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
|
|
|
