|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
63% (02:08) correct
37%
(02:23)
wrong
based on 397
sessions
History
Date
Time
Result
Not Attempted Yet
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. \(\frac{3}{16}\)
Source: Brilliant.com
A. \(\frac{11}{16}\)
B. \(\frac{11}{32}\)
C. \(\frac{7}{16}\)
D. \(\frac{3}{8}\)
E. \(\frac{3}{16}\)
Source: Brilliant.com
04 Mar 2018, 09:01
Kudos
Bookmarks
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 \(\frac{1}{2}\):
\(P\left(\text{heads}>\text{tails}\right) = \frac{1}{2} \cdot \frac{11}{16} = \frac{11}{32}\)
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 \(\frac{1}{2}\):
\(P\left(\text{heads}>\text{tails}\right) = \frac{1}{2} \cdot \frac{11}{16} = \frac{11}{32}\)
General Discussion
03 Mar 2018, 17:05
Kudos
Bookmarks
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
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
