Question

Flip a coin until heads shows, assume that the probability of heads on one flip is \(\large \frac{3}{5}\). We define a RV (random variable) X = the # of flips.

1. What are the possible values of X?
2. Find the probability distribution function for X: give the first four values then find a general formula for the probability that X = n.

Answer 1

this is a geometric distribution

http://en.wikipedia.org/wiki/Geometric_distribution

X = 1,2,3,4 ...

\[P(X=n) = (\frac{3}{5}) (\frac{2}{5})^{n-1}\]

Answer 2

So "possible values of X" just wants a generic formula? Thank you.

If I wanted to sum this up using the formula for the sum of a geometric series, would this work?

\(\huge \frac{3}{5}*\frac{1-\frac{2}{5}^n}{1-\frac{2}{5}}\)?

Answer 3

yes that would work :)

Answer 4

n or n-1?

Answer 5

you can simplify that though .... 1-2/5 = 3/5
the 3/5 cancel

Answer 6

n

Answer 7

ah, so we're left with \(\huge \frac{3}{5}\frac{\frac{3}{5}^n}{\frac{3}{5}}=\frac{3}{5}*\frac{3}{5}^{n-1}\)?

Answer 8

that's probably wrong because of pemdas bleh

Answer 9

oh careful

\[1- a^n \ne (1-a)^n\]

Answer 10

also if you go to wiki link
the sum should equal the cumulative distribution

----> 1 - (2/5)^n

Answer 11

We have to prove that the sum of all the probabilities is 1.

\(\large \cancel{\frac{3}{5}}*\frac{1-\frac{3}{5}^n}{\cancel{\frac{3}{5}}}\)

Answer 12

\(\large \cancel{\frac{3}{5}}*\frac{1-\frac{2}{5}^n}{\cancel{\frac{3}{5}}}\) fixed

Answer 13

oh ok thats easy, the sum of all probabilities is the infinite sum of the series
let n ->infinity , what happens?

Answer 14

1-0? oh wow, thanks

Answer 15

yep, yw :)