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A fair coin with sides marked heads and tails is to be [#permalink]
 08 Dec 2007, 09:36
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Question Stats:
70% (02:29) correct
30% (01:06) wrong based on 10 sessions
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
Please provide your explanation
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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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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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Re: PS: Another Probability [#permalink]
16 Jan 2008, 23:30
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
Please provide your explanation 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: 6 tails 2 heads 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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Re: PS: Another Probability [#permalink]
17 Jan 2008, 03:13
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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Re: PS: Another Probability [#permalink]
27 Jan 2008, 17:40
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
for 7 heads: 8/256
for 8 heads: 1/256
sum is 37/256
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Re: PS: Another Probability [#permalink]
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
Please provide your explanation 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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Re: PS: Another Probability [#permalink]
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
Please provide your explanation 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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Re: PS: Another Probability [#permalink]
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
Please provide your explanation 8C6 *1/2^8 + 8C7 *1/2^8 + 8C8 *1/2^8 = 28+8+1/256= 37/256
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Re: PS: Another Probability [#permalink]
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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Re: PS: Another Probability [#permalink]
02 Oct 2009, 01:12
I think the solution should be: (C(8,6) + (C8,7) + (8,8)) / 2^8
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Re: PS: Another Probability [#permalink]
03 May 2011, 07:41
1/2^ (8c6 + 8c7 + 8c8) 37/256
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Re: PS: Another Probability
[#permalink]
03 May 2011, 07:41
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