more probability

L

luvugodm

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May 17, 2007
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If I roll two dice and am trying to find event a to equal at least once die shows a 4 would it be 1/12 and if I want event b to equal the sum of the dice is greater than 5 is it 23/36
and then how do I find what p(a) is do I do 4x3x2
 
Hello, luvugodm!

If I roll two dice and event A is "at least one die shows a 4", would \(\displaystyle P(A)\,=\,\frac{1}{12}\) ? .No

If event B = "the sum of the dice is greater than 5", would \(\displaystyle P(B)\,=\,\frac{23}{36}\) ? .No
Click to expand...

For a pair-of-dice problem, there is no neat formula.
We must list the 36 possible outcomes.

. . \(\displaystyle \begin{array}{cccccc}(1,1)&(1,2)&(1,3)&(1,4)&(1,5)&(1,6) \\
(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6) \\
(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6) \\
(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6) \\
(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\
(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)
\end{array}\)



There are eleven outcomes in which at least one die shows a 4:

. . \(\displaystyle \begin{array}{cccccc}- & - & - & (1,4)& - & - \\
- & - & - & (2,4) & - & - \\
- & - & - &(3,4)& - & - \\
(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6) \\
- & - & - &(5,4)& - & - \\
- & - & - & (6,4) & -& -
\end{array}\)

Therefore: \(\displaystyle \L\,P(A)\:=\:\frac{11}{36}\)



There are twenty-six outcomes with a sum greater than 5.

. . \(\displaystyle \begin{array}{cccccc} - & - & - & - &(1,5)&(1,6) \\
- & - & - &(2,4)&(2,5)&(2,6) \\
- & - &(3,3)&(3,4)&(3,5)&(3,6) \\
- &(4,2)&(4,3)&(4,4)&(4,5)&(4,6) \\
(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\
(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)
\end{array}\)

Therefore: \(\displaystyle \L\,P(B)\:=\:\frac{26}{36}\:=\:\frac{13}{18}\)

 
For the first one, the chance that at least one die shows a 4 is the complement of the chance that no die shows a 4, which is (5/6)^2 = 25/36.
 
Soroban

Thank you so much I think I am getting it .

Now if I want to find P(a and b) do I and them together???
 
Just count these:
\(\displaystyle \begin{array}{cccccc}\; & \; & \; & \; & \; & \; \\
\; & \; & \; & (2,4) & \; & \; \\
\; & \; & \; &(3,4)& \; & \; \\
\; &(4,2)&(4,3)&(4,4)&(4,5)&(4,6) \\
\; & \; & \; &(5,4)& \; & \; \\
\; & \; & \; & (6,4) & \;& \;
\end{array}\)
 
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