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;
\(P=\frac{28}{256}+\frac{8}{256}+\frac{1}{256}=\frac{37}{256}\).
Answer: A.
Hope it's clear.
Could you please clarify why are we multiplying by (1/2)^8.
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68% (02:12) correct 
