![Verify the odds of winning in the above. NUMBERS MATCHED s white +1 Power Ball /I69,5/c(26,1)]1-1/292,201,338 5 White+0 Power Bal [C(25,1)1/(292 Million) 1/11,688,053 4 White+1 Power Ball C(S,4)C(64,1)1/292 M)-1/913,129 4 White+0 Power Ball (C(5,4)C(64,1)1/1292 M)-1/36,525 3 White 1 Power Ball [CI5,3)C(64,2)1/(292 M)-1/14,494 3 White+0 Power Ball [C(5,3)C(64,2)C(25,1)/(292 M)-1/580 2 White + 1 Power Ball [C(5,1)C(64,4)1/(292 M)-1/701 1 White+1 Power Ball [C(5,1)C(64,4)1/(292 M)-1/91.977 0 White 1 Power Ball](http://img.homeworklib.com/questions/cc1921a0-7156-11ea-96aa-71cf745c51f5.png?x-oss-process=image/resize,w_560)
![A five-card hand is random dealt from a standard deck. We can clasily five-card hand into following possblites: 1. Calculate the number possible five-card hands that are . Two denominations: o Four of a kind: four cards from one denomination, one from another denomination C(13,1)C(14,4)C(12,1)C(4,1)-624 Full house: three cards from one denomination, two from another denomination C 13,1)C(4,3)C(12,1)C(4,2)-3,744 o Three denominations: Three of a kind: three cards from one denomination, the other two from two different denominations o 13, 1)C(4,3 С(12,2)Ic(4,1r2)549 12 Two pairs: two pairs and the fifth card is from a denomination other than the two pairs C(13,2)IC(4,2)A2]C(11,1)0(4,1)-123,552 o Four denominations: Exactly one pair: one pair and the other three cards are from three different denominations o C(13,1)0(4,2)C(12,3)[C(4,1)A3]-1,098,240 . Five denominations: o No pairs: five cards from five different denominations C(13,5)IC(4,1)5]-1,317,888 2. Show that the total counts in (1) is equal to C(52, 5). 624 +3,744+54,912+123,552+ 1,098, 240+ 1,317,888-2588,960 OR Cl 52.5) Circle your final answers on the above 19](http://img.homeworklib.com/questions/cce39040-7156-11ea-9ae5-cd928acfbe9d.png?x-oss-process=image/resize,w_560)
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Supplemental Worksheet 7.1 Group No. Names Counts of Arrangements (Permutations) and Sets (Combinations) HOW TO PLAY...
Can someone do 2,6,8,12 34 Exercises 187 EXAMPLE 7 GETTING FIVE HEARTS Find the probability of being dealt five bearts...
Can someone do 2,6,8,12
34 Exercises 187 EXAMPLE 7 GETTING FIVE HEARTS Find the probability of being dealt five bearts SOLUTION The sample is the same as in Example 5. The event consists of all pos- hands that include five hearts and no non-hearts This involves cltegories (hearts and non-hearts), so we will use the Fundamental Count- hinciple and multiply the number of ways of getting five hearts and the sible number of ways of getting no non-hearts There are...
Complete the following assignment with the tester class included. *If the text is too small in...
Complete the following assignment with the tester class
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Can someone do 2,6,8,12
34 Exercises 187 EXAMPLE 7 GETTING FIVE HEARTS Find the probability of being dealt five bearts SOLUTION The sample is the same as in Example 5. The event consists of all pos- hands that include five hearts and no non-hearts This involves cltegories (hearts and non-hearts), so we will use the Fundamental Count- hinciple and multiply the number of ways of getting five hearts and the sible number of ways of getting no non-hearts There are...
Complete the following assignment with the tester class
included.
*If the text is too small in the pictures to read, open a new
window and copy and paste the address below into the search bar,
and go to that page for the same instructions in the pictures
below:
This is the address for the same instructions in the
pictures below:
https://www.chegg.com/homework-help/questions-and-answers/purpose-purpose-lab-design-write-many-complex-methods-working-arrays-array-list-objects-ma-q37244441?trackid=6abm6xXV
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public class Tester
{
/**
* main() method
*/
public static void main(String[] args)
{
// No...
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