Two dice are thrown. let e be the event that the sum of the dice is
even, let f be the event that at least one of the dice lands on 6 and
let g be the event that the numbers on the two dice are equal. find
p(e), p(f), p(g), p(ef), p(f g), p(eg).

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

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation: