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

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

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New post 03 Mar 2018, 15:15
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A
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Question Stats:

59% (02:09) correct 41% (02:22) wrong based on 180 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)
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Re: If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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New post 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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If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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New post 04 Mar 2018, 00:11
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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
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Re: If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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New post 04 Mar 2018, 09:01
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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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If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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New post 20 Mar 2018, 17:30
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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)
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If 6 fair coins are flipped, what is the prob  [#permalink]

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

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New post 05 May 2018, 06:13
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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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Re: If 6 fair coins are flipped, what is the prob  [#permalink]

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New post 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.

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

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New post 05 May 2018, 06:40
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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:
1) More Heads
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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New post 05 May 2018, 10:57
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Re: If 6 fair coins are flipped, what is the probability that there are...  [#permalink]

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New post 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
Answer is B

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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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