How can I compute $\Bbb{P}(M_n=m_n|M_1=m_1,...M_{n-1}=m_{n-1})$?

Question

Let me define $X_n$ an i.i.d. sequence of random variables which are Bernoulli distributed with parameter $p$. Then for $M_n=|\{k:1\leq k\leq n, X_k=0\}|$ I need to compute

$$\Bbb{P}(M_n=m_n|M_1=m_1,...M_{n-1}=m_{n-1})$$

But now could maybe someone help me how to compute the probability since I don't see how to work with the cardinality in this case. I think it should be equal to $\Bbb{P}(M_n=m_n|M_{n-1}=m_{n-1})$ but this is only intuitively.

Thanks for your help.

Answer 1

Initial observation is that $M_n$ counts the number of 'fails' in the $n$ Bernoulli $X_{i}'s$. Thus $M_{n}$ can either equal $m_{n-1}$ or $m_{n-1} + 1$

Let's start with smaller n and examine what we're dealing with. For n = 2,

If $M_2=m_1 + 1$

$$\Bbb{P}(M_2=m_2|M_1=m_1) = \Bbb{P}({X_2=0}) = 1-p$$

If $M_2=m_1$

$$\Bbb{P}(M_2=m_2|M_1=m_1) = \Bbb{P}({X_2=1}) = p$$

Can you take it from there