Hypothesis in Bayes theorem

Question

For eg :A Pack of playing card was found to contain 51 cards.I f the first 13 cards are examined all red,thrn probability that missing card is red. Can you tell me how to identify the components in this case.

Also in Bayes theorem,does the new relevant information changes the probability of $\mathbf p$$\mathbf r$$\mathbf i$$\mathbf o$$\mathbf r$ and that is what we calculate or is it the probability of $\mathbf p$$\mathbf r$$\mathbf i$$\mathbf o$$\mathbf r$ under the consideration of the new evidence

Answer 1

"Prior" and "posterior" are defined only relative to a certain piece of evidence; "prior" is what you'd believe without the new evidence (if you're thinking chronologically, think about what you believed before the evidence came to light), and "posterior" is what you believe with the evidence (chronologically "now"). We can only identify what "prior" and "posterior" are one we've identified what the relevant piece of evidence is.

Let's go through identifying the components in your example first, then make some comments about your general question.

What we want to answer is: What is the probability that the missing card is red? Two events happen: (a) the deck is found to contain $51$ cards, and (b) the first $13$ cards are found to be red. Now (a) cannot be the relevant piece of evidence, because asking a question about the missing card makes no sense if we don't yet know there is a missing card. The question has to be relevant both prior to and posterior to the evidence. So the evidence in question is (b): the first $13$ cards are found to be red. The prior probability should be thought of as follows: before we looked at the first $13$ cards (but after we noticed a card was missing; as noted above, prior to (b) but not prior (a)), what did we believe was the probability that the missing card was red? The posterior probability: now that we know the first $13$ cards are red, what do we believe is the probability that the missing card is red?

What we really want is the posterior probability, since it takes into account the evidence we know to be true. But to calculate it, we need to "step back in time" and consider the prior probability.

If I understood your last question correctly, you asked whether the prior probability changes? Once we take "prior" to be relative to a specific piece of evidence, the answer is no: the prior probability relative to evidence (b) is always what we believed the probability was before evidence (b) came to light. However, the posterior probability can become the prior probability relative to a new piece of evidence. For example, suppose in your example, we now examined the next $5$ cards and found them to be black. Call this evidence (c). Now one way to calculate the new probability is to view our answer from before -- the probability posterior to (b) -- as the probability prior to (c), and from there calculate the probability posterior to (c). From a chronological standpoint, this is what happens in most situations: a piece of evidence comes to light, you calculate a new posterior probability, which then becomes your prior probability relative to a new piece of evidence, which forces you to recalculate a new posterior probability. So "prior" and "posterior" are well-defined only relative to a specific piece of evidence -- in the broad picture, your justified belief of the probability can be changing several times.

Answer 2

For eg :A Pack of playing card was found to contain 51 cards. If the first 13 cards are examined all red, thrn probability that missing card is red.

Let $A$ be the event that the first thirteen cards are red, and $B$ be the event that the missing card is red.   We therefore seek: $\mathsf P(B\mid A)$.

How will I solve this question based on the components given above and identify each component in this example.

Then $\mathsf P(B) = 26/52 = 1/2\\\mathsf P(A\mid B)=\left.\tbinom{25}{13}\middle/\tbinom{51}{13}\right.\\\mathsf P(A\mid B^\complement) = \left.\tbinom{26}{13}\middle/\tbinom{51}{13}\right.$

And so by the Law of Total Probability: $\mathsf P(A) = \mathsf P(A\mid B)\,\mathsf P(B)+\mathsf P(A\mid B^\complement)\,\mathsf P(B^\complement)$

Thus you seek: $~\mathsf P(B\mid A) = \dfrac{\mathsf P(A\mid B)~\mathsf P(B)}{\mathsf P(A\mid B)\,\mathsf P(B)+\mathsf P(A\mid B^\complement)\,\mathsf P(B^\complement)}$

Which evaluates to a nice little ratio. $\phantom{1/3}$

Answer 3

At this problem we need to define to random variables X,Y as follows:

First we generalize the problem ,meanwhile assume that n cards are removed among those ,say $Y=y$ cards are red. Also we show red color with number $1$ and black color with number $0$. $$X=Color \quad of \quad the\quad missing\quad card$$ $$Y=Number\quad of\quad red\quad cards\quad in\quad the \quad removed \quad n \quad cards$$ With these definitions we obviously have: $$\large Pr(X=1)=Pr(X=0)={1\over 2}$$ If we know that the missing card is red i.e. $X=1$ then there exist $25$ red cards and $26$ black card. Therefore we can obtain: $$\large Pr(Y=y|X=1)={{{\binom{25}{y}}{\binom{26}{n-y}}}\over{\binom{52}{n}}}$$ and similarly $$\large Pr(Y=y|X=0)={{{\binom{26}{y}}{\binom{25}{n-y}}}\over{\binom{52}{n}}}$$ and by incorporating we have: $$\large Pr(Y=y|X=x)={{{\binom{26-x}{y}}{\binom{25+x}{n-y}}}\over{\binom{52}{n}}}$$

Finding this probability now enables us to find $Pr(X=x|Y=y)$ using Bayes rule. First note that: $$\large Pr(Y=y)={1\over 2}{{{{\binom{25}{y}}{\binom{26}{n-y}}+{\binom{26}{y}}{\binom{25}{n-y}}}\over{\binom{52}{n}}}}$$ Finally we obtain the final equation: $$\Large Pr(X=x|Y=y)={{{{\binom{26-x}{y}}{\binom{25+x}{n-y}}}}\over{{\binom{25}{y}}{\binom{26}{n-y}}+{\binom{26}{y}}{\binom{25}{n-y}}}}$$