# SOLUTION: There is one more that I can't get. Roll a pair of dice 20 times and count the number of times you get a 7 and the number of times you get an 9. I did this but I can't figur

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: There is one more that I can't get. Roll a pair of dice 20 times and count the number of times you get a 7 and the number of times you get an 9. I did this but I can't figur      Log On

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 Click here to see ALL problems on Probability-and-statistics Question 44523: There is one more that I can't get. Roll a pair of dice 20 times and count the number of times you get a 7 and the number of times you get an 9. I did this but I can't figure out the following: a. Compute Pe(7) b Compute Pe(8) c Compute Pe (7 or 8) d Compute Pe (7) + pe (8) e Explain why the results of parts c and d are as they are.Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!There is one more that I can't get. Roll a pair of dice THERE ARE 6 POSSIBILITIES ON ONE DICE AND ANOTHER 6 ON THE SECOND DICE HENCE TOTAL NUMBER OF POSSIBILITIES=6*6=36 WE CAN GET 7 IN THE FOLLOWING WAYS 1+6,2+5,3+4,4+3,5+2,6+1....TOTAL..6 WAYS HENCE P7=6/36=1/6 SIMILARLY FOR 9 3+6,4+5,5+4,6+3.....3 WAYS P9=3/36=1/12 20 times and count the number of times you get a 7 and the number of times you get an 9.....HENCE WE EXPECT THAT IN 20 TRIALS... 7 TO TURN UP IN 20*1/6 TIMES=3 TO 4 TIMES TIMES 9 TO TURN UP IN 20*1/12 TIMES =1 TO 2 TIMES 8 TO TURN UP IN [2+6,3+5,4+4,5+3,6+2=5 WAYS]20*5/36=2 TO 3 TIMES I did this but I can't figure out the following: a. Compute Pe(7)=20*1/6 b Compute Pe(8)=20*5/36 c Compute Pe (7 or 8)20*{1/6 +5/36}=20*11/36 d Compute Pe (7) + pe (8)20*11/36 e Explain why the results of parts c and d are as they are. OR IS INCLUSIVE OR EITHER 7 OR 8 OR BOTH