For a particular set of decks, you observe that a player wins 30% of the time. Moreover, you observe that, when a player wins, 90% of the time the winning card is drawn from deck A (meaning an A was picked from deck X). On the other hand, when a player loses, 40% of the time the losing card is drawn from deck A (meaning an A was picked from deck X).
Since picking an A from deck X is so much more likely to occur during a win than it is to occur during a loss, you wonder if picking an A from deck X is a good indication that a player will win. In particular, you wonder if picking A indicates that one has a larger than 50% chance of winning.

What conditional probability are you interested in estimating?

What three probabilities are you given in the statement of the problem?

$= 0.3$

$= 0.9$

$= 0.4$

Write Bayes' Theorem for the probability that you want to estimate.

$＿=$

Of the three probabilities needed for Bayes' Theorem, you already have values for two of them. Which of these probability is not yet specified?

In many cases, this probability from the denominator of Bayes' Theorem can be the trickiest to obtain. Recall how, when calculating conditional probabilities, we would often divide by a row or column sum of a contingency table. The denominator of Bayes' theorem plays the same role as this row or column sum.

The information to calculate the denominator $＿$ is already specified in the problem statement. It just takes a few steps to calculate.

First, what is the probability of losing the game? $P(L)=$

Second, from $P(A\,|\,W)$ and $P(W)$, calculate $P(A,W)$. From $P(A\,|\,L)$ and $P(L)$, calculate $P(A,L)$. Enter these values into this column of a contingency table.

Third, the total of the column is the overall probability of picking an $A$ from deck X: $P(A)=P(A,W)+P(A,L)=$

. This is the value for the denominator of Bayes' Theorem.

Now, you can fill in all the numbers for Bayes' Theorem.

$P(W\,|\,A) = $

$\times$

$/$

$=$

It turns out that picking A from deck X

indicate that one has at least a 50% chance of winning.

For a different set of the four decks, you observe the probabilities for each separate deck. For deck X, you observe that $P(A)=0.7$, $P(B)=0.1$, and $P(C)=0.2$. For the three lettered decks, you observe that $P(W\,|\,A) = 0.1$, $P(L\,|\,A) = 0.9$, $P(W\,|\,B) = 0.5$, $P(L\,|\,B) = 0.5$, $P(W\,|\,C) = 0.7$, and $P(L\,|\,C) = 0.3$.
Let's say you wanted to calculate the conditional probability $P(B\,|\,W)$, the probability, given that a winning card was drawn, that the win was obtained from deck B. Write down Bayes' Theorem for $P(B\,|\,W)$.

$P(B\,|\,W) =$

Again the denominator, the probability of winning $P(W)$, is the probability not explicitly known. But, you can combine six of the given probabilities to determine $P(W)$. (You calculated this same probability before in the context of conditional probabilities.)

$P(W)=$

Given that a player won, what is the probability that the player had selected a $B$? $P(B\,|\,W)=$