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:
5600.014
= 5000 + 600 + 0.010 + 0.004
i am a mathematics teacher. if anything to ask please pm me
Step-by-step explanation:
 
        
             
        
        
        
Answer:
169.5 yd²
Step-by-step explanation:
See attachment.
In situations like this, ALWAYS (!) make as sketch.
Divide the areas by adding dotted lines on sensible places, and write in the missing numbers for the correct distances.
The only possible difficult one is the triangle, which is the half of a rectangle. So you are dealing with a series of areas of rectangles. That is really easy if you understand what you are doing.
Total area = 
area 1 + area 2 + area 3 + area 4
10*4 + 4*7 + 3*13 + (0.5 * 7*17)
40 + 28 + 42 + 59.5
Total area = 169.5 yd²
 
        
             
        
        
        
Answer:34.93
Step-by-step explanation:
Using a^2+b^2=c^2we can substitute a and b in which is 34^2+8^2=c^21156+64=c^21220 = c^2Now we need to square both sides√1220 = √c^234.9284983931 ----> 34.9334.93 = cc = 34.93