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605-655 Level|   Probability|      
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aiming4mba
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A fair coin with sides marked heads and tails is to be tosse 01 Sep 2010, 11:15
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A
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45% (medium)

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68% (02:07) correct 32% (02:26) wrong based on 488 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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A
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Bunuel
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A fair coin with sides marked heads and tails is to be tosse 01 Sep 2010, 11:44
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aiming4mba
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
Show more

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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A fair coin with sides marked heads and tails is to be tosse 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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A fair coin with sides marked heads and tails is to be tosse 19 Aug 2013, 03:54
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aiming4mba
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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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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A fair coin with sides marked heads and tails is to be tosse 08 Dec 2015, 20:23
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Bunuel
aiming4mba
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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Hi Bunuel,

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

Thanks
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A fair coin with sides marked heads and tails is to be tosse 14 Apr 2017, 15:31
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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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A fair coin with sides marked heads and tails is to be tosse Updated on: 29 Jul 2025, 07:36
Originally posted by BrushMyQuant on 16 Oct 2022, 11:07.
Last edited by BrushMyQuant on 29 Jul 2025, 07:36, edited 1 time in total.
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[quote="aiming4mba"]

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

Coin is tossed 8 times => Total number of cases = \(2^8\) = 256

P(More than 5 Tails) = P(Getting 6 Tails) + P(Getting 7 Tails) + P(Getting 8 Tails)

P(Getting 6 Tails)

Out of the 8 places lets pick the 6 places where Tail will come. This can be in 8C6 ways
= \(\frac{8!}{6!*(8-6)!}\) = \(\frac{8*7*6!}{6!*2!}\) = 28 ways

P(Tail) = P(head) = \(\frac{1}{2}\) in each case

=> P(Getting 6 Tails) = Number of ways * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) = 28 * \((\frac{1}{2})^8\) = \(\frac{28}{256}\)

P(Getting 7 Tails) = P(Getting 1 Head)

Out of the 8 places lets pick the 1 place where Head will come. This can be in 8C7 ways = 8 ways

=> P(Getting 7 Tails) = Number of ways * \((\frac{1}{2})^8\) = 8 * \((\frac{1}{2})^8\) = \(\frac{8}{256}\)

P(Getting 8 Tails)

There is only on way in which we can get 8 Tails

=> P(Getting 8 Tails) = 1 * \((\frac{1}{2})^8\) = \(\frac{1}{256}\)

P(More than 5 Tails) = P(Getting 6 Tails) + P(Getting 7 Tails) + P(Getting 8 Tails) = \(\frac{28}{256}\) + \(\frac{8}{256}\) + \(\frac{1}{256}\) = \(\frac{37}{256}\)

So, Answer will be A
Hope it helps!

Playlist on Solved Problems on Probability here

Watch the following video to MASTER Probability with Coin Toss Problems

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A fair coin with sides marked heads and tails is to be tosse 30 Jul 2024, 04:38
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