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A fair 2 sided coin is flipped 6 times. What is the

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A fair 2 sided coin is flipped 6 times. What is the  [#permalink]

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Updated on: 28 Nov 2013, 06:51
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Question Stats:

58% (02:07) correct 42% (02:33) wrong based on 332 sessions

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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!

Originally posted by brentbrent on 06 Dec 2009, 18:50.
Last edited by Bunuel on 28 Nov 2013, 06:51, edited 2 times in total.
Edited the question and added the OA
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Posts: 49496
Re: Combination problem - Princenten Review 2009 Bin 4 Q2  [#permalink]

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06 Dec 2009, 19:14
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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: $$\frac{1}{2^6}=\frac{1}{64}$$

$$P=1-(\frac{1}{64}+\frac{6}{64}+\frac{1}{64})=\frac{56}{64}=\frac{7}{8}$$

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]

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06 Dec 2009, 20:14
2
1
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

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]

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06 Dec 2009, 20:37
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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Joined: 02 Sep 2009
Posts: 49496
Re: Combination problem - Princenten Review 2009 Bin 4 Q2  [#permalink]

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07 Dec 2009, 03:42
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 $$6$$.
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Re: A fair 2 sided coin is flipped 6 times. What is the  [#permalink]

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11 Mar 2013, 13:50
1
"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]

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28 Nov 2013, 07:25
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: A fair 2 sided coin is flipped 6 times. What is the  [#permalink]

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21 Sep 2014, 21:05
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: $$\frac{1}{2^6}=\frac{1}{64}$$

$$P=1-(\frac{1}{64}+\frac{6}{64}+\frac{1}{64})=\frac{56}{64}=\frac{7}{8}$$

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

Hi Bunuel,

I understand the numerator part.
2C6 + 3C6 + 4C6 + 5C6 = 56

but how to calculate denominator part. I mean how can i count total no of combinations. I am not getting 64 .
Like in normal cases if we calculate for 6 ball, we take 6! as total no of combinations.
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Re: A fair 2 sided coin is flipped 6 times. What is the  [#permalink]

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22 Sep 2014, 01:26
1
him1985 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: $$\frac{1}{2^6}=\frac{1}{64}$$

$$P=1-(\frac{1}{64}+\frac{6}{64}+\frac{1}{64})=\frac{56}{64}=\frac{7}{8}$$

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

Hi Bunuel,

I understand the numerator part.
2C6 + 3C6 + 4C6 + 5C6 = 56

but how to calculate denominator part. I mean how can i count total no of combinations. I am not getting 64 .
Like in normal cases if we calculate for 6 ball, we take 6! as total no of combinations.

Each coin can land on heads or tails, so 2 ways. We have 6 coins, so total number of outcomes is 2*2*2*2*2*2 = 2^6.
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Re: If a fair two-sided coin is flipped 6 times, what is the probability  [#permalink]

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10 Sep 2016, 09:42
azamaka wrote:
If a fair two-sided coin is flipped 6 times, what is the probability that tails is the result at least twice but at most 5 times?

A) 5/8
B) 3/4
C) 7/8
D) 57/64
E) 15/16

Atleast twice but atmost 5 times could be written as 1 - P(No + Exactly once + All)

P(No time)=$$1/2^8$$
P(Exactly Once) = $$6/2^8$$
P(All) = $$1/2^8$$

So, Required P = 1- $$8/2^8$$ = 7/8. Hence, C
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Re: A fair 2 sided coin is flipped 6 times. What is the  [#permalink]

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21 Feb 2018, 22:53
Hi All,

In probability questions, there are two results that you can calculate - what you WANT to have happen or what you DON'T want to have happen. Since there are so many different ways to flip 2, 3, 4 or 5 tails, it will be easier for us to calculate what we DON'T want (0, 1 or 6 tails).

Since each toss has 2 possible outcomes (heads or tails), there are 2^6 = 64 different results for 6 coin flips.

Of those 64 options...

0 tails -->
HHHHHH = 1 option

1 tail -->
THHHHH
HTHHHH
HHTHHH
HHHTHH
HHHHTH
HHHHHT = 6 options

6 tails -->
TTTTTT = 1 option

1 + 6 + 1 = 8 options (of the 64) that we DON'T want...

Thus 64/64 - 8/64 = 56/64 = 7/8 that we DO want.

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Re: A fair 2 sided coin is flipped 6 times. What is the  [#permalink]

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15 Jun 2018, 01:38
brentbrent wrote:
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

Total # of Outcomes on flipping a coin for 6 times $$= 2^6 = 64$$

Favorable outcomes = getting a tails at least twice but not more than 5 times

Two Tails: TTHHHH = 6!/4!2! = 15

Three Tails: TTTHHH = 6!/3!3! = 20

Four Tails: TTTTHH = 6!/4!2! = 15

Five Tails: TTTTTH = 6!/5! = 6

Total # of favorable outcomes = 15 + 20 + 15 + 6 = 56

Required Probability = 56/64 = 7/8

Thanks,
GyM
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Re: A fair 2 sided coin is flipped 6 times. What is the  [#permalink]

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19 Jun 2018, 06:13
Thank you so much for posting the question.
Re: A fair 2 sided coin is flipped 6 times. What is the &nbs [#permalink] 19 Jun 2018, 06:13
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