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

More than 5 times tails = 6times +7times+8times = 8C6 + 8C7 + 8C8 = 37

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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