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

Question

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)

drexxie · Accepted Answer

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(	ext{heads}>	ext{tails}ight) = \frac{1}{2} \cdot \frac{11}{16} = \frac{11}{32}

alexmonnin · Answer

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 =  *(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

vineet6316 · Answer

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

Therefore the required probability = (6C4+6C5+6C6)/(2^6) = 11/32

hoang221 · Answer

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)

Gladiator59 · Answer

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

SonalSinha803 · Answer

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

GMATBusters · Answer

Nice explanation:

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

GMATBusters · Answer

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.

Bunuel · Answer

Merging topics. Please check the discussion above.

push12345 · Answer

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

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards