2
votes
Tossing two coins with unknown success probability
The probability is not hard to compute. I will be using the same argument as here. Simulate, all independently,
$$
\begin{align}
X_1, X_2, \dots, X_{N+1} &\sim \text{Unif}[0,1), \quad p_1:=X_1,\\
...
- 6,032
1
vote
Probability of a white ball chosen from a bag if there are two bags.
It's easier not to use Bayes thereom to calculate $P(W|A)$. You are being confused by the notation of $|A$. But that just says if your universe of options is pulling marbles from a bag that is $A$ ...
- 119k
1
vote
Accepted
Finding $P(Y \leq y | Z \leq z)$ when both are conditional on a binary random variable $X$
It would be the second one. Let $A = \{Y \leq y\}$ and $B = \{Z\leq z\}$ and for $i\in\{0,1\}$ let $C_i = \{X=i\}$. Then you want to compute $P(A|B)$ (assuming $P(B\cap C_i)>0$ for each $i$). Then,
...
- 2,223
1
vote
Accepted
MGF from a conditional distribution
We can first write conditional MGF of bernoulli random r.v.$(I\sim BN(p))$, $MGF_{I}=pe^{t}+1-p$ then given that binomial r.v. is $B(n,p)=\sum_{k=1}^{n}I_{k}$
the MGF of binomial r.v. $B(n,p)$ is,
\...
1
vote
MGF from a conditional distribution
Starting with the definition is the right step. Insert in the sum the expression for $B(x,n,p)$ ,considering for the moment $p$ to be fixed: you'll get a $M_x(t,p,n)$, that is the mgf subject to being ...
- 33.7k
1
vote
Accepted
Likelihood of knowing the answer given y correct answers on multiple choice test
The likelihood of $\theta$ is $\mathbb P(Y=y | \theta)$ (the probability of the data given the parameter). But $Y$ is just a binomial random variable given $\theta$, where the success probability is ...
- 1,781
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