# probability of rolling a die...

• Apr 16th 2011, 06:16 PM
slapmaxwell1
probability of rolling a die...
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?
• Apr 16th 2011, 07:15 PM
TKHunny
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(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?
• Apr 16th 2011, 07:22 PM
slapmaxwell1
the or means both, but im not understanding what you want me to ponder on? or i should ask, what 4 items do you want me to ponder on?
• Apr 16th 2011, 08:39 PM
slapmaxwell1
Quote:

Originally Posted by TKHunny
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(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 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(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

or:

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
• Apr 17th 2011, 05:10 AM
Soroban
Hello, slapmaxwell1!

Quote:

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 $\cup$ B) .= .P(A) + P(B) - P(A $\cap$ B)

We have: .P(6) = 1/6
. . . . . . . .P(H) = 1/2
. . . . . P(6 $\cap$ H) = (1/6)(1/2) = 1/12

Therefore:

. . P(6 $\cup$ H) .= .P(6) + P(H) - P(6 $\cap$ H)

. . . . . . . . .= .(1/6) + (1/2) - (1/12)

. . . . . . . . .= . 7/12
• Apr 17th 2011, 07:52 AM
slapmaxwell1
Quote:

Originally Posted by Soroban
Hello, slapmaxwell1!

You are expected to know this formula:

. . P(A $\cup$ B) .= .P(A) + P(B) - P(A $\cap$ B)

We have: .P(6) = 1/6
. . . . . . . .P(H) = 1/2
. . . . . P(6 $\cap$ H) = (1/6)(1/2) = 1/12

Therefore:

. . P(6 $\cup$ H) .= .P(6) + P(H) - P(6 $\cap$ H)

. . . . . . . . .= .(1/6) + (1/2) - (1/12)

. . . . . . . . .= . 7/12

um thanks i know the formula now. :O)