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# A fair coin with sides marked heads and tails is to be

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A fair coin with sides marked heads and tails is to be [#permalink]

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08 Dec 2007, 09:36
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Difficulty:

55% (hard)

Question Stats:

65% (01:51) correct 35% (01:42) wrong based on 250 sessions

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A fair coin with sides marked heads and tails is to be tossed eight times. What is the probability that the coin will land tails side up more than five times?

A. 37/256
B. 56/256
C. 65/256
D. 70/256
E. 81/256

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-fair-coin-with-sides-marked-heads-and-tails-is-to-be-tosse-100222.html
[Reveal] Spoiler: OA

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Director
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08 Dec 2007, 10:18
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Is it A?

Landing on tails more than 5 times means tails has to hit 6, 7 or 8 times.

8!/6!2! = 7*4 = 28
8!/7!1! = 8
8!/8! = 1

28 + 8 + 1 = 37/256

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16 Jan 2008, 23:21
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tarek99 wrote:
OA is A.

A is correct:

P= nCm * p^m * q^n-m= 8C6*1/2^6 * 1/2^2 + 8C7 * 1/2^7 *1/2 + 8C8 1/2^8=28+8+1/256= 37/256

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16 Jan 2008, 23:30
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tarek99 wrote:
A fair coin with sides marked heads and tails is to be tossed eight times. What is the probability that the coin will land tails side up more than five times?

a) 37/256

b) 56/256

c) 65/256

d) 70/256

e) 81/256

This actually is not that bad at all. I did it in less than 2 min in my head. Just gotta think about it logically.

List the winning scenarios:
1/2^8 we have 8!/6!2! ways to arrage TTTTTTHH so 28/256
7 tails 1 head 8!/1!7! ---> 8/256

8 tails 8!/8! --> 1/256

28+8+1 = 37 --> 37/256

A

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17 Jan 2008, 03:13
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Expert's post
my 2 cents

I tried to go from the other side and made a mistake.

the wrong logic: the probability more than 4 tails is $$\frac12$$ (due to symmetry), the probability for 5 tails is $$\frac{C^8_5}{2^8}=\frac{56}{256}$$. therefore, $$p=\frac12-\frac{56}{256}=\frac{72}{256}$$

the right logic: we have 9 variants for the number of tails: 0,1,2,3,4,5,6,7,8. Therefore, correct formula with symmetry approach is:
$$p=\frac12-\frac12*\frac{C^8_4}{2^8}-\frac{C^8_5}{2^8}=\frac{128}{256}-\frac12*\frac{70}{256}-\frac{56}{256}=\frac{37}{256}$$
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27 Jan 2008, 17:40
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getting the hang of using binomial theorem for these types of questions:

answer will be sum of probabilities of getting 6 , 7 or 8 heads

for 6 heads: (8C6)*(1/2)^6*(1/2)^2 = 28/256

sum is 37/256

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27 Jan 2008, 19:45
tarek99 wrote:
A fair coin with sides marked heads and tails is to be tossed eight times. What is the probability that the coin will land tails side up more than five times?

a) 37/256

b) 56/256

c) 65/256

d) 70/256

e) 81/256

First mistake which I did, I calculated it as "at least 5 times", but question clearly states "more" than five times! I should be more careful in reading the questions!

This is basically Bernoulli's formula.

$$8C6*(1/2)^6*(1/2)^2 + 8C7*(1/2)^7*(1/2) + 8C8*(1/2)^8$$

(1/2^8)*[ 8C6 + 8C7 + 8C8 ]

= 37/2^8

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27 Jan 2008, 22:46
LM wrote:
tarek99 wrote:
A fair coin with sides marked heads and tails is to be tossed eight times. What is the probability that the coin will land tails side up more than five times?

a) 37/256

b) 56/256

c) 65/256

d) 70/256

e) 81/256

First mistake which I did, I calculated it as "at least 5 times", but question clearly states "more" than five times! I should be more careful in reading the questions!

This is basically Bernoulli's formula.

$$8C6*(1/2)^6*(1/2)^2 + 8C7*(1/2)^7*(1/2) + 8C8*(1/2)^8$$

(1/2^8)*[ 8C6 + 8C7 + 8C8 ]

= 37/2^8

How my mistake the same to your! I start off by 5 coins landing tail. And one more, I calculate 2^8 is only 64. Comparing with Answer series, my god, I made a stupid mistake.

Cheer!
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25 Aug 2008, 12:38
tarek99 wrote:
A fair coin with sides marked heads and tails is to be tossed eight times. What is the probability that the coin will land tails side up more than five times?

a) 37/256

b) 56/256

c) 65/256

d) 70/256

e) 81/256

8C6 *1/2^8 + 8C7 *1/2^8 + 8C8 *1/2^8
= 28+8+1/256= 37/256
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27 Sep 2009, 20:50
A fair coin with sides marked heads and tails is to be tossed eight times. What is the probability that the coin will land tails side up more than five times?

a) 37/256

b) 56/256

c) 65/256

d) 70/256

e) 81/256

Soln: For more than 5 times it can be broken into
= Prob(that tails appears 6 times) + Prob(that tails appears 7 times) + Prob(that tails appears 8 times)
using formula nCr * p^r * q^(n-r)
= 8C6 * (1/2)^6 * (1/2)^2 + 8C7 * (1/2)^7 * (1/2)^1 + 8C8 * (1/2)^8 * (1/2)^0
= 37/(2^8)

A it is

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02 Oct 2009, 01:12
I think the solution should be:

(C(8,6) + (C8,7) + (8,8)) / 2^8
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03 May 2011, 07:41
1/2^ (8c6 + 8c7 + 8c8)
37/256
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Re: A fair coin with sides marked heads and tails is to be [#permalink]

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09 Dec 2013, 02:57
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Re: A fair coin with sides marked heads and tails is to be [#permalink]

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09 Dec 2013, 03:27
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A fair coin with sides marked heads and tails is to be tossed eight times. What is the probability that the coin will land tails side up more than five times?
A. 37/256
B. 56/256
C. 65/256
D. 70/256
E. 81/256

The probability that the coin will land tails side up more than five times equals to the sum of the probabilities that coin lands 6, 7 or 8 times tails side up.

$$P(t=6)=\frac{8!}{6!2!}*(\frac{1}{2})^8=\frac{28}{256}$$, we are multiplying by $$\frac{8!}{6!2!}$$ as the scenario tttttthh can occur in $$\frac{8!}{6!2!}=28$$ # of ways (tttttthh, ttttthth, tttthtth, ... the # of permutations of 8 letters tttttthh out of which 6 t's and 2h's are identical);

$$P(t=7)=\frac{8!}{7!}*(\frac{1}{2})^8=\frac{8}{256}$$, the same reason of multiplication by $$\frac{8!}{7!}=8$$;

$$P(t=8)=(\frac{1}{2})^8=\frac{1}{256}$$, no multiplication as the scenario tttttttt can occur only in one way;

$$P=\frac{28}{256}+\frac{8}{256}+\frac{1}{256}=\frac{37}{256}$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-fair-coin-with-sides-marked-heads-and-tails-is-to-be-tosse-100222.html
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Re: A fair coin with sides marked heads and tails is to be   [#permalink] 09 Dec 2013, 03:27
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