ok slight complicated; but will do it……………; let me know to find the basic formulas for the no. properties?

thanks

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Re: Combination problem - Princenten Review 2009 Bin 4 Q2
[#permalink ]
28 Nov 2013, 06:25