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

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)


\(Prabability = \frac{37}{2^8} = \frac{37}{256}\)

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

[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!

Watch the following video to learn How to Solve Probability with Coin Toss Problems


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