Tosses |
Probability |
1 |
p |
2 |
(1-p)p |
3 |
(1-p)(1-p)(p) |
4 |
(1-p)(1-p)(1-p)p |
. |
. |
. |
. |
. |
. |
. |
. |
n |
$(1-p)^{n-1} \times p$ |
$E = 1 \times p + 2 \times p(1-p) + 3 \times (1-p)^{2}p + ......... $
$E = \Large \sum_{n=1}^{\infty} n \times (1-p)^{n-1} \times p $
$E = \Large p \sum_{n=1}^{\infty} n \times (1-p)^{n-1} $
$E = \Large p \sum_{n=0}^{\infty} (n+1) \times (1-p)^{n} $
and we know that,
$\Large \sum_{k=0}^{\infty} (k+1)x^k = 1 + 2x + 3x^2 + 4x^3+..... = \frac{1}{(1-x)^2}$
Hence,
$E = \frac{1}{(1-(1-p))^2} \times p$
$E = \frac{1}{(p)^2} \times p$
$E \large = \frac{1}{p}$