$n$ biased coins, with $m_{th}$ coin having probability of throwing head equal to $\frac{1}{2m+1}$
$(m = 1, 2, ….., n)$, are tossed once. The probability of getting an odd numbers of heads, if
results for each coin are independent is ?
Let
$H_i$ = the event of getting a head on $i_{th}$ coin
$H$= getting odd number of heads on throwing n coins once
Coin number: 1 2 3 … m …. n
$P(H_i) =1/3,1/5, 1/7 …. 1/(2m+1)..1/(2n+1) …1$
$P(\bar{H_i})=2/3,4/5,6/7…2m/(2m+1)…2n/(2n+1) …2$
Let $P(H)$ denotes the sum of series which has each term consisting of product of an odd number of terms
from (1) and even number of terms from (2) such that total number of factors in each term is n
My book states :$P(\bar{H})-P(H)=1/(2n+1)$ and I don't understand this.
$P(\bar{H})+P(H)=1$
Hence $P(H)=n/(2n+1)$
Related:Probability of getting an odd number of heads if n biased coins are tossed once.
NOTE:The above question doesn't answers my question the way I want to do,They have used recursion.
I want to understand this question by my method.Please consider adding details.
Best Answer
Consider the product
$\prod_{i=1}^{n}{\left(\frac{-1}{2i+1}+\frac{2i}{2i+1}\right)}$
Notice that the sum of the negative terms is the probability of getting odd number of heads and the sum of the positive terms is the probability of getting even number of heads. Furthermore
$\prod_{i=1}^{n}{\left(\frac{-1}{2i+1}+\frac{2i}{2i+1}\right)}=\prod_{i=1}^{n}{\left(\frac{2i-1}{2i+1}\right)}=\frac{1}{2n+1}$
Therefore
$P(even)-P(odd)=\frac{1}{2n+1}$
And
$P(even)+P(odd)=1$
There you go