We administer a drug only once, after the onset of symptoms of an illness. if k days after the first symptoms, we give $2^k$ ml of medicine. Treatment begins on the day the patient goes to the doctor for the first time. The probability of going to the doctor on any given day is $\frac{3}{4}$.Calculate the expected ml that the doctor will >administer.
If it is on the first day of symptoms k = 1 with probability 0.75
If it is on the second day of symptoms k = 2 with probability of (1-0.75)*0.75
If it is on the third day of symptoms k = 3 with probability of (1-0.75)*(1-0.75)*0.75
If it is on the fourth day of symptoms k = 4 with probability of $(1-0.75)^3*0.75$
If it is on the fifth day of symptoms k = 5 with probability of $(1-0.75)^{k-1}*0.75$
So far is it correct?
How do I find K?
I assume P=1 and then $(1-0.75)^{k-1}*0.75 = 1$ and then use log?
Thanks for any help.
Do it till infinity.
$E = \frac{3}{4}*2 + \frac{1}{4}\frac{3}{4}*2^2 + (\frac{1}{4})^2\frac{3}{4}*2^3 + ... = \frac{3}{4}(2 + 1 + 1/2 + 1/4 + ... ) = 3/4 * 4 = 3$