+1995 A hat contains four balls. The balls are numbered 2, 4, 4, and 7. One ball is randomly selected and not replaced, and then a second ball is selected. The numbers on the two balls are added together.
A fair decision is to be made about which of three sizes of ice cream cone will be ordered, using the sum of the numbers on the balls. The sizes are small, medium and large.
Which description accurately explains how a fair decision can be made in this situation?
A) If the sum is 6 or 9, a small cone will be ordered.
If the sum is 8, a medium cone will be ordered.
If the sum is 11, a large cone will be ordered.
B) If the sum is 6 or 8, a small cone will be ordered.
If the sum is 9, a medium cone will be ordered.
If the sum is 11, a large cone will be ordered.
C) If the sum is 6, a small cone will be ordered.
If the sum is 8 or 11, a medium cone will be ordered.
If the sum is 9, a large cone will be ordered.
D)If the sum is 6, a small cone will be ordered.
If the sum is 8 or 9, a medium cone will be ordered.
If the sum is 11, a large cone will be ordered.
+129852 Here's my attempt
P(6) = P( 2 on first roll) * P ( 4 on second roll) + P(4 on first roll) *P(2 on second roll) =
(1/4)(2/3) + ( 1/2) (1/3) =
2/12 + 1/6 =
4/12 =
1/3
P (8) = P(4 on both rolls) = (1/2) ( 1/3) = 1/6
P(9) = P( 2 on first roll) * P ( 7 on second roll) + P(7 on first roll) *P(2 on second roll) =
(1/4)(1/3) + ( 1/4) (1/3) =
1/12 + 1/12 =
2/12 =
1/6
P(11) = P( 4 on first roll) * P ( 7 on second roll) + P(7 on first roll) *P(4 on second roll) =
(1/2)(1/3) + ( 1/4) (2/3) =
1/6 + 2/12 =
2/12 + 2/12 =
4/12 =
1/3
So...
P(6) = 1/3
P (8 or 9) = 1/6 + 1/6 = 1/3
P(11) = 1/3
So....D seems to provide a fair division
D)If the sum is 6, a small cone will be ordered.
If the sum is 8 or 9, a medium cone will be ordered.
If the sum is 11, a large cone will be ordered.
+1995 A lot of information , will defentely need to take notes. I agree with everything and I would say D would be correct as well. Thanks for your help.
