In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are [tex]\binom44\binom{48}1=48[/tex] possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by [tex]\binom{13}1=13[/tex]. So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability
[tex]\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024[/tex]
There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing [tex]\binom41\binom{48}4=778,320[/tex] possible hands. Exactly 2 aces are drawn in [tex]\binom42\binom{48}3=103,776[/tex] hands. And so on. This gives a total of
[tex]\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656[/tex]
possible hands containing at least 1 ace, and hence B occurs with probability
[tex]\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412[/tex]
The product of these probability is approximately 0.000082.
A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. [tex]P(A\cap B)=P(A)P(B)[/tex]. This happens if
The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So [tex]A\cap B[/tex] consists of 96 possible hands, which occurs with probability
[tex]\dfrac{96}{\binom{52}5}\approx0.0000369[/tex]
and so the events A and B are NOT independent.
