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7 votes
7 votes

Let $C$ be a biased coin such that the probability of a head turning up is $p.$ Let $p_n$ denote the probability that an odd number of heads occurs after $n$ tosses for $n \in \{0,1,2,\ldots \},$ Then which of the following is TRUE ?

  1. $p_n=\dfrac{1}{2}\text{ for all }n \in \{0,1,2,\ldots \}.$
  2. $p_n=(1-p)(1-p_{n-1})+p.p_{n-1}\text{ for }n\geq 1\text{ and }p_{0}=0.$
  3. $p_{n}=\sum_ {i=1}^{n}p(1-2p)^{i-1} \text{ for }n\geq 1.$
  4. $\text{If } p=\dfrac{1}{2},\text{ then } p_{n}=\dfrac{1}{2} \text{ for all }n \in \{0,1,2,\ldots\}$.
  5. $p_{n}=1 \text{ if } n \text{ is odd and } 0 \text{ otherwise}.$
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2 Answers

Best answer
8 votes
8 votes
$p_0 = 0$

$p_1 = p$

$p_2 = {}^2C_1 p(1-p)$

$p_3 = {}^3C_1 p(1-p)^2 +{}^3C_3 p^3$

$\vdots$

$p_n = {}^nC_1p(1-p)^{n-1} + {}^nC_3p^3(1-p)^{n-3}+ \ldots $

$\quad =\frac{1}{2} \left[((1-p) + p)^n - ( (1-p) -p)^n \right]$
(Since $(a+b)^n = {}^nC_0a^n +{}^nC_1a^{n-1}b+\ldots+{}^nC_nb^n)$ and $(a-b)^n = {}^nC_0a^n -{}^nC_1a^{n-1}b+\ldots+ (-1)^n {}^nC_nb^n)$

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

From option C,

$\sum_ {i=1}^{n}p(1-2p)^{i-1}  = p \sum_ {i=1}^{n}(1-2p)^{i-1}$

$= p \frac{1 - (1-2p)^{n}}{2p}$ (Sum to $n$ terms of GP with $a = 1, r = (1-2p)$

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

So, correct option: C.
2 votes
2 votes
option C

$P_n$ is event that odd number of heads after occur $n$ tosses.

$P_0=0$

$P_1=p (H)$

$P_2=2p(1-p) (HT, TH)$

$P_3=3p(1-p)(1-p)+p^3 (HTT,HHH,THT,TTH)$

$p_4=4(p^{3})(1-p)+4p(1-p)^{3}$ (HHHT(4 permutations), HTTT(4 permutations))

Clearly option A,E is false, option B is false for $P_1,P_2$...(check by substituting values)

Take $P_2$, using given equation in option B we get $P_2=2p^2-3p+2$ but actual answer is $2p(1-p)$

option D is false as $P_0=0$.
Answer:

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