Understanding the Biased Die Problem

This problem involves calculating probabilities for a special type of six-faced die. The die is 'biased', meaning the outcomes (numbers 1 through 6) do not have equal probabilities. We are given specific conditions about the probabilities of prime versus non-prime numbers.

Defining Prime and Non-Prime Numbers

First, let's identify the numbers on a standard six-faced die and classify them:

• The numbers on the die are: {1, 2, 3, 4, 5, 6}.
• Prime Numbers: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. In our set, the prime numbers are {2, 3, 5}.
• Non-Prime Numbers: These are the numbers that are not prime. In our set, the non-prime numbers are {1, 4, 6}. (Note: 1 is neither prime nor composite).

Setting Up Probability Variables

Let's denote the probability of rolling a non-prime number as '$p$'. According to the question, the probability of rolling a prime number is twice that of a non-prime number. So:

• Probability of rolling a non-prime number = $P(\text{non-prime}) = p$.
• Probability of rolling a prime number = $P(\text{prime}) = 2p$.

This means:

• The probability for each non-prime face (1, 4, 6) is $p$. So, $P(1) = p$, $P(4) = p$, $P(6) = p$.
• The probability for each prime face (2, 3, 5) is $2p$. So, $P(2) = 2p$, $P(3) = 2p$, $P(5) = 2p$.

Calculating the Value of '$p$'

The sum of the probabilities of all possible outcomes must equal 1. Let's sum the probabilities:

$P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1$

Substitute the probabilities in terms of '$p$':

$p + 2p + 2p + p + 2p + p = 1$

Combine the terms:

$9p = 1$

Solve for '$p$':

$p = \frac{1}{9}$

Determining Individual Probabilities

Now we can find the specific probability for each face:

• $P(\text{non-prime}) = p = \frac{1}{9}$ (for faces 1, 4, 6)
• $P(\text{prime}) = 2p = 2 \times \frac{1}{9} = \frac{2}{9}$ (for faces 2, 3, 5)

Let's summarize this in a table:

Calculating the Probability of an Odd Number

The question asks for the probability that an odd number will show up. The odd numbers on the die are {1, 3, 5}.

The probability of rolling an odd number is the sum of the probabilities of rolling a 1, a 3, or a 5:

$P(\text{odd}) = P(1) + P(3) + P(5)$

Substitute the calculated probabilities:

$P(\text{odd}) = \frac{1}{9} + \frac{2}{9} + \frac{2}{9}$

Add the fractions:

$P(\text{odd}) = \frac{1 + 2 + 2}{9} = \frac{5}{9}$

Conclusion

Therefore, the probability that an odd number will show up when this biased die is tossed is $\frac{5}{9}$.