SOLUTION: Please help me solve this question. A coin is biased so that it is twice as likely to show heads as tails. a) What is the probability that the coin will show heads when it is tos

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Question 1111734: Please help me solve this question.
A coin is biased so that it is twice as likely to show heads as tails.
a) What is the probability that the coin will show heads when it is tossed?
b) Andy tosses the coin n times. Find the least value of n for which the probability that the coin shows heads each time is less than 0.01.
so for part a) i got two thirds as the probability of getting heads and a third as probability of getting tails. I'm now stuck on part b)
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
twice as likely to show heads as tails would be 2/3, as you stated.

i think for part b, you would have the equation (2/3)^n = .01

you would then solve for n.

this equation will tell you the value of n that gets exactly .01.

take the log of both sides of the equation to get:

log((2/3)^n) = log(.01)

since log((2/3)^n) is equal to n * log(2/3), your equation becomes:

n * log(2/3) = log(.01)

divide both sides of the equation by log(2/3) to get:

n = log(.01) / log(2/3)

solve for n to get:

n = 11.35774717

when n = 11.35774717. (2/3)^n = .01

therefore, when n = 12, (2/3)^n will be less than .01.

this assumes n has to be an integer and that the result must less than .01

(2/3)^11 = .011... > .01
(2/3)^12 = .007... < .01

use of logs makes this easier to solve.

i'm not sure how you would do it without the use of logs, except for trial and error, either going up sequentially, or trying to narrow it down using some other logic.

if you graphed the equation, your graph would look like this.