Understanding Probability with a Biased Die

This problem involves a biased six-faced die. A standard die has an equal probability (1/6) for each face (1, 2, 3, 4, 5, 6). However, this die is different because odd numbers are three times more likely to appear than even numbers.

The numbers on the die are:

• Odd numbers: 1, 3, 5 (Total 3 odd numbers)
• Even numbers: 2, 4, 6 (Total 3 even numbers)

Let \( P(\text{Odd}) \) be the probability of rolling an odd number and \( P(\text{Even}) \) be the probability of rolling an even number on a single throw.

According to the question, \( P(\text{Odd}) \) is thrice more likely than \( P(\text{Even}) \). This means:

\( P(\text{Odd}) = 3 \times P(\text{Even}) \)

We know that the sum of probabilities of all possible outcomes must equal 1. For this die, the only possible outcomes are either odd or even numbers. So:

\( P(\text{Odd}) + P(\text{Even}) = 1 \)

Now, substitute the first equation into the second one:

\( (3 \times P(\text{Even})) + P(\text{Even}) = 1 \)

\( 4 \times P(\text{Even}) = 1 \)

So, the probability of rolling an even number is:

\( P(\text{Even}) = \frac{1}{4} \)

And the probability of rolling an odd number is:

\( P(\text{Odd}) = 3 \times P(\text{Even}) = 3 \times \frac{1}{4} = \frac{3}{4} \)

We can verify this: \( P(\text{Odd}) + P(\text{Even}) = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1 \). This confirms our probabilities for a single throw.

Probability of Specific Outcomes

Assuming each odd number (1, 3, 5) has the same probability and each even number (2, 4, 6) has the same probability:

• Probability of rolling a specific odd number (1, 3, or 5) is \( P(\text{Odd}) / 3 = (3/4) / 3 = 3/12 = 1/4 \). So, \( P(1) = P(3) = P(5) = 1/4 \).
• Probability of rolling a specific even number (2, 4, or 6) is \( P(\text{Even}) / 3 = (1/4) / 3 = 1/12 \). So, \( P(2) = P(4) = P(6) = 1/12 \).

Sum of Two Throws is Even

The die is thrown twice. Let the outcome of the first throw be \( T_1 \) and the outcome of the second throw be \( T_2 \). We want to find the probability that the sum \( T_1 + T_2 \) is even.

The sum of two numbers is even if and only if both numbers are even, or both numbers are odd.

There are two possible scenarios for the sum to be even:

1. The first throw is Even AND the second throw is Even.
2. The first throw is Odd AND the second throw is Odd.

Since the two throws are independent events, we can multiply their probabilities.

Scenario 1: First throw Even, Second throw Even

• Probability of the first throw being Even is \( P(\text{Even}) = 1/4 \).
• Probability of the second throw being Even is \( P(\text{Even}) = 1/4 \).

The probability of this scenario is \( P(\text{Even on 1st}) \times P(\text{Even on 2nd}) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \).

Scenario 2: First throw Odd, Second throw Odd

• Probability of the first throw being Odd is \( P(\text{Odd}) = 3/4 \).
• Probability of the second throw being Odd is \( P(\text{Odd}) = 3/4 \).

The probability of this scenario is \( P(\text{Odd on 1st}) \times P(\text{Odd on 2nd}) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \).

The total probability that the sum is even is the sum of the probabilities of these two mutually exclusive scenarios:

\( P(\text{Sum is Even}) = P(\text{Even on 1st & Even on 2nd}) + P(\text{Odd on 1st & Odd on 2nd}) \)

\( P(\text{Sum is Even}) = \frac{1}{16} + \frac{9}{16} \)

\( P(\text{Sum is Even}) = \frac{1 + 9}{16} = \frac{10}{16} \)

Simplify the fraction:

\( P(\text{Sum is Even}) = \frac{10 \div 2}{16 \div 2} = \frac{5}{8} \)

Therefore, the probability that the sum of the numbers in the two throws is even is 5/8.

Sum of probabilities for even sum: \( \frac{1}{16} + \frac{9}{16} = \frac{10}{16} = \frac{5}{8} \).

Revision Table: Probability Concepts

Additional Information: Probability and Biased Events

Understanding biased events is important in probability. Unlike fair events (like flipping a fair coin or rolling a fair die), where each outcome is equally likely, biased events have unequal probabilities. This bias significantly impacts the calculation of probabilities for combined events.

In this problem, the bias towards odd numbers fundamentally changed the probabilities of rolling an odd (3/4) versus an even (1/4) number. If the die were fair, the probability of rolling an odd would be 3/6 = 1/2 and an even would be 3/6 = 1/2. For a fair die thrown twice, the probability of the sum being even would be \( (1/2 \times 1/2) + (1/2 \times 1/2) = 1/4 + 1/4 = 1/2 \).

The method used here involves breaking down the desired outcome (sum is even) into simpler, mutually exclusive events (Even + Even, Odd + Odd) and calculating their probabilities using the probabilities from a single throw. This approach is standard for compound probability problems involving independent events.

Key steps:

1. Determine the probabilities of the basic outcomes (odd or even) considering the bias.
2. Identify the combinations of these outcomes over two throws that result in the desired overall outcome (sum is even).
3. Calculate the probability of each combination using the rule for independent events.
4. Sum the probabilities of all combinations that meet the criteria (sum is even).

This systematic approach helps solve probability problems for both fair and biased scenarios.