Question

Independent trials that result in a success with probability $p$and a failure with probability $1-p$are called Bernoulli trials. Let $P_n$denote the probability that n Bernoulli trials result in an even number of successes ($0$ being considered an even number). Show that

$P_n=p\left(1-P_{n-1}\right)+(1-p)P_{n-1}n\ge 1$

Short answer

Condition upon the outcome of the first trial. Independence make the following trials, $n-1$Bernoulli trials.

For the step of the induction substitute explicit formula for $P_n$into recursive formula.

Step-by-step solution

conditions for the problems

Conditioning to the first trial can solve the problem.

If and only if one of the two disjoint situations below is true, we will get an even number of successes:

1. The first trial was a success, and in $n-1$trials, an odd number of successes occurs.
2. The first trial ended in failure, and $n-1$trials result in an even number of successes.

Independence between the trials

Because of the recursive relationship, the trials are completely independent of one another. in other words,

$P_n=p\left(1-P_{n-1}\right)+q{P}_{n-1},\forall n\ge 1$

$n=1$is included since we anticipated $0$to be an even integer, hence $P_0=1$.

Derivation by Induction

In the other hand, the demonstration by induction goes like this,

$n=1,\frac{1+1-2p}{2}=q=P_1$

$n$, For a universal natural number $n$, we suppose that the equality holds.

$n+1$, Substituting$P_n=\frac{1+(1-2p{)}^n}{2}$