Question

For the Experiment, flip a coin until heads shows, assume that the probability of heads on one flip is 3/5 we define a RV X= the number of flip 1)what are the possible values of X. Find the probability distribution function for X: gives the first four values, and then find a general formula for the probability that x=n

Answer 1

X can take any integer 1, 2, 3, 4, 5, ... until infinity
In fact this r.v is often said to have a Geometric distribution

Because essentially you can have
X = 1: then you got heads on the first flip
X= 2: then you got tails on 1st flip, then heads on the 2nd flip
X=3: you got tails on the first 2 flips, and heads on the 3rd flip
...
And continue like that. While it's very unlikely that you won't encounter heads within a "reasonable" number of flips in an actual experiment, you can't actually determine an absolute upper bound on which flip this will occur. It could theoretically only occur on the 1000000th flip. So that's why we say X has value 1, 2, 3, 4, 5, .... with no upper bound

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Probabilities for 1st 4 values:
since":
X = 1: then you got heads on the first flip -> prob heads = 3/5
X= 2: then you got tails on 1st flip, then heads on the 2nd flip -> prob. tails = 2/5
X=3: you got tails on the first 2 flips, and heads on the 3rd flip -> Prob(tails) and Prob(tail) and Prob(heads)
\[P(X=1)=\left( \frac{3}{5}\right)\\
P(X=2)=\left( \frac{2}{5}\right)\left( \frac{3}{5}\right)\\
P(X=3)=\left( \frac{2}{5}\right)\left( \frac{2}{5}\right)\left( \frac{3}{5}\right)=\left( \frac{2}{5}\right)^2\left( \frac{3}{5}\right)\\
P(X=4)=\left( \frac{2}{5}\right)\left( \frac{2}{5}\right)\left( \frac{2}{5}\right)\left( \frac{3}{5}\right)=\left( \frac{2}{5}\right)^3\left( \frac{3}{5}\right)\\
...
P(X=n)=\left( \frac{2}{5}\right)^{n-1}\left( \frac{3}{5}\right)\]

Answer 2

Thanks alot..now they are asking me to prove the sum of all the probabilities is 1 using the geometric formula..

Answer 3

Recall that \[ \sum_{n=0}^{\infty}ar^n=a\frac{1}{1-r}, \text{ if }|r|<1\]
Now your distribution starts at n=1, and goes to infinity, so we sum from n=1 to infinity:
\[ \sum_{n=1}^{\infty}\left( \frac{2}{5}\right)^{n-1}\left( \frac{3}{5}\right)\]
Re-index this to get a sum from n=0 to infinity:
\[ \sum_{n=0}^{\infty}\left( \frac{2}{5}\right)^{n}\left( \frac{3}{5}\right)=\frac{3}{5}\frac{1}{1-\frac{2}{5}}\]