0

$n$ 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.

user69608
  • 858
  • "I want to understand this question by my method" Your method being constructing a sum with $2^{n-1}$ different summands, each summand being a product of an odd number of terms from (1) and whose total number of terms is $n$? Such a sum is ugly and unwieldy. I can not in good conscience suggest you continue with that approach for this problem. – JMoravitz Aug 08 '20 at 14:49
  • 1
    It is ...fine... for small values of $n$. But for general values of $n$, you really need to be looking for a better argument like the one for recursion. – JMoravitz Aug 08 '20 at 14:54
  • @JMoravitz may be once it is somehow solvable it may result in an elegant way.and in the linked question even user Lulu states " Of course, the simplicity of the final form suggests that there might be a direct argument for it." – user69608 Aug 08 '20 at 14:54
  • And may be that direct argument was meant for this method. – user69608 Aug 08 '20 at 14:55

2 Answers2

1

$$P(not H)-P(H)=\left(\frac23-\frac13\right)\left(\frac45-\frac15\right) \cdots \left(\frac{2n}{2n+1}-\frac1{2n+1}\right)$$
When there is an odd number of heads, the term is negative; with an even number of heads the term is positive.

Empy2
  • 50,853
1

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

acat3
  • 11,897
  • why P(H) took even number of terms from equation no. 2? – user69608 Aug 08 '20 at 15:34
  • Do you mean probability of even number of heads being the positive terms? Because we make the probability of getting head the negative terms in the line after “consider the product”. Product of odd number of negative terms is negative, product of even number of negative terms is positive – acat3 Aug 08 '20 at 15:48
  • no i meant why P(H) took even number of terms from equation no. 2, "equation 2 in my post" – user69608 Aug 08 '20 at 15:49
  • here "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" – user69608 Aug 08 '20 at 15:50
  • not sure about that, i think the only requirement is odd number of heads, we can have odd or even number of tails. – acat3 Aug 08 '20 at 16:01