# Tag Info

0

What are you asking is what is the probability of drawing two cards, without putting them back. Probability of drawing first card is 1/52. Now, there are 51 cards left. Therefore the probability of drawing second card is 1/51. Therefore PR(A & B) = PR(A)*PR(B).

1

Hint: You still have three cards to draw out of $50$. How many combinations of three cards result in a straight flush? How many total draws are there?

1

I would say that the difference is merely that the first one denotes a probability mass function and the latter denotes just a probability, that is a value. For instance $$P(A|B)=\begin{cases} 1/3 \mbox{ if A is true}\\ 2/3 \mbox{ if A is false}\end{cases}$$ and hence $$P(A=t | B)=1/3.$$ It does not appear to have anything to do with Bayesian theory.

1

The difference is in the prior expectation. The pure frequentists doesn't make any. Suppose I show you a coin. Before I even toss it, what would you expect the bias to be? The pure frequentist approach says that the probability is completely indeterminate, $0/0$. The approach only uses sample data to and cannot make claims without it. Divide by zero ...

2

A frequentist maximum likelihood estimate is akin to the mode of the likelihood function, so in a Bayesian context with uniform prior, the MLE is like a mode of the posterior; whereas the expectation of the posterior is going to be influenced by the overall shape of the distribution. Indeed, the mode of the posterior is called the "maximum a posteriori" ...

0

Assuming everything is positive, the first inequality is equivalent to $\alpha\gamma<\lambda\beta$ and the second to $(\alpha+1)\gamma<(\lambda+1)\beta$, or to $\alpha\gamma+\gamma<\lambda\beta+\beta$. So the question is: If $$\alpha\gamma<\lambda\beta,\tag 1$$ then does it follow that $\alpha\gamma+\gamma<\lambda\beta+\beta$, or ...

0

Bayes' theorem does not depend on the probability distribution, only on the definition of conditional probability. In particular, the distribution is not required to be normal, and the theorem still applies to log-normal distributions.

1

$$|T\hat{V}_{22} + T(E_2 - \hat{E}_2)(E_2 - \hat{E}_2)'| = |T\hat{V}_{22}| |I + (E_2 - \hat{E}_2)(E_2 - \hat{E}_2)' \hat{V}_{22}^{-1}| \\ = |T\hat{V}_{22}| |1 + (E_2 - \hat{E}_2)' \hat{V}_{22}^{-1}(E_2 - \hat{E}_2)|$$

1

Firstly, I'd like to apologize because my comments definitely provided more criticism than constructive commentary. (Although hopefully a healthy blend of both). Secondly, I think the general answer to your question is: Yes, Bayesian inference can be used in this manner where you take known quantities $P($Goalie is Sick$)$, $P($Goal Keeper Is Sick | USA ...

Top 50 recent answers are included