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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: 12
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
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Question Stats:

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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)
[Reveal] Spoiler: OA
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
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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
Joined: 15 Oct 2016
Posts: 34
If 6 fair coins are flipped, what is the probability that there are... [#permalink]

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04 Mar 2018, 00:11
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
Posts: 12
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
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 $$\frac{1}{2}$$:

$$P\left(\text{heads}>\text{tails}\right) = \frac{1}{2} \cdot \frac{11}{16} = \frac{11}{32}$$
Re: If 6 fair coins are flipped, what is the probability that there are...   [#permalink] 04 Mar 2018, 09:01
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