91Ó°ÊÓ

Mathematical induction is a type of mathematical proof. It is mostly used to demonstrate that a proposition $P\left(n\right)$ holds for every natural number $n=0,1,2,3,4,5,...,$ i.e., that the overall assertion is a series of infinitely many examples.

$P_n$, After January $1$, the weather will be dry for a few days.

Define the probabilities:

$P\left(A_n\right)=P_n$

$P\left(A_nâˆ\pounds A_{n-1}\right)=p$

$P\left(A_n^câˆ\pounds A_{n-1}^c\right)=p$

$P_0=1$

Total probability formula:

$P\left(A_n\right)=P\left(A_nâˆ\pounds A_{n-1}\right)P\left(A_{n-1}\right)+P\left(A_nâˆ\pounds A_{n-1}^c\right)P\left(A_{n-1}^c\right)$ $nâ‰\yen 1$

first term on right hand is namely p

formula for the probability of a complement gives:

$$\begin{gathered} P\left(A_{n-1}^c\right)=1-P\left(A_{n-1}\right) \\ =1-P_{n-1} \end{gathered}$$

$$\begin{gathered} P\left(A_nâˆ\pounds A_{n-1}^c\right)=1-P\left(A_n^câˆ\pounds A_{n-1}^c\right) \\ =1-p \end{gathered}$$

Transferring total probability formula into:

$P_n=p{P}_{n-1}+(1-p)\left(1-P_{n-1}\right)nâ‰\yen 1$

which is equal to

$P_n=(2p-1)P_{n-1}+(1-p)nâ‰\yen 1$

$P_n=\frac{1}{2}+\frac{1}{2}×(2p-1{)}^n,nâ‰\yen 0$

For, $n=0$

$P_0=\frac{1}{2}+\frac{1}{2}×(2p-1{)}^0=1$

By the first half of the exercise, it is true for$n+1$that:

$P_{n+1}=(2p-1)P_n+(1-p)nâ‰\yen 0$

and because the formula holds for$n$

$P_{n+1}=(2p-1)\left[\frac{1}{2}+\frac{1}{2}×(2p-1{)}^n\right]+(1-p)$

The result of algebraic multiplication is,

$P_{n+1}=\frac{1}{2}×(2p-1{)}^{n+1}+\frac{1}{2}$

This formula is valid for every $n∈N$since this statement is true.