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Question: supplemental worksheet 71 group no names counts of arrangements permutations...

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Supplemental Worksheet 7.1 Group No. Names Counts of Arrangements (Permutations) and Sets (Combinations) HOW TO PLAY POWERBALL Powerbal tickets cost $2 per play. Pick flve numbers between 1 and 69 and one Powerball number between 1 and 26. . Quick Pick your numbers by letting the computer select or play your own lucky numbers using a playslp avalable wherever Lottery is sold How To Wirn Powerbat has nine winning ball comoinatons. The more numbers that match the numbers drawn the more you wih. NUMBERS MATCHED PR AMOUNT oDDS OF WINNING White 1 Power Ball 1 in 292 Million JACKPOT White 0 Power Ball 1 in 11.7 Milion $1,000,000 4 White 1 Power Ball $50.000 1 in 913 Thousand 4 White +O Power Ball $100 1 in 36.5 Thousand 3 White + 1 Power Bal 5100 1 in 14.5 Thousand 3 White 0 Power Ball $7 1 in 580 2 White 1 Power Ball $7 1 in 701 1 White 1 Power Ball $4 1 in 920 0 White 1 Power Boll $4 in 38.3 17
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
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
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