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

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

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New post 01 Sep 2010, 11:15
1
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Question Stats:

68% (02:10) correct 32% (02:24) wrong based on 300 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

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Re: Probability question  [#permalink]

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New post 01 Sep 2010, 11:44
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aiming4mba 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


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;

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

Answer: A.

Hope it's clear.
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Re: Probability question  [#permalink]

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New post 19 Aug 2013, 02:09
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\(Prabability = \frac{Desired Outcomes}{Total Outcomes}\)


Total Outcomes = Coin has to be tossed 8 times and Each time it tossed will give any one of two results i.e. Head or tail. So Total Possible Outcomes = 2 X 2 X ......8 times = \(2^8\)


Desired Outcomes = We want Coin be landed with head up more than 5 times i.e. 6 times or 7 times or 8 times. It is same as choosing 6 places from 8 places OR choosing 7 places from 8 places OR choosing 8 places from 8 places = 8C6 + 8C7 + 8C8 = 28 + 8 + 1 = 37 (Remember AND=Multiplication, OR=Addition)


\(Prabability = \frac{37}{2^8} = \frac{37}{256}\)
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Re: A fair coin with sides marked heads and tails is to be tosse  [#permalink]

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New post 19 Aug 2013, 03:54
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aiming4mba 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

More than 5 times tails = 6times +7times+8times = 8C6 + 8C7 + 8C8 = 37

- - - - - - - -
2 2 2 2 2 2 2 2

2^8 times total events and 37 events where tails side up .

So probability = 37/2^8 = 37/256 (Answer A)
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Re: A fair coin with sides marked heads and tails is to be tosse  [#permalink]

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New post 08 Dec 2015, 20:23
Bunuel wrote:
aiming4mba 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


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;

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

Answer: A.

Hope it's clear.


Hi Bunuel,

Could you please clarify why are we multiplying by (1/2)^8.

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

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New post 14 Apr 2017, 15:31
8C6 * (1/2)^8 = 28 * (1/2)^8
8C7 * (1/2)^8 = 8 * (1/2)^8
8C8 * (1/2)^8 = 1 * (1/2)^8

28+8+1 = 37.

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Re: this will probably be fairly easy for some of you, but i'm  [#permalink]

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Re: this will probably be fairly easy for some of you, but i'm   [#permalink] 30 Sep 2019, 06:05
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