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

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Manager
Joined: 20 Nov 2009
Posts: 126
A fair coin with sides marked heads and tails is to be tosse  [#permalink]

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01 Sep 2010, 10:15
8
00:00

Difficulty:

45% (medium)

Question Stats:

68% (02:12) correct 32% (02:18) wrong based on 276 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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Joined: 02 Sep 2009
Posts: 53063

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01 Sep 2010, 10:44
8
5
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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19 Aug 2013, 01:09
4
1
$$Prabability = \frac{Desired Outcomes}{Total Outcomes}$$

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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19 Aug 2013, 02:54
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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Asif vai.....

Manager
Joined: 10 Mar 2014
Posts: 188
Re: A fair coin with sides marked heads and tails is to be tosse  [#permalink]

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08 Dec 2015, 19:23
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;

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

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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14 Apr 2017, 14:31
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.

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

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18 Oct 2018, 13:04
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Re: A fair coin with sides marked heads and tails is to be tosse   [#permalink] 18 Oct 2018, 13:04
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