One die probability questions

Question

I was given this exercise:

If a standard die has twice the probability of rolling an odd number than an even number. Find the probability of rolling a non-prime number with this die.

My attempt:

P(even) = 1/3

P(odd) = 2/3

P(non-prime) = P(1)+P(4)+P(6)

         = (1/6)(2/3) + (1/6)(1/3) + (1/6)(1/3)

         = (4/18) = (2/9)

Answer 1

I interpret your problem statement to say that you have a standard (6-sided) die, but it's non-standard in that the probability of rolling an odd number is twice the probability of rolling an even number.

$P($non-prime$)=P($rolling a 1, 4, or 6$)$

$P($rolling $1)=P($odd$)\cdot P($that odd number is 1$)=(2/3)(1/3)=2/9$

$P($rolling $4)=P($even$)\cdot P($that even number is 4$)=(1/3)(1/3)=1/9$

$P($rolling $6)=P($even$)\cdot P($that even number is 6$)=(1/3)(1/3)=1/9$

Thus, $P($non-prime$)=(2/9)+(1/9)+(1/9)=4/9$.

Answer 2

I am assuming by a standard dice you mean one that has six sides and contains the numbers 1,2,3,4,5,6.

If P(even) = 1/3 and P(odd) = 2/3 then to find the chance of rolling a non-prime we must find the chance of rolling a non-prime that is even or odd then adding the probability.

First let us consider the amount of even non-primes from 1-6. There is only two even non-primes and those are 4 and 6.

Because 4 and 6 are two of three possible numbers and there is a 2/3 chance you randomly select these numbers from the three possible choices those numbers. Then you multiply that with the probability of rolling an even number.

(2/3)*(1/3) = (2/9)

Then let us consider the probability of rolling a non-prime odd number. In the number 1-6 there is only one non-prime odd number out of the three number 1-6. So the chance of randomly choosing a non-prime odd number out of the three choices is 1/3. Then multiply that with the chance of rolling a odd number.

(1/3)*(2/3)=(2/9)

The answer is the sum of rolling both an non-prime odd and a non-prime even.

(2/9)+(2/9)=(4/9)

Thus P(non-prime) = (4/9)

Answer 3

If we interprete the question to require that: The probability of any particular odd number is twice that of any particular even number, but its an otherwise standard die, so odd numbered results are equiprobable, and even numbered results also.

There is $3$ of each kind, and a $2:1$ weight ratio of odd:even, but equally weighted within the kinds.

Meaning that: $\Bbb P(X=x)=\begin{cases} (1/3)(2/3) & :x\in \{1,3,5\}\\ (1/3)(1/3) & :x\in\{2,4,6\} \\ 0 & : \text{elsewhere}\end{cases}$

Now you can find: $\Bbb P(X\in\{1,4,6\})=4/9$