# dice probability

• January 15th 2010, 07:02 AM
thereddevils
dice probability
A fair die is tossed threee times . Events A, B are defined as follows .

A=Total of the scores is odd

B=Total score is 13 .

Find p(A) and p(B)

i hv problem in counting the number of ways of both event A and B happen .
• January 15th 2010, 09:03 AM
For A, with 3 tosses,

odd=odd+even=(odd+even)+even, or odd+(odd+odd).

We must add two even numbers and an odd or three odd numbers.

Therefore the answer is $P(EvOdOd)+P(OdEvOd)+P(OdOdEv)+P(OdOdOd)$.

These can then be calculated.

For B,

three 1 to 6 digits sum to 13....

If the 1st digit is 1, the next two must both be 6....
If the 1st digit is 2, the next two must be 6,5 or 5,6..
If the 1st digit is 3, the next two must be 6,4, or 5,5 or 4,6.
With each step, there is an extra solution.

Hence the number of combinations adding to 13 is 1+2+3+4+5+6=21

Then the answer can be calculated.
• January 15th 2010, 09:59 AM
Soroban
Hello, thereddevils!

Quote:

A fair die is tossed threee times.
Events $A, B$ are defined as follows:

. . $A$ = Total of the scores is odd.

. . $B$ = Total score is 13.

Find: . $p(A)\text{ and }\,p(B)$

i hv problem in counting the number of ways of both events $A$ and $B$ happen.
Why? . . . It doesn't ask for it.

As Archie pointed out, an odd sum means: 3 Odds . . . or One odd, two Evens.

$p(\text{3 Odds}) \:=\:\left(\frac{1}{2}\right)^3 \:=\:\frac{1}{8}$

$p(\text{1 Odd, 2 Evens}) \:=\:{3\choose2}\left(\frac{1}{2}\right)\left(\fra c{1}{2}\right)^3 \:=\:\frac{3}{8}$

Therefore: . $p(\text{odd sum}) \:=\:\frac{1}{8} + \frac{3}{8} \:=\:\frac{4}{8} \:=\:\boxed{\frac{1}{2} \;=\;p(A)}$

For a sum of 13, there are only five combinations and their permutations.

. . $\begin{array}{cc}
\text{Combination} & \text{Permutations} \\ \hline
(1,6,6) & 3 \\
(2,5,6) & 6 \\
(3,4,6) & 6 \\
(3,5,5) & 3 \\
(4,4,5) & 3 \\ \hline
\text{Total:} & 21 \end{array}$

There are 21 ways to get a sum of 13.
There are $6^3 = 216$ possible outcomes.

Therefore: . $p(\text{sum of 13}) \:=\:\frac{21}{216} \:=\:\boxed{\frac{7}{72} \:=\:p(B)}$

• January 15th 2010, 09:32 PM
thereddevils
thanks both of you, but there is another event in this question , event C where the number 6 appears at the first toss and event A is where the total scores is odd as mentioned previously .

Find P (A|C)

How do i do this ? Is P(C)=1/6 ?
• January 16th 2010, 02:27 AM
Yes, as all values 1 to 6 are equally likely for a fair die.

However, you only need count the permutations for which 6 occurs first.
You only need calculate P(total score is odd given that 6 occurs first).

6 happens first and the sum is 13 for

661
652
643
634
625
616

Hence, the probability is ?
• January 16th 2010, 05:18 AM
thereddevils
Quote:

Yes, as all values 1 to 6 are equally likely for a fair die.

However, you only need count the permutations for which 6 occurs first.
You only need calculate P(total score is odd given that 6 occurs first).

6 happens first and the sum is 13 for

661
652
643
634
625
616

Hence, the probability is ?

thanks , but why 13 only , total score is odd can also be 11 (6+4+1) , and so on ?
• January 16th 2010, 10:58 AM
Oh yeah!

That's just the probability of getting a sum of 13 given that the first one is 6.

The probability of getting an odd sum given that the first one is 6,
requires that the second and third throws have one even and one odd.
Since 6 is even...the only ways the sum will be odd is if the next two have an even and an odd.

Therefore you can multiply probabilities as you were ready to do....
because P{6** and sum is odd} is

P{1st is 6 and 2nd is even and 3rd is odd..... or 1st is 6 and 2nd is odd and 3rd is even}.

What you must calculate then is P{6,odd,even}+P{6,even,odd}

6 has a $\frac{1}{6}$ chance of being first.

Since the die always has the digits 1 to 6, there is a $\frac{1}{2}$ chance the 2nd will be odd and the same chance the 3rd will be even.

We get the exact same calculations for the 2nd being even and the 3rd odd.

We have now allowed for all possible scenarios,
hence the probability is the sum of these two possibilities.

P{1st is 6 and sum is odd}=2{ $\frac{1}{6}\ \frac{3}{6}\ \frac{3}{6}$}