Question

Diane tossed a coin 9 times and got 9 tails. Assume that Diane's coin is fair and answer each of the following questions:
a) What is the probability of the event of tossing a tail on the 10th toss?
b) What is the probability of the event of tossing 10 more tails in a row?
c) What are the odds against tossing 10 more tails in a row?

Answer 1

a) Since the coin is fair, then each trial is independent of each other (i.e. previous trials will not affect the outcome of other trials). So, the probability on the 10th toss of getting a tail is still 1/2

Answer 2

b) The probability of getting 10 more tails in a row... That means on the next 10 trials, you only tails on every flip. So:
1st flip: probabillity of tails is 1/2
2nd flip: probability of tails on the 1st and 2nd flip is 1/2*1/2
3rd flip: probability of tails on 1st, 2nd and 3rd flip is (1/2)^3
... etc. until 10 flips

Answer 3

To find the odds, I know the equation for odds in favor is P(E)=m/m+n and odds against is P(E)=n/m+n. What numbers do I plug in to find this?

Answer 4

The odds in favour of 10 more tails is exactly the result in b. You can thing of the odds against 10 more tails as being the complement of result in b) (i.e. do 1 minus the result in b))

It works since
\[ \frac{m}{m+n}+\frac{n}{m+n}=1\], so the odds in favour and agaisnt are complement events

Answer 5

can think of*

Answer 6

Okay, so the odds against getting 10 tails in a row would be 1-.0009765625, correct?

Answer 7

yes