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Manager  Joined: 20 Nov 2009
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A fair coin with sides marked heads and tails is to be tosse  [#permalink]

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Difficulty:   45% (medium)

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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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;

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

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

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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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Senior Manager  Joined: 10 Jul 2013
Posts: 289
Re: A fair coin with sides marked heads and tails is to be tosse  [#permalink]

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1
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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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;

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

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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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.

A
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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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