another probability dice

[mathproblems]

If you roll aloaded die where the even numbers come up twice as often as the odd numbersfind the probability of each of:
Can someone check my answers please: Thank you.
{1 22 3 44 5 66} 9 outcomesP(1)=1/9 P(2)=2/9
P(3)=1/9 P(4)=2/9
P(5)=1/9 P(6)=2/9
P(5 | odd)= not sure ?? 1/9 {rolling a 5 given that theroll was an odd number}

 

[soroban]

Hello, mathproblems!

If you roll a loaded die where the even numbers come up twice as often as the odd numbers,
find the probability of:

(Can someone check my answers please? Thank you.)

\(\displaystyle \{1,2,2,3,4,4,5,6,6\},\,\text{ 9 outcomes}\)

. . \(\displaystyle \begin{array}{cccccc}P(1) \:=\:\frac{1}{9} & P(2) \:=\:\frac{2}{9} \\ \\ P(3) \:=\:\frac{1}{9} & P(4) \:=\:\frac{2}{9} \\ \\ P(5) \:=\:\frac{1}{9} & P(6) \:=\:\frac{2}{9} \end{array}\)

Correct!

\(\displaystyle P(5\,|\text{ odd})\)

Bayes' Theorem: .\(\displaystyle P(5\,| \text{ odd}) \;=\; \dfrac{P(5\,\wedge\,\text{odd})} {P(\text{odd})}\)

We have: .\(\displaystyle P(5\,\wedge\text{ odd}) \:=\:\frac{1}{9},\;\; P(\text{odd}) \:=\:\frac{3}{9}\)

Therefore: .\(\displaystyle P(5\,|\text{ odd}) \;=\;\dfrac{\frac{1}{9}}{\frac{3}{9}} \:=\:\dfrac{1}{3}\)

 

[mathproblems]

thank you so much!

could you also help me understand
If you roll two fair dice, one red and one green. What is the probability that you get a sum of 7 given that one die is an even number?
Thank you.

 

[tkhunny]

Let's learn a little etiquette.

1. Don't post the same problem in multiple locations. This wastes the time fo valunteers.
2. Don't post ANY problem without showing ANY work. Just don't do it.
3. Try not to allow other people to do your homework for you.
4. Build a "Dice Table". It is simple. You can then read off multiple answers.
5. Why is this problem popping up? Is it a contest on some website?
6. Don't ask multiple questions on a single thread. Start a new one.