Question

What is the probability that  tails in a row with a coin for which the probability of heads is ? Given a fraction , with 0 <  < 1, show that the number of flips one must perform to be  sure of obtaining at least one head is  =    .

Answer 1

P(T in a row) = (.5) ^n where n is the number of flips. so if P(H) = x = 1 - (.5)^n solving that equation will give you n.

Answer 2

The probability of no heads is:
\[(1/2)^{n}\]
where n is the number of flips
The probability that a head is obtained is:
\[1-(1/2)^{n}\]
Let the probability that a head is obtained be equal to 0.99 to be "sure" that a head is obtained.
Find n so that:\[1-(1/2)^{n}>0.99\]
\[(1/2)^{n}<0.01\]
n log (1/2) < log (1/100)
-n log 2 < -log 100
n > (log 100)/(log 2) = 6.64
Therefore 7 flips are needed to be sure of obtaining at least one head.