a. Why is it possible to estimate `P(A|x1 = Y, x2 = Y, x3 = N, x4 = N)` from this data but not `P(A|x1 = Y, x2 = Y, x3 = Y, x4 = N)`?

It is possible to estimate `P(A|x1 = Y, x2 = Y, x3 = N, x4 = N)`
from the data because there is an event that shows this characteristic in
the data, while there is no evidence which shows
`P(A|x1 = Y, x2 = Y, x3 = Y, x4 = N)` in the data.

b. Determine $P(A)$ and $P(B)$, the prior probabilities that an individual belongs to class A and class B, respectively.

```{r Q1b}
#P(A)
P_A <- 6/10
P_A
#P(B)
P_B <- 4/10
P_B
```

c. Find the "score" for class A for the individual in question (x1 = Y, x2 = Y, x3 = Y, x4 = N). It's between 0.03 and 0.04.

```{r Q1c}
#P(x1 = Y, x2 = Y, x3 = Y, x4 = N|A)*P(A)
#P(x1 = Y, x2 = Y, x3 = Y, x4 = N|A) = P(x1 = Y|A) *P(x2 = Y|A) *
#P(x3 = Y|A)*P(x4 = N|A)
# P(x1 = Y|A) = 4/6
# P(x2 = Y|A) = 3/6
# P(x3 = Y|A) = 2/6
# P(x4 = N|A) = 3/6
score_A <- (4/6*3/6*2/6*3/6)*6/10
score_A
```

d. Find the "score" for class B for the individual in question (x1 = Y, x2 = Y, x3 = Y, x4 = N). It's between 0.01 and 0.02.

```{r Q1d}
#P(x1 = Y, x2 = Y, x3 = Y, x4 = N|B)*P(B)
#P(x1 = Y, x2 = Y, x3 = Y, x4 = N|B) = P(x1 = Y|B) *P(x2 = Y|B) *
#P(x3 = Y|B)*P(x4 = N|B)
# P(x1 = Y|B) = 2/4
# P(x2 = Y|B) = 1/4
# P(x3 = Y|B) = 2/4
# P(x4 = N|B) = 3/4
score_B <- (2/4*1/4*2/4*3/4)*4/10
score_B
```

(5 points) Problem 21.6. Let Y 2 Gamma(a,B). Find the density of X...