Probability question concerning maximums and minimums There are marbles in a jar for digits 1-10. For each digit, there is that corresponding number of marbles in the jar (1 labeled 1; 2 labeled 2, and so on). So, there are 55 total marbles in the bucket. 
P(A) is that the probability of the maximum is 8, and P(B) is the probability that the minimum is 3. 
For 5 draws with replacement
It asks to determine P(A): I calculated 8^5/10^5 - 7^5/10^5., or .1596.
P(B) I calculated to be (1-6^5/10^5) - (1-7^5/10^5), or .0903.
It then asks for P(A|B). I'm not sure how to proceed. Assuming that I did the other calculations correctly. I know the formula is P(Aint.B)/P(B), but I'm not certain how to find the intersection. 
 A: Comment: Mostly for verification.
I simulated a million 5-draw experiments. Results for $P(A)$ and $P(B)$ should be accurate
to two or three places. (Argument pr=1:10 sets the proportions of
each number; R turns them into probabilities. Argument repl=T means
sample with replacement. The m-vectors a and b are logical, 
containing elements FALSE and TRUE; the mean of a logical vector is its
proportion of TRUE's.)
m = 10^6;  n = 5;  sm = lg = numeric(m)
for (i in 1:m) {
  x = sample(1:10, n, repl=T, pr=1:10)
  sm[i] = min(x);  lg[i] = max(x) }
a = (lg==8);  b = (sm == 3)
mean(a);  mean(b);  mean(a & b); mean(a[b])
## 0.085816    # aprx P(A)
## 0.193684    # aprx P(B)
## 0.020432    # aprx P(A and B)
## 0.1054914   # aprx P(A|B)

(36/55)^5 - (28/55)^5 
## 0.08594705   # your P(A)
(52/55)^5 - (49/55)^5
## 0.194182     # your P(B)


A: In general we have $$P(A\,|\,B)=\frac {P(A\cap B)}{P(B)}$$ We know $P(B)\sim .19418$ So we just need $P(A\cap B)$.  
To compute $P(A\cap B)$:  We define $P_1$, the probability that the max is $8$ assuming we don't draw any numbers below $3$.  There are $52$ marbles with a value of $3$ or greater, $33$ of those have values $≤8$ and $25$ have value $≤7$  so $$P_1=\left(\left(\frac {33}{52}\right)^5-\left(\frac {25}{52}\right)^5\right)\times \left(\frac {52}{55}\right)^5\sim .05835621$$  
Now we define $P_2$, the probability that the max is $8$ assuming we don't draw any numbers below $4$. There are $49$ marbles with a value of $3$ or greater, $30$ of those have values $≤8$ and $22$ have value $≤7$  so $$P_2=\left(\left(\frac {30}{49}\right)^5-\left(\frac {22}{49}\right)^5\right)\times \left(\frac {49}{55}\right)^5\sim .03804284$$    Then $$P(A\cap B)=P_1-P_2\sim .02031337$$  Finally we get $$P(A\,|\,B)=\frac {P(A\cap B)}{P(B)}=\frac {.02031337}{.19418}\sim .10460992$$
Note:  Many thanks to @BruceET for simulating the process and identifying an error in the previous version of the post.  For those who are interested, the error was in the computation of the $P_i$.  The first draft failed to multiply by the probability of missing the lower values.  
