Answer:
- P(E) = 1/2
- P(F) = 11/32
- P(G) = 1/6
- P(EF) = 5/52
- P(FG) = 1/32
- P(EG) = 1/6
Step-by-step explanation:
For the sum to be even, both dice can be odd, or both even. The probability of a dice being odd is 1/2 and the same is for it to be even. Since the result of the dices are independent, we have that
P(E) = (1/2)² + (1/2)² = 1/2
Out of the 36 possible outcomes for the dice (assuming that you can distinguish between first and second dice), there are 11 cases in which one dice is a 6 (if you fix 1 dice as 6, there are 6 possibilities for the other, but you are counting double 6 twice, so you substract one and you get 6+6-1 = 11). Since all configurations for the dices have equal probability, we get that
P(F) = 11/32
The probability for the second dice to be equal to the first one is 1/6 (it has to match the same number the first dice got). Hence
P(G) = 1/6
for EF, you need one six and the other dice even. For each dice fixed as 6 we have 3 possibilities for the other. Removing the repeated double six this gives us 5 possibilities out of 32 total ones, thus
P(EF) = 5/32
If one dice is 6 and both dices are equal, then we have double six, as a result there is only one combination possible out of 32, therefore
P(FG) = 1/32
If both dices are equal, in particular the sum will be even, this means that G= EG, and as a consecuence
P(EG) = P(G) = 1/6
