You must first decide what "or" means. Does it include both or is it exclusive?
p(6) = d
p(not six) = 1-d = f
p(heads) = h
p(not heads) = 1-h = j
Ponder this (d+f)*(h+j) = d*h + d*j + f*h + f*j
Four items to ponder. Which fit the given criteria?
ok this problem is a little different from the others...
what is the probability of either a die roll of a six or flipping a coin once and getting a heads?
ok so this means i will use both laws together? first multiply and then add?
so the probability of a die roll is (1/6) times the probability of getting a heads (1/2)
so this is 1/12 + (1/6)(1/2)=1/6?
did i do this right?
ok i have to be honest, but i have no idea what you just said to me....but nonetheless i figured out the answer to the problem...here is my solution.
P(rolling a 6) = 1/6
P(flipping a head) = 1/2
P(rolling a 6 AND flipping a head) = 1/6 * 1/2 = 1/12
P(rolling a 6 OR flipping a head) = 1/6 + 1/2 - 1/12 = 7/12
1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T
Of those, the ones that are either a 6 or a head include
1H, 2H, 3H, 4H, 5H, 6H, and 6T
That's 7 out of 12
What is the probability of either a die roll of a 6
or flipping a coin once and getting a Head?
You are expected to know this formula:
. . P(A B) .= .P(A) + P(B) - P(A B)
We have: .P(6) = 1/6
. . . . . . . .P(H) = 1/2
. . . . . P(6 H) = (1/6)(1/2) = 1/12
. . P(6 H) .= .P(6) + P(H) - P(6 H)
. . . . . . . . .= .(1/6) + (1/2) - (1/12)
. . . . . . . . .= . 7/12