To solve the problem step by step, we will calculate the probabilities of A and B winning the game based on the given conditions.
### Step 1: Determine the events for A and B
- **Event E1**: A wins if he gets tails on both coins.
- **Event E2**: B wins if he rolls a prime number on the die.
### Step 2: Calculate the sample space for A's toss
When A tosses two coins, the possible outcomes are:
- HH (Heads, Heads)
- HT (Heads, Tails)
- TH (Tails, Heads)
- TT (Tails, Tails)
Thus, the sample space \( S_1 \) has 4 outcomes.
### Step 3: Calculate the probability of A winning (E1)
The event E1 occurs only when A gets TT (Tails, Tails). Therefore:
- \( P(E1) = \frac{1 \text{ (favorable outcome)}}{4 \text{ (total outcomes)}} = \frac{1}{4} \)
### Step 4: Calculate the probability of A not winning (E1')
The probability that A does not win is:
- \( P(E1') = 1 - P(E1) = 1 - \frac{1}{4} = \frac{3}{4} \)
### Step 5: Calculate the sample space for B's roll
When B rolls a die, the possible outcomes are:
- 1, 2, 3, 4, 5, 6
The prime numbers in this range are 2, 3, and 5.
### Step 6: Calculate the probability of B winning (E2)
The event E2 occurs when B rolls a prime number. Therefore:
- \( P(E2) = \frac{3 \text{ (favorable outcomes: 2, 3, 5)}}{6 \text{ (total outcomes)}} = \frac{1}{2} \)
### Step 7: Calculate the probability of B not winning (E2')
The probability that B does not win is:
- \( P(E2') = 1 - P(E2) = 1 - \frac{1}{2} = \frac{1}{2} \)
### Step 8: Calculate the overall probability of A winning
A starts the game. The probability of A winning can be calculated as follows:
- A can win on the first turn: \( P(E1) = \frac{1}{4} \)
- If A does not win, B must also not win for A to have another chance:
- Probability of A not winning: \( P(E1') = \frac{3}{4} \)
- Probability of B not winning: \( P(E2') = \frac{1}{2} \)
Thus, the probability that A wins in subsequent rounds can be represented as:
- \( P(A \text{ wins}) = P(E1) + P(E1') \cdot P(E2') \cdot P(A \text{ wins}) \)
Let \( P(A \text{ wins}) = p \):
- \( p = \frac{1}{4} + \frac{3}{4} \cdot \frac{1}{2} \cdot p \)
- Rearranging gives:
\( p = \frac{1}{4} + \frac{3}{8}p \)
- Multiplying through by 8 to eliminate the fraction:
\( 8p = 2 + 3p \)
- Simplifying gives:
\( 5p = 2 \)
\( p = \frac{2}{5} \)
### Step 9: Calculate the overall probability of B winning
Using a similar approach:
- \( P(B \text{ wins}) = P(E1') \cdot P(E2) + P(E1') \cdot P(E2') \cdot P(A \text{ loses}) \)
- Let \( P(B \text{ wins}) = q \):
- \( q = \frac{3}{4} \cdot \frac{1}{2} + \frac{3}{4} \cdot \frac{1}{2} \cdot q \)
- Rearranging gives:
\( q = \frac{3}{8} + \frac{3}{8}q \)
- Multiplying through by 8:
\( 8q = 3 + 3q \)
- Simplifying gives:
\( 5q = 3 \)
\( q = \frac{3}{5} \)
### Final Result
- The probability of A winning is \( \frac{2}{5} \).
- The probability of B winning is \( \frac{3}{5} \).
