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If 6 fair coins are flipped, what is the probability that there are...
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03 Mar 2018, 15:15
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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...
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03 Mar 2018, 17:05
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...
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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



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Re: If 6 fair coins are flipped, what is the probability that there are...
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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 \(\frac{1}{2}\):
\(P\left(\text{heads}>\text{tails}\right) = \frac{1}{2} \cdot \frac{11}{16} = \frac{11}{32}\)



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If 6 fair coins are flipped, what is the probability that there are...
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20 Mar 2018, 17:30
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
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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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Re: If 6 fair coins are flipped, what is the prob
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05 May 2018, 06:13
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
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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 HHHHHTEach 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
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05 May 2018, 06:40
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 tailsNow 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) =1As, 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...
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05 May 2018, 10:57



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Re: If 6 fair coins are flipped, what is the probability that there are...
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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 Answer is B Posted from my mobile device
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