MATH GENIUS PLZ HELP ME~!!!

P(product=prime) = one dice must be 1 and second dice is either 2,3,5
= P(dice=1)x P(dice=2,3,5) x 2!
= 1/6 x 1/2 x 2!
= 1/6
therefore P(product=composite) = 1-1/6 = 5/6

Expected value of Player A = 60 x 1/6 x 10 = 100
Expected value of Player B = 60 x 5/6 x 1 = 50

P(product of 2 numbers will be prime) = P(X)
= P(at least one of the numbers is 1)
= 1 - P(none of the numbers is 1)
= 1 - (5/6)^2
= 11/36

rolled over 60 times:
E(A) = P(X) * 60 = 18 and 1/3
E(B) = P(not X) * 60 = 41 and 2/3

By the way, the answer to the last question you asked was wrong. The correct answer is:
number of 3 digit numbers = 999-100 + 1 = 900
numbers that have at least one 1:
set 1:
100-199 = 100 numbers

set 2 (numbers with 1 in 2nd digit not already included in set 1):
210,310,410,510, ... , 910
212, 312, 412, 512, ..., 912
213, 313, 413, 513, ... , 913
...
219, 319, 419, 519, ..., 919 = 9 * 8 = 72 numbers

set 3 (numbers with 1 in 3rd digit not already included in set 1 and 2):
221, 231, 241, 251, ..., 291
321, 331, 341, 351, ..., 391
421, 431, 441, 451, ... , 491
...
921, 931, 941, 951, ..., 991 = 8 * 8 = 64 numbers

set 1 + set 2 + set 3 = 100 + 72 + 64 = 236
therefore probability = 236/900 (which is not equal to 1/4 in the previous answer)

2007-06-07 10:22:34 補充：
P(product of 2 numbers will be prime) = P(X)= P(one dice is 1)*P(one dice is 1,2,3,5)*2(*2 because it doesn't matter which dice is which)P(X) = 1/6 * 4/6 * 2 = 2/9rolled over 60 times:E(A) = 13.333333E(B) = 46.666666666

P(product=prime number)
=P(dice1=1 or dice2=1)
=P(dice1=1)+P(dice2=1)
=1/6+1/6
=1/3
so P(product NOT=prime number)=1-1/3=2/3

Rolled 60 times
Expected points of player A=60*1/3*10=200
Expected points of player B=60*2/3*1=40