I asked a question on this site about the same scenario:

The probability of a head occurring when a biased coin is tossed is $p$, where $p < 1$. Let the random variable $X$ represent the number of tosses up to and including the first toss on which a tail occurs.

So I found out that the probability is $p^{n-1} (1-p)$

I now need to find out the $E(X)$ is (which I think is the same as the mean).

I tried tabulating the values and adding them but got nowhere. How do I get the $E(X)$ value?

The answer is meant to be $\frac{1}{1-p}$.

  • $\begingroup$ Yes $E(X)$ is the expectation, or also called the mean, of the RV $X$ and is computed for discrete positive integer-valued random variables by the formula (which ought to be in your textbook, notes, on wikipedia etc) $E(X)=\sum_{n=1}^\infty n P(X=n)$. $\endgroup$ Jun 3, 2019 at 1:55
  • $\begingroup$ Additionally, if this specific sum is giving you trouble, here is a hint: review geometric series from calculus. $\endgroup$ Jun 3, 2019 at 1:57
  • $\begingroup$ Is it correct to use the a1 / 1 - r formula? I tried this by substituting 1-p as a and p as r, but I could not get the right answer. $\endgroup$ Jun 3, 2019 at 2:23
  • $\begingroup$ the generous hint in the answer given below should be enough to conclude with some thought, what do you think? $\endgroup$ Jun 3, 2019 at 16:09

3 Answers 3


A bit more machinery than perhaps required. Using the law of total expectation we have that $$ \begin{align} EX&=E(X\mid X>1)P(X>1)+E(X\mid X=1)P(X=1)\\ &=(1+EX)p+1(1-p)\tag{0}\\ &=1+pEX \end{align} $$ where in (0) we used the fact that $P(X=1)=1-p$ (tails on the first try), $E(X\mid X=1)=E(1)=1$ and that $P(X>1)=p$ since the event $(X>1)$ corresponds to heads on the first toss. Finally $E(X\mid X>1)=1+EX$ since we failed to get a tail on the first toss and the process starts anew thereafter. Hence $$ EX=\frac{1}{1-p}. $$


Hint: To compute $\sum_{n=1}^\infty np^{n-1}$, try to use the fact: $$(\sum_{n=1}^\infty x^n)’=\sum_{n=1}^\infty nx^{n-1},$$ for $x\in [0,1)$.


$$P(X = k) = p^{k-1}\times (1-p)$$

This is a Hypergeometric Distribution.

$$E(X) = \sum_{k=1}^{\infty} k \times P(X=k)$$ $$\implies E(X) = \sum_{k=1}^{\infty} k \times p^{k-1}\times (1-p) \quad (1)$$

$$p\times E(X) = \sum_{k=1}^{\infty} k \times p^{k}\times (1-p) \quad (2)$$

Now do $(1)-(2)$.

$$(1-p) E(X) = (1-p)\times \left[\sum_{k=0}^{\infty} p^k\right]$$

$$\implies E(X) =\frac{1}{1-p}$$


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