Probability: 3 dice

[Josephine]

Three dice, one red, one blue and one green are rolled simultaneously. The events S and T are as follows:
S: the sum of the numbers on the red dice and the green dice is 4.
T: the total of the numbers on the three dices is 5.

Find P(T) and P(S/T).

The answers are 1/36 and 1/2.

How do I get P(T)? I've tried this so far:

Red+Green+Blue
1+2+2
2+2+1
3+1+1

It goes on and on. Please show me how to solve this. Thanks!

 

[galactus]

For P(T):

When you roll 3 dice, there are 6^3=216 possible combinations.

How many combinations add to 5?:

1,2,2
2,2,1
2,1,2
1,3,1
3,1,1
1,1,3

6\216=1/36.

 

[soroban]

Hello, Josephine!

Three dice, one red, one green and one blue are rolled simultaneously.
The events S and T are as follows:
S: the sum of the numbers on the red die and the green die is 4.
T: the total of the numbers on the three dice is 5.

Find P(T) and P(S|T).

The answers are 1/36 and 1/2.

As Galactus pointed out, there are: \(\displaystyle 6^3\,=\,216\) possible outcomes.

To get a total of 5, there are six ways:
\(\displaystyle \;\;(R,G,B)\;=\;(1,1,3),\;(1,3,1),\;(3,1,1)\;(1,2,2),\;(2,1,2),\;(2,2,1)\)

Therefore: \(\displaystyle \,P(\text{sum is 5})\:=\:\frac{6}{216}\:=\:\frac{1}{36}\)


We want: \(\displaystyle P(S|T)\) . . . the probability that the sum of the red and green dice is 4,
\(\displaystyle \;\;\)given that the sum of all three dice is 5.

Since the sum of the three dice is 5, we have only six outcomes:
\(\displaystyle \;\;\;(R,G,B)\:=\1,1,3),\;(1,3,1),\;(3,1,1),\;(1,2,2),\;(2,1,2),\;(2,2,1)\)

Of these, three satisfy statement \(\displaystyle S:\;\;\)(1,3,1), (3,1,1),(2,2,1)

Therefore: \(\displaystyle \,P(S|T)\:=\:\frac{3}{6}\:=\:\frac{1}{2}\)

 

[Josephine]

I get it now!

Oh thanks for your help! I understand now. Thanks!