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# If 6 fair coins are flipped, what is the probability that there are...

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Intern
Joined: 03 Oct 2016
Posts: 13
GMAT 1: 550 Q42 V25
If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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03 Mar 2018, 15:15
1
6
00:00

Difficulty:

65% (hard)

Question Stats:

68% (01:25) correct 32% (02:02) wrong based on 168 sessions

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If 6 fair coins are flipped, what is the probability that there are more heads than tails?

A: $$\frac{11}{16}$$

B: $$\frac{11}{32}$$

C: $$\frac{7}{16}$$

D: $$\frac{3}{8}$$

E: $$\frac{3}{16}$$

OA will be provided shortly!

(Hint: Think of symmetry; Source: brilliant)
Intern
Joined: 19 Dec 2017
Posts: 40
Location: United States (OH)
Concentration: General Management, Strategy
GPA: 3.79
WE: Engineering (Manufacturing)
Re: If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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03 Mar 2018, 17:05
1
The probability that there are more heads than tails can be translated to the probability of 4, 5, or 6 heads.

Probability of 4 Heads: 6C4*(1/2)^6 = [6*5*4*3/(4*3*2*1)]*(1/2)^6 = 15/64
Probability of 5 Heads: 6C5*(1/2)^6 = 6*(1/2)^6 = 6/64
Probability of 6 Heads: 6C6*(1/2)^6 = 1/64

15/64 + 6/64 + 1/64 = 22/64 = 11/32 Answer is B
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Intern
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Posts: 31
If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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04 Mar 2018, 00:11
1
drexxie wrote:
If 6 fair coins are flipped, what is the probability that there are more heads than tails?

A: $$\frac{11}{16}$$

B: $$\frac{11}{32}$$

C: $$\frac{7}{16}$$

D: $$\frac{3}{8}$$

E: $$\frac{3}{16}$$

OA will be provided shortly!

(Hint: Think of symmetry; Source: brilliant)

For more heads, we need 4 Hs OR 5 Hs OR 6 Hs.

Therefore the required probability = (6C4+6C5+6C6)/(2^6) = 11/32
Intern
Joined: 03 Oct 2016
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GMAT 1: 550 Q42 V25
Re: If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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04 Mar 2018, 09:01
3
OAE (brilliant):

The probability that there are the same number of heads and tails is:

$$\frac{\binom{6}{3}}{2^6} = \frac{5}{16}$$

There are $$6C3$$ ways to construct a sequence with 3 heads and 3 tails. The probabilities for the three options must sum to 1, so:

$$1- \frac{5}{16} = \frac{11}{16}$$

However, since this probability includes the two cases 'either the number of heads > tails or the number of tails > heads', we need, by symmetry, to multiply this probability by $$\frac{1}{2}$$:

$$P\left(\text{heads}>\text{tails}\right) = \frac{1}{2} \cdot \frac{11}{16} = \frac{11}{32}$$
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Concentration: Finance
GMAT 1: 710 Q50 V36
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If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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20 Mar 2018, 17:30
2
drexxie wrote:
If 6 fair coins are flipped, what is the probability that there are more heads than tails?

A: $$\frac{11}{16}$$

B: $$\frac{11}{32}$$

C: $$\frac{7}{16}$$

D: $$\frac{3}{8}$$

E: $$\frac{3}{16}$$

OA will be provided shortly!

(Hint: Think of symmetry; Source: brilliant)

Because number of heads is more than tails, there are 3 scenario that can happen:

4 Head vs 2 Tails: The probability in this scenario will be: $$\frac{6!}{4!*2!} * \frac{1}{(2^6)} = \frac{15}{64}$$
5 Head vs 1 Tails: The probability in this scenario will be: $$\frac{6!}{5!} * \frac{1}{(2^6)} = \frac{6}{64}$$
6 Head vs 0 Tails: The probability in this scenario will be: $$\frac{1}{(2^6)} = \frac{1}{64}$$

Total: $$\frac{15}{64} + \frac{6}{64} + \frac{1}{64} = \frac{22}{64} = \frac{11}{32}$$ => Answer (B)
Manager
Joined: 16 Sep 2016
Posts: 225
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If 6 fair coins are flipped, what is the prob  [#permalink]

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05 May 2018, 05:43
If 6 fair coins are flipped, what is the probability that there are more heads than there are tails?

A. 1/2
B. 11/16
C. 6C3/2^6
D. 11/32
E. 6/13

Source: Brilliant.com
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Joined: 14 Feb 2018
Posts: 379
Re: If 6 fair coins are flipped, what is the prob  [#permalink]

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05 May 2018, 06:13
2
1
Following are the possibilities -
HHHHTT or HHHHHT or HHHHHT

Each can also be arranged in the following respective ways -

6!/(4! 2!) or 6!/(5!) or 1 which sums up to 22.

Total no of possibilities = 2^6 = 64.

Thus, required probability = 22/64 = 11/32.

Thus, IMO D.

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Location: India
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Re: If 6 fair coins are flipped, what is the prob  [#permalink]

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05 May 2018, 06:27
Nice explanation:

There is a typo in your explanation,
Possibilities shall be
"HHHHTT or HHHHHT or HHHHHH"

SonalSinha803 wrote:
Following are the possibilities -
HHHHTT or HHHHHT or HHHHHT
Each can also be arranged in the following respective ways -

6!/(4! 2!) or 6!/(5!) or 1 which sums up to 22.

Total no of possibilities = 2^6 = 64.

Thus, required probability = 22/64 = 11/32.

Thus, IMO D.

Sent from my Lenovo K53a48 using GMAT Club Forum mobile app

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Re: If 6 fair coins are flipped, what is the prob  [#permalink]

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05 May 2018, 06:40
1
1
Hi
I would like to present a different approach to solve this question:

Since there are even number of coins, we can have three cases:
2) More tails
3) Equal number of Heads & tails

It is very clear that the probability of getting more heads = probability of getting more tails
Now first we need to find the third case when number of heads = number of tails
we will have 3 heads, 3 tails : No of cases = 6!/(3!*3!) = 20
total no of cases = $$2^6$$= 64
Probability (number of heads = number of tails) =20/64 = 5/16

Now P(number of heads = number of tails) +P(getting more heads) +P(probability of getting more tails) =1
As, probability of getting more heads = probability of getting more tails = P (say)

$$5/16$$ + P+ P =1
P = $$11/32$$

Why is this approach important?
If number of coins had been 60 (larger number) in place of 6, then doing this question by the other method would be quite tedious.
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Re: If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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05 May 2018, 10:57
If 6 fair coins are flipped, what is the probability that there are more heads than there are tails?

A. 1/2
B. 11/16
C. 6C3/2^6
D. 11/32
E. 6/13

Source: Brilliant.com

Merging topics. Please check the discussion above.
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Posts: 599
Re: If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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06 May 2018, 13:04
4H2T. ,5H1T. ,6H

Probability of 4 Heads two tail: 6C4*(1/2)^6 = 15/64
Probability of 5 Heads one tail: 6C5*(1/2)^6 = 6/64
Probability of 6 Heads no tail: 6C6*(1/2)^6 = 1/64

15/64 + 6/64 + 1/64 = 22/64 = 11/32

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Re: If 6 fair coins are flipped, what is the probability that there are... &nbs [#permalink] 06 May 2018, 13:04
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