I'm assuming a 5-card hand being dealt from a standard 52-card deck, and that there are no wild cards.
A full house is made up of a 3-of-a-kind and a 2-pair, both of different values since a 5-of-a-kind is impossible without wild cards.
Suppose we fix both card values, say aces and 2s. We get a full house if we are dealt 2 aces and 3 2s, or 3 aces and 2 2s.
The number of ways of drawing 2 aces and 3 2s is

and the number of ways of drawing 3 aces and 2 2s is the same,

so that for any two card values involved, there are 2*24 = 48 ways of getting a full house.
Now, count how many ways there are of doing this for any two choices of card value. Of 13 possible values, we are picking 2, so the total number of ways of getting a full house for any 2 values is

The total number of hands that can be drawn is

Then the probability of getting a full house is

Answer:
Row ships
Step-by-step explanation:
They helped because a lot of travel took place on the Mediterranean Sea which was a massive body of water that led to the coasts of trading locations.
Use the quadratic formula, x=

(don't know how to type the "-"sign in the formula, so there is only the "+" in the formula)
in this case, a=1, c=34, b is unknown
from the roots, we can tell that -b+

=2(5+3i)=10+6i
Note: the original equation already give b as -b, so -(-b)=10, b=10
using b^2-4ac=(6I)^2,
b^2-4*34=-36
b^2=100
you will also get b=10
Answer:
all of dem
Step-by-step explanation:
Answer:
-pi over 6 or B.
Step-by-step explanation:
I plugged it into desmos and this is what I got