An urn initially contains $1$red and $1$blue ball. At each stage, a ball is randomly withdrawn and replaced by two other balls of the same color. (For instance, if the red ball is initially chosen, then there would be $2$red and $1$blue balls in the urn when the next selection occurs.) Show by mathematical induction that the probability that there are exactly $i$ red balls in the urn after $n$ stages have been completed is$\frac{1}{n+1},1\le i\le n+1$

1 red and 1 blue ball in such a vase

Procedure: Take a ball as whim and restore two balls of such a colors here to vase.

Prove

The likelihood there'll be$i\in \{1,2,\dots ,n+1\}$red balls as in urn at draws can condensed as "likelihood that there would be $i$red orbs during $n$entries."

$P(i,n)=\frac{1}{n+1}\mathrm{\forall }i\in \{1,2,\dots ,n+1\},\mathrm{\forall }n\in \mathrm{\mathbb{N}}$

$B_n-$The blue ball is hit picked finally( $n-th$draw) within the game.

$R_n=B_n^c-$The red ball is picked finally ($n-th$draw)

Unless there are $i$red balls with in urn for $n-1$shows:

$P_i\left(R_n\right)=\frac{i}{n+1}{P}_i\left(B_n\right)=1-P\left(R_n\right)=\frac{n+1-i}{n+1}$

As indicated earlier, the the worth $i$signifies likelihood function unless there are $i$balls during $n-1$sweeps.

When included balls, one from each raffle, however with the opening are $n+1$balls as in urn, if $i$of them would be red, with both the randomized provide its indicated possibility.

If $n=1$

$P\left(R_1\right)=\frac{1}{2}$

$i\in \{1,2\}\text{probability is}P(i,1)=\frac{1}{2}$

$P(2,1)=P\left(R_1\right)=\frac{1}{2}$

$P(2,1)=P\left(B_1\right)=\frac{1}{2}$

$n_0\le n-1$

$i=\{1,2,\dots (n-1)+1\}$

$P(i,n)=P_1(i,n)P(1,n-1)+P_2(i,n)P(2,n-1)+\dots +P_n(i,n)P(n,n-1)$

$P_k(i,n)=0\text{if}k\ne i/i-1$

$k=i/i-1$

$P_i(i,n)=P_i\left(B_n\right)P_{i-1}(i,n)=P\left(R_n\right)$

The improved significantly estimate has been shortened to:

$P(i,n)=P_{i-1}\left(R_n\right)P(i-1,n-1)+P_i\left(B_n\right)P(i,n-1)$

The likelihood on $B_n\text{and}{R}_n$ have already got , and $P(i-1,n-1)$and $P(i,n-1)$already had been included supposition:

$P(i,n)=\frac{i-1}{n+1}\times \frac{1}{n-1+1}+\frac{n+1-i}{n+1}\times \frac{1}{n-1+1}$

$=\frac{p-1+n+1-p}{n+1}\times \frac{1}{n}$

$=\frac{1}{n+1}$

As a result, the argument stands for $n$then every potential $i$(boundary $i-s$are self-evident).

The proposed assertion is accurate with all $n\in \mathrm{\mathbb{N}}$as per the principles of mathematical induction.