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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\(Prabability = \frac{Desired Outcomes}{Total Outcomes}\)


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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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)

Hi Bunuel,

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

Thanks

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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